Download MA3056: Metric Spaces and Topology

MA3056: Metric Spaces and Topology
Course Notes
Stephen Wills
Contents
0 Background Set Theory
Basic constructions . . . . . .
Functions . . . . . . . . . . .
Partitions and onto functions
Exercises . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1
1
2
4
5
1 Metric Spaces
Definition and examples . . . . . .
Convergence and continuity . . . .
Properties of open and closed sets
Exercises . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
6
6
8
13
16
.
.
.
.
.
19
19
21
24
28
33
3 Compactness
Compactness for topological spaces; the Heine-Borel Theorem . . . . . .
Compactness for metric spaces: sequential compactness . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
36
41
43
4 Connectedness
Equivalent definitions; subintervals of R . . . . . . . . . . . . . . . . . .
Path-connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
45
48
49
5 Completeness and Uniformity
Uniform convergence and continuity . . . . . . . . . . .
Complete metric spaces . . . . . . . . . . . . . . . . . .
Applications of completeness: fixed points and category
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .
50
50
53
56
61
.
.
.
.
.
.
.
.
2 Topological Spaces
Definitions and examples . . . . .
Sequences; separation axioms . . .
Subspaces and product spaces . . .
Quotient spaces; homeomorphisms
Exercises . . . . . . . . . . . . . .
6 Hints to the Exercises
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
63
0
Background Set Theory
Topological arguments depend heavily on having a reasonable grasp of the basics
of set theory. This section gathers together all of the facts required for the course.
Rather than taking the fully axiomatic approach to the subject of set theory,
we shall instead take a slightly more naive path, assuming from the outset that
the notions of “set”, and “object” or “element” are understood. Indeed, the set
theorists would argue that such ideas cannot be defined, but rather must be given
as axioms.
Basic constructions
We shall generally start with a given nonempty set X. That is X is some collection
of objects, and given any object x we can decide if x belongs to the set X (denoted
x ∈ X), or if x does not belong to the set (denoted x ∈
/ X). Then we shall focus
on subsets of X, that is, sets whose elements are taken from X. These are often
specified by a rule which determines whether or not a given element of X belongs
to the particular subset, and this is written Y = {x ∈ X : P (x)} where P (x)
denotes the rule. For example (0, 1] = {x ∈ R : 0 < x ≤ 1} is the subset of the
real line consisting of all those real numbers lying between 0 and 1 together with
the right end point.
So a set Y is a subset of X if every element of Y also belongs to X, and is a
proper subset if in addition Y is not equal to X, i.e. there is at least one element
of X that is not in Y , so that X is strictly larger (in some sense. . . ) than Y . There
are at least two notational conventions for this which are:
(a) Y ⊂ X denotes that Y is a subset of X, and Y $ X denotes that Y is a
proper subset of X, or
(b) Y ⊆ X denotes that Y is a subset of X, and Y ⊂ X denotes that Y is a
proper subset of X.
In this notes I use the former, but it is worth noting that we shall almost always
be dealing with not necessarily proper subsets, and that proper subsets will be
explicitly mentioned when they arise.
Given sets Y and Z, the set difference {x ∈ Y : x ∈
/ Z} (i.e. those elements
in Y that are not in Z) is written either as Y \ Z or Y − Z. We will use the first
notation. The set {x ∈ Y : x ∈ Z} = {x ∈ Z : x ∈ Y } = {x : x ∈ Y and x ∈ Z}
is written Y ∩ Z, and the set {x : x ∈ Y or x ∈ Z} is written Y ∪ Z. However,
we can need to take intersections and unions of larger families of sets. If we are
givenTtwo nonempty
and for each
S sets X and Λ, T
S λ ∈ Λ we have a subset Yλ ⊂ X,
then λ∈Λ Yλ and λ∈Λ Yλ (or just λ Yλ and
T λ Yλ ) denote the intersection and
union of this entire family
{Y
}
.
That
is
λ∈Λ Yλ consists of every x ∈ X that
S λ λ∈Λ
lies in all of the Yλ , and λ∈Λ Yλ consists of every x ∈ X that lies in at least one of
the Yλ . In this situation Λ is called an indexing set. Note that with this notation
we have De Morgan’s laws which assert that
X\
S
λ Yλ
=
T
λ (X
\ Yλ )
and
1
X\
T
λ Yλ
=
S
λ (X
\ Yλ ).
(1)
It is possible to define and deal with arbitrarily large families of subsets without
recourse to index sets. For instance given a nonempty set X and some point
x0 ∈ X we can consider the family F = {Y ⊂ XT: x0 ∈ Y } consisting of those
subsets of X that contain this x0 . Note then that Y ∈F T
Y is the set of all x ∈ X
that belong S
to every Y ∈ F, from which it follows that Y ∈F Y = {x0 }. On the
other hand, Y ∈F Y = X.
The class F is in turn a subset of the collection P(X) of all subsets of X.
This is known as the power set of X, and is a cause of notational annoyance,
since we would normally like to use lowercase letters to denote the elements of
sets, which will be denoted by uppercase letters, but now our subsets are elements
of the set P(X). Note that if X is a finite set, then we can list its elements
X = {x1 , x2 , . . . , xn }, and P(X) is also finite with 2n members. Perhaps partly
for this reason the power set is sometimes denoted 2X . Furthermore, if X is
infinite (that is, it can be put into one-to-one correspondence with a proper subset
of itself), then P(X) is clearly infinite. On the other hand, even if X is countably
infinite (i.e. we can find a bijection f : N → X), the power set P(X) will contain
an uncountable number of elements. We shall not generally be too concerned with
whether an infinite set is countable or uncountable, but it is worth noting that a
large part of measure theory resembles parts of this course, except that infinite
unions and intersections are only considered for countable families {Yλ }λ∈Λ . That
is the index set Λ is countable, and so we may may instead consider sequences of
sets {Yn }∞
n=1 .
The Cartesian product of sets Y and Z consists of all ordered pairs (y, z) where
y ∈ Y and z ∈ Z. This generalises readily to the product of any finite number of
sets, and in particular we shall denote by Y n the Cartesian product of Y with itself
n times. That is, Y n = {(y1 , y2 , . . . , yn ) : yi ∈ Y }. More technically challenging is
the product of an infinite number of sets. If {Yλ }λ∈Λ is a family of sets for which
Λ 6= ∅ and Yλ 6= ∅ for all λS∈ Λ, then the product is denoted Πλ∈Λ Yλ , and consists
of all functions f : Λ → λ Yλ for which f (λ) ∈ Yλ for all λ ∈ Λ. (You should
convince yourself that this gives the same thing for Y1 × Y2 as the first definition.)
That such a function f exists in general is not actually provable within set theory,
and so the fact that Πλ Yλ is a nonempty set has to be taken as an axiom, the
so-called Axiom of Choice. This is a somewhat controversial axiom since some
equivalent forms of the axiom are immensely counterintuitive. On the other hand,
if we consider set theory without the Axiom of Choice then it possible to come
to some equally counterintuitive conclusions, so whether or not one should use it
becomes a matter of taste. We shall only touch on such infinite products briefly,
and so the development of our theory will be consistent with whichever version of
set theory you prefer — providing you are happy with proof by contradiction. . .
Functions
A function or map (the terms are interchangeable) between two sets X and Y is
written f : X → Y , and is a rule that assigns to each element of X an element
of Y . The set X is called the domain or source space of f , and the set Y the
codomain or target space. From a set theory point of view it is actually preferable
to think of f in terms of its graph: Gf = {(x, y) ∈ X × Y : y = f (x)}. It is not
2
hard to see that if we start with our intuitive idea of a function and look at the
graph, then Gf satisfies the following:
Each x ∈ X appears as the first element in exactly one pair from Gf .
Conversely, if G ⊂ X × Y is any subset that satisfies the above property then G
defines a function g : X → Y by setting, for each x ∈ X, g(x) ∈ Y to be the
unique element of Y such that (x, g(x)) ∈ G. This is the traditional set theoretic
definition of a function.
A function f : X → Y is injective or one-to-one if whenever x1 , x2 ∈ X
are distinct points, so are their images f (x1 ) and f (x2 ). That is, if x1 6= x2
then f (x1 ) 6= f (x2 ). Equivalently, f is injective if whenever x1 , x2 ∈ X satisfy
f (x1 ) = f (x2 ), then we must have x1 = x2 . The function f : X → Y is surjective
or onto if for each y ∈ Y there is some x ∈ X such that f (x) = y. That is, every
point in the target space Y is the image of some point x from the source space.
A function that is both injective and surjective is called bijective. Given such a
function f : X → Y there is a unique function g : Y → X such that g ◦ f = IX
and f ◦ g = IY , where IX : X → X is the identity function defined by IX (x) = x
for all x, and similarly for IY . This function g is usually denoted f −1 , the inverse
function of f . The function f is bijective if and only if it is invertible; note also
that in this case f −1 is invertible with (f −1 )−1 = f .
If A ⊂ X is a subset, and f : X → Y any map, then the image of A under f
is the subset
f (A) = {f (a) : a ∈ A} = {y ∈ Y : y = f (a) for some a ∈ A}
of Y . If B ⊂ Y is a subset of Y , then the preimage or inverse image of B under
f is
f −1 (B) = {x ∈ X : f (x) ∈ B},
that is, the subset of those points in X that are mapped by f into B. For example
if we take X = Y = R and let f : R → R be defined by f (x) = x2 , then
f ([2, 3]) = {x2 : 2 ≤ x ≤ 3} = [4, 9], but f −1 ([4, 9]) = [−3, −2] ∪ [2, 3]. Note that
the above definition does not require f to be invertible, and this certainly was
not the case for our example. If, however, f is invertible, then f −1 (B) could be
taken to mean either the preimage of B under f , or the image of B under the map
f −1 : Y → X. Fortunately these sets turn out to be the same thing, and so our
notation cannot lead to confusion.
A more extreme case of when f is not invertible is to take any two sets X and
Y , fix a y0 ∈ Y , and consider the constant map f : X → Y given by f (x) = y0 for
all x ∈ X. Then we have
(
(
∅
if A = ∅,
∅ if y0 ∈
/ B,
f (A) =
and f −1 (B) =
{y0 } if A 6= ∅,
X if y0 ∈ B.
It is not hard to prove the following identities for any family {Aλ }λ∈Λ of subsets
of X and any family {Bγ }γ∈Γ subsets of Y :
T
T
S
S
f ( λ∈Λ Aλ ) ⊂ λ∈Λ f (Aλ ),
f ( λ∈Λ Aλ ) = λ∈Λ f (Aλ )
T
T
S
S
f −1 ( γ∈Γ Bγ ) = γ∈Γ f −1 (Bγ ),
f −1 ( γ∈Γ Bγ ) = γ∈Γ f −1 (Bγ )
(2)
f −1 (Y \ Bγ ) = X \ f −1 (Bγ )
3
Furthermore it is possible to find examples where the first of the identities is in
fact an equality, and other examples when the set on the left is a proper subset
of the set on the right. Thus operation of taking preimages is “more compatible”
with the operations of intersection and union than the operation of taking images.
Further identities include
f f −1 (B) ⊂ B and A ⊂ f −1 f (A) ,
(3)
and again both of these can either be equalities or strict inclusions depending on
the choice of f , A and B.
If f : X → Y is a function from the set X to the set Y , and if Z is a third
set and g : Y → Z is a function from Y to Z, then the function g ◦ f : X → Z is
defined by setting (g ◦ f )(x) = g f (x) for each x ∈ X. Given subsets A ⊂ X and
C ⊂ Z we have
(g ◦ f )(A) = g f (A)
and (g ◦ f )−1 (C) = f −1 g −1 (C) .
(4)
Partitions and onto functions
One important operation on topological spaces that we shall consider briefly involves taking a given space and identifying points, or gluing them together, to form
a new space. For example if we join the ends of the unit interval [0, 1] together we
get a circle. This operation of identifying or gluing involves grouping together the
points in the original space into a family of nonoverlapping subsets, i.e. forming a
partition. More formally a partition of a set X is any family {Aλ }λ∈Λ of nonempty
subsets of X that satisfy
S
and Aλ ∩ Aγ = ∅ whenever λ 6= γ.
λ Aλ = X
Recall that there is a well-known one-to-one correspondence between partitions
of X and equivalence relations on X. The gluing process, however, is defined via
another way of looking at partitions. First given any partition {Aλ }λ∈Λ of X we
can define a function f : X → Λ by
f (x) = λ if x ∈ Aλ ,
since every x ∈ X sits inside precisely one of the sets Aλ . Moreover since each Aλ
is nonempty it follows that the map f is onto.
On the other hand, suppose that Y is a set and that g : X → Y is an onto
function. Then the family of subsets
−1
g ({y}) y∈Y
is a partition of X since each g −1 ({y}) is nonempty (because g is onto), the sets
are disjoint by an application of the third identity from (1), and x ∈ g −1 {g(x)}
for all x ∈ X, so that these preimages cover X.
Unfortunately there is not quite a one-to-one correspondence between partitions of X and onto functions with X as domain, since the target spaces could be
totally disjoint. However, if two onto functions do generate the same partition,
then the target spaces must be related according to the following proposition.
4
Proposition
−10.1. Let
f : X →−1Y and g : X → Z be onto maps. Then the
partitions f ({y}) y∈Y and g ({z}) z∈Z are the same if and only if there is
a bijective map h : Y → Z such that the following diagram commutes:
~~
~~
~
~
~~ ~
f
Y
X@
h
@@ g
@@
@@
@
/Z
That is, if and only if such a map h : Y → Z exists that satisfies h ◦ f = g.
This result allows us to say that there is a one-to-one correspondence between
the collection of partitions the set X and the set of onto maps from X up to
bijective equivalence of their ranges.
For example, if we split the integers Z into the subsets of even and odd integers,
Z = E ∪ O = {0, ±2, ±4, . . .} ∪ {±1, ±3, ±5, . . .},
then we could define onto maps from Z to {−1, 1} and to {0, 1} by
f : Z → {−1, 1},
f (n) = (−1)n
and
g : Z → {0, 1},
g(n) = sin(nπ/2)|.
Then f −1 ({−1}) = O = g −1 ({1}) and f −1 ({1}) = E = g −1 ({0}), so the required
map h : {−1, 1, } → {0, 1} is given by h(−1) = 1 and h(1) = 0.
Exercises
1. Prove De Morgan’s Laws — equation (1).
2. Prove the identities in equations (2–4). Also find an example where the first
inclusion in (2) is strict, and another example where it is an equality. Repeat this
for the two inclusions in (3).
3. Prove Proposition 0.1.
5
1
Metric Spaces
Definition and examples
Recall what it means for a function f : R → R to be continuous at a point a ∈ R:
∀ε > 0 ∃δ > 0 s.t. |x − a| < δ ⇒ |f (x) − f (a)| < ε
In words this says that for any degree of error ε we can find some positive number
δ such that if x is within distance δ of a then f (x) is less than ε from f (a). The
same thing is true if we consider a function g : R2 → R; this is continuous at the
point (a, b) if
∀ε > 0 ∃δ > 0 s.t. |(x, y) − (a, b)| < δ ⇒ |g(x, y) − g(a, b)| < ε
p
where here |(x, y) − (a, b)| = (x − a)2 + (y − b)2 is the Euclidean distance of
(x, y) from (a, b). So again we have that g(x, y) will be as close to g(a, b) as we
like, provided we take (x, y) sufficiently close to (a, b). The basic idea of a metric
space is to abstract this idea of distance between points in Euclidean space.
Definition 1.1. Let X be a nonempty set. A metric on X is a function d :
X × X → R that satisfies:
M1 d(x, y) ≥ 0
∀x, y ∈ X, with d(x, y) = 0 if and only if x = y.
M2 d(x, y) = d(y, x) ∀x, y ∈ X.
(symmetry)
M3 d(x, y) ≤ d(x, z) + d(z, y) ∀x, y, z ∈ X.
(triangle inequality)
The pair (X, d) is called a metric space, and we say that d(x, y) is distance between
x and y.
Example 1.2.
(a) X = R, the real line, and d(x, y) = |x − y|.
(b) X = C, the complex plane, and d(z, w) = |z − w|.
where in both cases | · | denotes the usual modulus function. In both cases M1 and
M2 are easily seen to hold; for (b) note that we have
|z + w| ≤ |z| + |w|
∀z, w ∈ C,
which is the inequality that is usually referred to as the triangle inequality. Given
any other u ∈ C it follows that
d(z, w) = |z − w| = |(z − u) + (u − w)| ≤ |z − u| + |u − w| = d(z, u) + d(u, w)
as required. This (partly) explains the terminology of M3. The inequality for
example (a) is verified in the same way.
Thus we see that the properties M1, M2 and M3 hold in R and in C. In fact
they continue to hold in Rn for any n when equipped with the usual Euclidean
distance, and careful inspection of the proofs of many results about continuous
functions Rm → Rn show that these are the only properties that are really used.
6
Example 1.3. Consider X = Rn together with the map d2 : X × X → [0, ∞)
given by
qPn
2
d2 (x1 , . . . , xn ), (y1 , . . . , yn ) =
i=1 (xi − yi ) .
This defines a metric on Rn (exercise — prove this; M3 requires making use of
the Cauchy-Schwarz inequality). The pair (Rn , d2 ) is known as n-dimensional
Euclidean space.
But d2 is by no means the only metric on Rn . Consider instead the metric d1
given by
P
d1 (x1 , . . . , xn ), (y1 , . . . , yn ) = ni=1 |xi − yi |.
This again is easily seen to be a metric on Rn (indeed, it is simpler to check M3
this time), but note that (Rn , d1 ) 6= (Rn , d2 ) whenever n > 1. For instance, if we
take x = (0, · · · , 0) and y = (1, · · · , 1) then we get
√
d1 (x, y) = n 6= n = d2 (x, y).
The last example showed that we have more than one metric on familiar spaces
such as Rn . In fact it is possible to define a metric on any nonempty set.
Example 1.4. Let X be a nonempty set and define d : X × X → [0, ∞) by
(
0 if x = y,
d(x, y) =
1 if x 6= y.
Then (X, d) is a metric space in which every point is equidistant from every other
point! This metric d is called the discrete metric.
Other natural examples of metrics are those defined on sets of functions.
Example 1.5. Let X = C[0, 1], the set of all continuous functions from [0, 1] to C.
This is an infinite dimensional vector space under the pointwise defined operations
(f + g)(x) = f (x) + g(x),
(tf )(x) = tf (x).
where f, g ∈ X, 0 ≤ x ≤ 1 and t ∈ C. Define
Z 1
|f (x) − g(x)| dx.
d1 (f, g) =
0
Then d1 is a metric on X. It obviously satisfies M2, and for M3 we have for any
h ∈ X that
Z 1
d1 (f, g) =
|f (x) − h(x) + h(x) − g(x)| dx
0
Z 1
≤
|f (x) − h(x)| + |h(x) − g(x)| dx
0
Z 1
Z 1
=
|f (x) − h(x)| dx +
|h(x) − g(x)| dx
0
0
= d1 (f, h) + d1 (h, g).
7
It is also easy to see that d1 (f, g) ≥ 0 for any f, g ∈ X and that d1 (f, f ) = 0.
Exercise: prove carefully that if d1 (f, g) = 0 then f = g.
Again, there are other natural metrics on this set X. For instance
d∞ (f, g) = sup |f (x) − g(x)|
0≤x≤1
defines a metric on X (prove this!). Again, we have d∞ 6= d1 . For instance take
f (x) = 0 and g(x) = x2 , then we have that
Z
d1 (f, g) =
1
x2 dx = [x3 /3]10 = 1/3
0
and
d∞ (f, g) = sup x2 = 1.
0≤x≤1
Note that the modulus function of R defines a metric on the subset Q of rational
numbers. That is, (Q, | · |) is a metric space. It is incomplete, unlike (R, | · |). This
is one example of how to make new metric spaces out of others.
Proposition 1.6. (a) Let (X, dX ) be a metric space and let Z be any nonempty
subset of X. Let dZ denote the restriction of the function dX to the subset Z × Z
of X × X. Then (Z, dZ ) is a metric space.
(b) Let (X, dX ) and (Y, dY ) be metric spaces, and let W = X × Y . Then the
following are three metrics on W :
d1 (x1 , y1 ), (x2 , y2 ) = dX (x1 , x2 ) + dY (y1 , y2 )
p
d2 (x1 , y1 ), (x2 , y2 ) = dX (x1 , x2 )2 + dY (y1 , y2 )2
d∞ (x1 , y1 ), (x2 , y2 ) = max dX (x1 , x2 ), dY (y1 , y2 )
Proof. Exercise.
The space (Z, dZ ) is a subspace of (X, dX ). The set W with any of the metrics
d1 , d2 or d∞ is a candidate for the product of (X, dX ) and (Y, dY ), and in a certain
sense (W, d1 ), (W, d2 ) and (W, d∞ ) are equivalent as we shall see later. The product
construction extends the product of any finite number of metric spaces. Products
of an infinite number of spaces require more care.
Convergence and continuity
Analysis is concerned with the study of limits and continuity. Now that we have
introduced the idea of a metric space with its notion of the distance between any
two points of an abstract set, we can give the obvious definitions for convergence
of sequences and continuity of functions in this setting.
Definition 1.7. Let (X, dX ) and (Y, dY ) be metric spaces, and let x ∈ X.
8
(a) Let (xn )n≥1 be a sequence of elements from X. The sequence converges to
x if for each ε > 0 there is some N ≥ 1 such that
n ≥ N ⇒ d(xn , x) < ε
This is denoted limn xn = x or xn → x.
(b) A function f : X → Y is continuous at x0 ∈ X if for each ε > 0 there is
some δ > 0 such that
dX (x0 , x) < δ ⇒ dY (f (x0 ), f (x)) < ε.
The function is said to be continuous (on X) if it is continuous at each point
of X.
Note. The sequence (xn ) converges to x if and only if dX (xn , x) → 0, where we are
now dealing with the familiar concept of convergence of sequences of real numbers.
In particular the above definition coincides with the usual one for sequences of
numbers, when R is equipped with its usual metric.
Example 1.8. The particular choice of metric can make a big difference as to
whether or not a sequence is convergent. For example consider X = C[0, 1] with
the metrics d1 and d∞ as given in Example 1.5, and consider the following sequence
of functions:
1
1
1
f1
1
1
2
1
f2
1
3
1
f3
1
We have d1 (fn , 0) = 2n
→ 0, and so fn → 0 with respect to d1 . However
d∞ (fn , 0) = 1 for all n, and so fn 6→ 0 with respect to the uniform metric.
On the other hand, let d denote the discrete metric. If (gn ) ⊂ C[0, 1] is convergent to some g then there should be some N ≥ 1 such that d(g, gn ) < 1 for all
n ≥ N . Hence d(g, gn ) = 0 for all n ≥ N — the sequence is eventually constant,
which is certainly not true of the fn above.
Having introduced a general notion of distance, we want to show that we can
do away with this when discussing continuity, rephrasing everything in terms of a
distinguished class of subsets. The idea of distance still does have an important
(even indispensable) role in other matters.
Definition 1.9. Let (X, d) be a metric space. For any x ∈ X and ε > 0 we write
B(x, ε) = {y ∈ X : d(x, y) < ε}.
The set B(x, ε) is called the open ball of radius ε centred on x.
9
Example 1.10. Equipping R with its usual metric we see that B(x, ε) is the open
interval (x − ε, x + ε). Conversely, if we take any bounded open interval (a, b) then
this is equal to B(x, ε) for x = (a + b)/2 and ε = (b − a)/2.
Example 1.11. Consider the set X = R2 together with the metrics d1 and d2
from Example 1.3, along with a third metric d∞ given by d∞ (x, y) = max{|x1 −
y1 |, |x2 − y2 |} (exercise: check this is a metric). The open balls of radius 1 about
the origin 0 have the following forms:
1
1
1
1
1
1
d1
d2
d3
Example 1.12. If a set X is a given the discrete metric then
(
{x} if ε ≤ 1,
B(x, ε) =
X
if ε > 1.
Definition 1.13. Let X be a metric space.
(a) A subset U of a metric space X is open (in X) if for each x ∈ U there is
some ε > 0 such that B(x, ε) ⊂ U . The ε is allowed to depend on the point
x.
PSfrag
(b) A subset R of X is a neighbourhood of the point x ∈ X if R is an open set
such that x ∈ R.
ε
x
B(x, ε)
R
Remarks. (i) It follows vacuously that the empty set ∅ is open — there are no
points x ∈ ∅ to check.
(ii) Some authors use a slightly more general definition of neighbourhood that
does not require R to be an open set. However it should still contain an open ball
centred on x.
The next result shows that all the various notions of open fortunately coincide.
10
Proposition 1.14. Let X be a metric space. Each open ball is an open set according the above definition.
y
δ
x
ε
Proof. Pick an open ball B(x, ε) and a point y ∈ B(x, ε). So in particular we have
d(x, y) < ε, hence δ := ε − d(x, y) > 0. Consider the ball B(y, δ). If z ∈ B(y, δ)
then
d(x, z) ≤ d(x, y) + d(y, z) < d(x, y) + δ = d(x, y) + ε − d(x, y) = ε.
Thus B(y, δ) ⊂ B(x, ε) as required.
Remark. We have shown that B(x, ε) is a neighbourhood of all of its points.
As an example of a subset of R that is not open, consider [a, b] for any a ≤ b.
(In particular we could take a = b = 0 to get the set {0}). Now for any point x
such that a < x < b we can find some ε > 0 such that (x − ε, x + ε) ⊂ [a, b]. For
example let ε = min{x−a, b−x}. But no such ε exists for a or b. For example, the
set B(a, ε) contains points to the left of a, and consequently outside the interval
[a, b]. Similarly at b. For the same sort of reason, neither (a, b] nor [a, b) is open,
where now we must take a < b.
Proposition 1.15. Let (X, d) be a metric space, x ∈ X and (xn ) a sequence in
X. The following are equivalent:
(i) xn → x as n → ∞.
(ii) For each neighbourhood R of the point x there is some N ≥ 1 such that
n ≥ N ⇒ xn ∈ R.
[This is written: the sequence (xn ) is eventually/ ultimately in R]
Proof. (i ⇒ ii): Let R be a neighbourhood of x, so R is an open set such that
x ∈ R. Hence there is some ε > 0 such that B(x, ε) ⊂ R. Now xn → 0, so there is
some N ≥ 1 such that
n ≥ N ⇒ xn ∈ B(x, ε) ⊂ R
as required.
11
(ii ⇒ i): Let ε > 0, then B(x, ε) is a neighbourhood of x, and so by definition
there is some N such that
n ≥ N ⇒ xn ∈ B(x, ε),
that is, xn → x, since ε was arbitrary.
The next result gives several characterisations of continuity at a point, but in
order to state these we must recall the notation given in Section 0 concerning maps
and subsets.
Definition 1.16. Let X and Y be sets and f : X → Y a map. Let A ⊂ X and
B ⊂ Y be subsets. Then
f (A) = {f (x) : x ∈ A}
f −1 (B) = {x ∈ X : f (x) ∈ B}
(the image of A)
(the inverse image or preimage of B)
That is, f (A) consists of all those elements of Y that are the image of some point
of A, and f −1 (B) consists of all those points of X that are mapped into B.
Note. We do not require f to be surjective, so there could be some points in B
that no element of X is mapped to. So we could have f −1 (B) = ∅ when B 6= ∅ —
consider f −1 ([2, 3]) when f : R → R is the map f (x) = sin x. On the other hand,
if A 6= ∅ then f (A) 6= ∅.
Proposition 1.17. Let f : X → Y be a map between metric spaces, and let x ∈ X.
The following are equivalent:
(i) f is continuous at x
(ii) For each ε > 0 there is some δ > 0 such that f B(x, δ) ⊂ B(f (x), ε).
(iii) For each neighbourhood S of f (x) there is some neighbourhood R of x such
that f (R) ⊂ S. [i.e. x ∈ R ⇒ f (x) ∈ S]
(iv) If (xn ) ⊂ X is any sequence such that xn → x, then f (xn ) → f (x).
Proof. The equivalence of (i) and (ii) is clear — we have just rewritten the definition of continuity at a point in terms of open balls and their images under
f.
(ii ⇒ iii): Let S be any neighbourhood of f (x), so then S is an open set such
that f (x) ∈ S. Thus there is some ε > 0 such thatB(f (x), ε) ⊂ S. Now by (ii) we
know that there is some δ > 0 such that f B(x, δ) ⊂ B(f (x), ε) ⊂ S, and B(x, δ)
is a neighbourhood of x.
(iii ⇒ iv): Let (xn ) be any sequence that is convergent to x, and let S be any
neighbourhood of f (x). Then there is a neighbourhood R of x such that f (R) ⊂ S.
But xn → x, and so there is some N ≥ 1 such that if n ≥ N then xn ∈ R. But
this implies that
n ≥ N ⇒ xn ∈ R ⇒ f (xn ) ∈ f (R) ⊂ S
12
and so f (xn ) → f (x) as required, by Proposition 1.15.
(iv ⇒ ii): Suppose that (ii) does not hold, so then there must be some ε > 0
such that
f (B(x, δ)) 6⊂ B(f (x), ε) ∀δ > 0.
In particular f B(x, 1/n) 6⊂ B(f (x), ε) for each integer n ≥ 1, and so we can
choose xn ∈ B(x, 1/n) for each n such that d f (x), f (xn ) ≥ ε.
So now (xn ) is a sequence in X with d(x, xn ) < n−1 → 0, so that xn → x. But
f (xn ) 6→ f (x), since d f (x), f (xn ) 6→ 0. Hence (iv) does not hold. Taking the
contrapositive gives (iv ⇒ ii) as required.
Corollary 1.18. Let f : X → Y be a map between metric spaces. The following
are equivalent:
(i) f is continuous
(ii) For each open subset V ⊂ Y , f −1 (V ) is open in X
Proof. (i ⇒ ii): Suppose V ⊂ Y is open and let x ∈ f −1 (V ). Now V is a neighbourhood of f (x) and f is continuous at x, so by part (iii) of our proposition there
is some neighbourhood R of x such that f (R) ⊂ V . This implies that R ⊂ f −1 (V ).
But R being a neighbourhood of x means that there is some δ > 0 such that
B(x, δ) ⊂ R ⊂ f −1 (V ), and so f −1 (V ) is open.
(ii ⇒ i): Let x ∈ X and let V ⊂ Y be open with f (x) ∈ V . That is, V is a
neighbourhood of f (x), and f −1 (V ) is open by hypothesis. Moreover x ∈ f −1 (V )
and f (f −1 (V )) ⊂ V . So f is continuous at x by condition (iii) of the proposition,
and since x was arbitrary, f is continuous on X.
Properties of open and closed sets
We have now characterised continuity of maps in terms of the behaviour of a
particular class of subsets under the operation of taking the inverse image. This
will be the basis of the definition of continuity later on in the context of topological
spaces. First we must investigate other useful properties of open sets, since this
will also inform these later definitions.
Proposition 1.19. Let X be a metric space. The subsets X and ∅ are open.
Moreover :
S
(a) If {Uλ }λ∈Λ is a family of open subsets of X then their union λ∈Λ Uλ is also
open.
n
(b) If
Tn{Vi }i=1 is a finite family of open subsets of X then their intersection
i=1 Vi is again open.
Proof. Let x ∈ X then B(x, ε) ⊂ X for all choices of ε > 0, hence X is open.
Similarly, to show ∅ is open we must find, for any given x ∈ ∅, an ε > 0 such that
B(x, ε) ⊂ ∅. ButSthis holds vacuously, since there are no points in ∅!
(a) Let x ∈ λ Uλ , then x ∈ Uλ0 for some λ0 ∈ Λ, the indexing
S set. Now Uλ0
is open and so there is some ε > 0 such that B(x, ε) ⊂ Uλ0 ⊂ λ Uλ . Hence the
union is open.
13
(b) It is enough to prove this for the intersection of two open sets, U and V
say, since the general case then follows by induction. So let x ∈ U ∩ V , then x ∈ U
and x ∈ V , and thus there are numbers εU > 0 and εV > 0 such that
B(x, εU ) ⊂ U and B(x, εV ) ⊂ V.
Put ε = min{εU , εV }, then ε > 0 and
B(x, ε) = B(x, εU ) ∩ B(x, εV ) ⊂ U ∩ V,
and so U ∩ V is open.
Remarks. (i) In part (a) the family {Uλ }λ∈Λ may contain an uncountable number
of sets. One way to think of this object is as a map from the index set Λ into the
power set P(X), whose range lies in the subclass of open sets.
(ii) Part (b) already begins to fail if we take the intersection of a countable
family of open sets. For example take a ≤ b in R and for each n ≥ 1 consider the
open interval
Un = (a − 1/n, b + 1/n).
These are all open (they are open balls), but
T
n≥1 Un = [a, b]
which is not open.
Although we shall generally define everything in terms of open sets in what
follows, an alternative ‘dual’ point of view is available, produced by taking the
complement of everything in sight.
Definition 1.20. Let X be a metric space. A subset F ⊂ X is closed if F c = X \F
is open.
Taking the complement of Proposition 1.19 and applying De Morgan’s Laws
yields:
Proposition 1.21. The intersection of any family of closed sets is again closed.
The union of a finite number of closed sets is again closed.
Proof. If {Fλ }λ∈Λ are closed then
T
S
X \ λ Fλ = λ (X \ Fλ ),
a union of open sets, hence open. The second part is proved similarly.
It is important to note that there are sets that are both open and closed, and
some that are neither. Indeed in any metric space X the sets X and ∅ are both
open and closed, and in many important examples they are the only two sets with
that property. Other examples of closed sets include the closed subintervals of R,
that is, sets of the form [a, b] for a ≤ b. That these sets are closed follows since
[a, b]c = (−∞, a) ∪ (b, ∞).
That is, the complement is the union of two open sets, so is itself open. Examples
of sets that are neither open nor closed are the intervals (a, b] and [a, b) (where
now we must take a < b to avoid obtaining the empty set).
14
Definition 1.22. Let X be a metric space and A ⊂ X a subset. The closure of
A, denoted A, is the smallest closed subset of X that contains A.
To see that this definition actually makes sense, consider the following family
of sets:
F = {F ⊂ X : F closed, F ⊃ A}.
Note that X
T ∈ F, no matter what choice of A we take, so the family F is nonempty.
Put A1 = F ∈F F , then A1 is closed by Proposition 1.21. Moreover, since A ⊂ F
for all F ∈ F, we have A ⊂ A1 . Conversely if E is any closed set that contains A
then E ∈ F and so A1 ⊂ E by construction of A1 . Thus A1 is the smallest closed
set containing A.
Proposition 1.23. Let A be a subset of a metric space X, and let x ∈ X. The
following are equivalent:
(i) x ∈ A.
(ii) R ∩ A 6= ∅ for all neighbourhoods R of x.
(iii) x = limn xn for some sequence (xn ) ⊂ A.
Proof. (i ⇒ ii): Suppose for a contradiction that there is some neighbourhood R
of x that satisfies R ∩ A = ∅. Then A ⊂ Rc , and Rc is closed since R is open.
Then, by definition of A,
A ⊂ Rc ⇒ A ⊂ Rc ⇒ A ∩ R = ∅.
Since x ∈ R we get x ∈
/ A, the required contradiction.
(ii ⇒ iii): For each integer n ≥ 1 we can take R = B(x, 1/n), and thus pick
some xn ∈ B(x, 1/n) ∩ A. Then (xn ) ⊂ A, and d(x, xn ) < 1/n for all n. Hence
(xn ) ⊂ A and xn → x.
(iii ⇒ i): Suppose that x = limn xn for some sequence (xn ) ⊂ A, but that
c
c
x ∈
/ A. Then x ∈ A , and A is open, hence a neighbourhood of x. But then,
c
by Proposition 1.15, we have xn ∈ A for all sufficiently large n, so that xn ∈
/ A,
hence xn ∈
/ A, which is impossible. Thus, by contradiction, x ∈ A.
Note that by definition A is closed, so if A ⊂ X satisfies A = A then it is a
closed subset. On the other hand, if A is known to be closed then it is clearly the
smallest closed subset that contains itself, and so A = A. That is, A = A if and
only if A is closed. In particular, since A is closed, the closure of the closure is
just A, that is A = A.
Corollary 1.24. Let A be a subset of a metric space X. Then A is closed if and
only if limn xn ∈ A for every sequence (xn ) ⊂ A that is convergent in X.
Proposition 1.25. Let X be a metric space.
(a) For any collection {Aλ }λ∈Λ of subsets of X we have
T
T
λ Aλ ⊂
λ Aλ .
15
(b) For any finite collection {Bi }ni=1 of subsets of X we have
S
i Bi
=
S
i Bi .
Proof. (a) Since Aλ ⊂ Aλ for all λ ∈ Λ we have
T
λ Aλ
⊂
T
λ Aλ .
But the set on the right hand side is closed by Proposition 1.21, so by definition
T
λ Aλ
⊂
T
⊂
S
λ Aλ .
(b) Similar reasoning gives
S
since our union is finite. So let x ∈
i Bi
S
i Bi ,
i Bi
then x ∈ Bj for some 1 ≤ j ≤ n. Thus
(R ∩ Bj ) 6= ∅
for all neighbourhoods R of x, and consequently,
R ∩ Bj ⊂
for every such R. Hence x ∈
S
S
i (R
i Bi
∩ Bi ) = (R ∩
S
i Bi )
6= ∅
as required.
It is impossible to get equality in (a) in general, even if we restrict ourselves
to only a finite number of sets. For examples consider the subsets A = (0, 1) and
B = (1, 2) of R. Then
A ∩ B = ∅ ⇒ A ∩ B = ∅;
A = [0, 1], B = [1, 2] ⇒ A ∩ B = {1}.
Similarly we must take a finite number of sets in (b); again working with subsets
of R we put Bq = {q} for each q ∈ Q, then Bq = {q}, and so
S
q∈Q Bq
= Q,
but
S
q∈Q Bq
= Q = R.
Exercises
1.
(a) Prove that the following inequality holds for any n ≥ 1 and ai , bi ∈ [0, ∞):
Pn
i=1 ai bi
≤
2 1/2
i=1 ai
Pn
2 1/2
i=1 bi
Pn
[Hint: consider the quadratic polynomial p(x) =
P
i (ai
− xbi )2 .]
Hence show that the map d2 defined as follows is a metric on Cn
Pn
2 1/2
d2 (z1 , . . . , zn ), (w1 , . . . , wn ) =
i=1 |zi − wi |
16
(‡)
(b) Prove that if f : [0, 1] → [0, ∞) is a continuous function then
Z 1
f (t) dt = 0 ⇔ f (t) = 0 ∀t ∈ [0, 1].
0
Hence show that the map d1 defined as follows is a metric on C[0, 1]
Z 1
|f (t) − g(t)| dt
d1 (f, g) =
0
[To show that the function d2 (f, g) =
requires an analogue of (‡).]
R
1
0 |f (t)
− g(t)| dt
1/2
is a metric on C[0, 1]
2. Find three different metrics on N, no two of which are multiples of each other.
3. Let (X, d) be a metric space. Show that d0 defined by
d0 (x, y) =
d(x, y)
1 + d(x, y)
is another metric on X. [Note that d0 (x, y) < 1 ∀x, y ∈ X.]
4. Consider the function d : R2 × R2 → R defined by
(
|y1 − y2 |
if x1 = x2 ,
d (x1 , y1 ), (x2 , y2 ) =
|y1 | + |x1 − x2 | + |y2 | if x1 6= x2 .
Show that d is a metric on R2 . Sketch the balls B (2, 0), 1 , B (1, 2), 1 and
B (1, 1), 2 .
5. Let X be a set and d a map X × X → R satisfying
M10 d(x, y) = 0 ⇔ x = y, and M20 d(x, y) ≤ d(x, z) + d(y, z) ∀x, y, z ∈ X.
Show that d is a metric on X.
6. Let (xn ) and (yn ) be two convergent sequences in a metric space X, with x =
limn xn and y = limn yn . Show that d(xn , yn ) → d(x, y).
7. Let X be a metric space and let f : X → R and g : X → R continuous functions.
Show that f + g, tf (t ∈ R), |f |, max{f, g}, min{f, g} and f g are continuous,
where
(f + g)(x) = f (x) + g(x),
(tf )(x) = tf (x),
|f |(x) = |f (x)|, etc.
8. Let X = C[0, 1] and let F : X → C denote the map F (f ) = f (0) (for f ∈ X). Is
F continuous when X is given (i) the d1 metric, or (ii) the d∞ metric?
9. Show that any open set in R (with the usual metric) is the union of a countable
collection of open intervals. Show that this collection can be chosen so that any
two distinct intervals are disjoint.
17
10. Prove that the set {x} is closed for any point x in any metric space X.
11. Prove that a subset of a metric space is open if and only if it is a union of open
balls.
12. Prove that every subset of a discrete metric space X is open. Hence show that
any map f : X → Y from X into any other metric space Y is continuous.
13. We have shown that a function f : X → Y between metric spaces is continuous if
and only if for every open set U in Y , the set f −1 (U ) is open in X. The analogous
statement involving images does not hold: find an example of metric spaces X and
Y and a continuous map f : X → Y such that f (V ) is not open for some open set
V ⊂ X.
14. Prove that Q ∩ (x, x + ε) 6= ∅ for all choices of x ∈ R and ε > 0. Hence deduce
that the closure of Q is R.
15. Let A be a subset of a metric space X. Show that if x ∈ A \ A then each open
ball centred on x contains an infinite number of (distinct) points of A. Does this
remain true if we take x ∈ A?
16. Let X and Y be metric spaces, and A a subset of X. If f : X → Y and g : X → Y
are continuous functions such that f (x) = g(x) for all x ∈ A, show that f (x) = g(x)
for all x ∈ A.
17. Let d1 , d2 and d∞ be the metrics on R2 given by
d1 (x1 , x2 ), (y1 , y2 ) = |x1 − y1 | + |x2 − y2 |,
1/2
d2 (x1 , x2 ), (y1 , y2 ) = |x1 − y1 |2 + |x2 − y2 |2
,
d∞ (x1 , x2 ), (y1 , y2 ) = max |x1 − y1 |, |x2 − y2 | .
Show that these metrics are Lipschitz equivalent. Show that the discrete metric
on R2 is not Lipschitz equivalent to any of the above metrics.
18. Prove that the metric d1 on C[0, 1] is not Lipschitz equivalent to the metric d∞ ,
where d∞ (f, g) = supt∈[0,1] |f (t) − g(t)|.
18
2
Topological Spaces
Definitions and examples
Having had a brief excursion through metric space theory, we have prepared the
way for a further generalisation, where now we can consider the theory of topological spaces. The idea is to do away totally with the concept of distance, and
deal only with subsets of our space.
Definition 2.1. A topological space is a pair (X, T) consisting of a nonempty
set X and a collection T of subsets of X that satisfy the following properties:
T1
∅, X ∈ T
T2
S
If {Uλ }λ∈Λ ⊂ T, then λ Uλ ∈ T.
T
If {Vi }ni=1 ⊂ T then i Vi ∈ T.
T3
The collection T is called a topology, and its elements of are called open sets.
The axioms state that the union of any collection of open sets is again open, but
we only require that T be closed under finite intersections.
So note that here we are supplied with our open sets, whereas in the case of
metric spaces it was the metric that was fundamental, and with which we had to
check if a given set was open or not.
Example 2.2. The following are all examples of topological spaces:
(i) The indiscrete topology on any (nonempty) set X is the collection {∅, X}.
(ii) The discrete topology on any (nonempty) set X is the power set P(X)
consisting of all subsets of X.
(iii) If (X, d) is a metric space, then let Td denoted the collection of subsets that
are open with respect to d, as in Definition 1.13. Then the pair (X, Td ) is a
topological space by Proposition 1.19.
If (Y, T) is a topological space for which there is some metric d on Y such that
T = Td , then the space (Y, T) is said to be metrizable. Not all topological
spaces occur this way — for example if |X| ≥ 2 then X equipped with the
indiscrete topology is not metrizable (why?). On the other hand, the discrete
topology is that induced by the discrete metric.
(iv) Given a set X, the cofinite or Zariski topology on X consists of ∅ together
with any subset A such that Ac = X \ A is finite. Exercise: show that this
does indeed define a topology on X.
Again, this is (rarely) a metrizable topological space, but is of great importance in algebraic topology since it is well-suited for the study of polynomial
equations. The Zariski topology on C has far fewer open sets than in the
usual metric topology. In fact, the only closed sets are those corresponding
to the zeros of polynomials.
19
Definition 2.3. Let (X, T) and (Y, S) be topological spaces, and let f : X → Y
be a map. Then f is continuous if
V ∈ S ⇒ f −1 (V ) ∈ T.
If we want to stress the topologies involved, we say that f is (T, S)-continuous.
Again, we should check that we have not lost anything from metric space during
this process of generalisation. That is, if f : X → Y is a map between metric spaces
(X, dX ) and (Y, dY ) then we would like it to be continuous according to the metric
space definition (Definition 1.7) if and only if it continuous with respect to the
topological space definition, where we equip X and Y with the topologies induced
by their metrics. But this is essentially the content of Corollary 1.18.
Also, we are now finally in a position to explain why the choice of metric is
sometimes not all that important.
Definition 2.4. Let X be a nonempty set, and let d1 and d2 be metrics on X.
They are Lipschitz equivalent if there are constants 0 < a ≤ b such that
ad1 (x, y) ≤ d2 (x, y) ≤ bd1 (x, y) ∀x, y ∈ X.
Examples of metrics that satisfy this condition include d1 , d2 and d∞ on Rn
as defined in Example 1.3 and Example 1.11. One can check that
d∞ (x, y) ≤ d1 (x, y) ≤
√
nd2 (x, y) ≤ nd∞ (x, y),
where the second inequality follows by an application of the Cauchy-Schwarz inequality.
Whenever we have Lipschitz equivalent metrics, we end up in the following
pleasant situation:
Proposition 2.5. Let X be a nonempty set, and let d1 and d2 be metrics on X
that are Lipschitz equivalent. Then the topologies that they induce on X are the
same. Consequently if (Y, d) is any other metric space, and g any map X → Y
or Y → X, then g is continuous with respect to the d1 metric if and only if it is
continuous with respect to the d2 metric.
Proof. This follows from the inclusions
B 1 (x, δ/b) ⊂ B 2 (x, δ)
and
B 2 (x, ε/a) ⊂ B 1 (x, ε)
Thus in many ways it is unimportant which metric we choose to work with
when dealing with Rn (or any subset of Rn ), and so we can choose whichever
is most convenient at that particular instant. Similar arguments show that the
three metrics given on the product X × Y of two metric spaces are also Lipschitz
equivalent, and hence lead to the same topology.
As an example of the economy gained by generalising to topological spaces, and
dealing with global rather than local continuity, considering the following result
that is standard in any course on real or complex analysis:
20
Proposition 2.6. Let (Xi , Ti ), i = 1, 2, 3 be topological spaces, and let f : X1 →
X2 and g : X2 → X3 be continuous maps. Then their composition g ◦ f : X1 → X3
is continuous.
Proof. For any subset A ⊂ X3 we have (g ◦ f )−1 (A) = f −1 (g −1 (A)), and so
U ∈ T3 ⇒ g −1 (U ) ∈ T2 ⇒ f −1 (g −1 (U )) = (g ◦ f )−1 (U ) ∈ T1 ,
since f and g are continuous. Hence g ◦ f is continuous.
Definition 2.7. Let (X, T) be a topological space. A subset F ⊂ X is closed if
F c = X \ F ∈ T.
Note that this is precisely the same definition as given in the special case of
metric spaces. So X and ∅ are closed, and once again, De Morgan’s laws lead
immediately to:
Proposition 2.8. The intersection of any family of closed sets in a topological
space is closed. The union of any finite number of closed sets is closed.
Topology can be developed in terms of closed rather than open sets, as the
following shows:
Proposition 2.9. Let f : X → Y be a map between topological spaces. Then f is
continuous if and only if for every closed subset F ⊂ Y , f −1 (F ) is closed in X.
Proof. This is an immediate consequence of the following identity:
f −1 (Y \ A) = X \ f −1 (A),
which is valid for any subset A ⊂ Y .
Thus continuity could have instead been defined by saying that the inverse
image of any closed subset is closed, rather than the given definition.
Sequences; separation axioms
The concepts of convergence of sequences, and of closure of sets carries over, if
we think about them in the correct way. For convergence of sequences we must
now make use of the alternative characterisation of convergence given in Proposition 1.15, since we no longer have any open balls to hand.
Definition 2.10. Let (xn ) be a sequence in a topological space X, and let x ∈ X.
Then the sequence converges to x if for every neighbourhood R of x there is some
N ≥ 1 such that
n ≥ N ⇒ xn ∈ R.
At this point where we can illustrate one way in which topological spaces can
be very far from our intuitive ideas coming from metric space theory. If X is any
set equipped with the indiscrete topology, then every sequence in X converges to
every point of X. This happens because T = {∅, X}, so given any x ∈ X and
(xn ) ⊂ X, the only open set containing x is X itself, and xn ∈ X for all n!
21
One way to avoid this embarrassment is to impose a further condition on the
sort of topologies that we shall work with. Such conditions fall under the heading
of separation axioms and involve making use of open sets to distinguish between
points. One of the most important examples is the following:
Definition 2.11. A Hausdorff space is any topological space (X, T) such that if
x, y ∈ X with x 6= y, then there are U, V ∈ T such that
x ∈ U, y ∈ V and U ∩ V = ∅.
That is, there are neighbourhoods of x and y that are disjoint.
One way of remembering what the Hausdorff condition does for you, is that
“distinct points can be housed off.” There are weaker conditions than this (e.g.
T1 -spaces) and stronger conditions (e.g. normal spaces). Simmons has more information on these. One useful effect of restricting our attention to Hausdorff spaces
is that sequences are better behaved.
Proposition 2.12. Any sequence in a Hausdorff space has at most one limit.
Proof. Suppose that X is a Hausdorff space and that (xn ) is a sequence in X
that converges to both x and y, with x 6= y. The Hausdorff condition implies the
existence of open sets U and V such that
x ∈ U, y ∈ V and U ∩ V = ∅.
But since xn → x and xn → y we have that the sequence must eventually be in
U , as well as eventually being in V , which is impossible.
The reason why we had not encountered such problems earlier is that the
topology induced by a metric automatically makes the space into a Hausdorff space.
(Exercise: prove this.) In certain books the authors will assume Hausdorff as part
of the definition of a topological space, in order to circumvent such pathologies as
we encountered above. But topologies such as the Zariski topology on an infinite
set are not Hausdorff (and hence cannot be metrizable), and yet have uses in
applications of this general theory.
Now turning to the concept of closures, the following is (almost) word for word
the same as Definition 1.22.
Definition 2.13. Let X be a topological space and A ⊂ X a subset. The closure
of A, denoted A, is the smallest closed subset of X that contains A.
Again to construct A it is enough to take the intersection of all closed subsets
that contain A. In Proposition 1.23 we characterised the elements of A in two
ways, one involving sequences, the other neighbourhoods and their intersection
with A. The following remains true:
Proposition 2.14. Let A be a subset of a topological space X, and let x ∈ X.
Then x ∈ A if and only if R ∩ A 6= ∅ for all neighbourhoods R of X.
We have already seen sequences behaving badly once. The next example shows
that we cannot use them to characterise the elements of closures:
22
Example 2.15. Let X = [0, 1] and let T be the family of subsets consisting of
∅ together with every subset U such that U c is countable. (Exercise: prove that
this Zariski-like collection of subsets is indeed a topology, sometimes called the
cocountable topology.) Now consider the set A = [0, 1), then we must have
A = [0, 1) or A = [0, 1]
c
since we always have A ⊂ A. However if A = [0, 1) then A = {1} should be open,
c
and so A = (A )c = [0, 1) must be countable. This certainly is not the case, so in
fact we must have that A = [0, 1].
Now let (an ) be any sequence in A, and set U = X \ {a1 , a2 , . . .}. By construction this U is an open set, being the complement of a countable set, and moreover
1 ∈ U , since we have only taken out elements of A. But now we see that an 6→ 1,
since it never gets into this particular neighbourhood of the point 1 ∈ A. So no
sequence from A converges to 1.
Definition 2.16. A subset A of a topological space X is dense if A = X. A
topological space is separable if it has a countable dense subset.
Example 2.17.
(i) We have shown previously that Q = R, when R is given its usual topology.
Thus R is separable since Q is a countable dense subset.
(ii) If a nonempty set X is given the indiscrete topology then the closure of any
nonempty subset A is the whole space X, since the only closed sets are ∅
and X. Thus in particular we have {x} = X for any x ∈ X, and so X is
separable.
(iii) If a nonempty set X is given the discrete topology then every set is open,
hence every set is also closed, and so A = A for every subset A ⊂ X. In
particular the only dense subset of X is X itself. Thus if we equip R with
the discrete topology then it is a nonseparable space.
We can equip a given set X with many different topologies in general. For instance we always have the discrete and indiscrete topologies, and these are distinct
unless |X| = 1. If T1 and T2 are topologies on the given set X, we say that T1 is
weaker or coarser than T2 if T1 ⊂ T2 . That is, every set that is T1 -open is also
T2 -open. Alternatively T2 is described as being stronger or finer than T1 . As an
example, the indiscrete topology is always coarser than the discrete topology.
If T1 and T2 are topologies on X that are related in this way, if (Y, S) is another
topological space and if f : X → Y is a map, then f is continuous with respect to
T2 if it is continuous with respect to T1 . Similarly if a given map g : Y → X is
continuous with respect to T2 then it is continuous with respect to T1 . That is,
strengthening the topology of the domain/source space does not disrupt continuity,
nor does weakening the topology of the codomain/target space.
For example, consider any set X equipped with a topology T, and let Ti and
Td denote the indiscrete topologies respectively. Now if we denote the identity
mapping X → X by i (that is i(x) = x), then i is always (T, T)-continuous, since
i−1 (A) = A for all A ⊂ X, so in particular if A ∈ T then i−1 (A) ∈ T. We also
have that i is
23
• (T, Ti )-continuous,
• (Td , T)-continuous, and
• (Td , Ti )-continuous,
• but not (Ti , Td )-continuous (unless |X| = 1).
Note also thatTif {Tλ }λ∈Λ is any family of topologies on a set X, then so is
their intersection λ∈Λ Tλ — this collection will always contain X and ∅.
Subspaces and product spaces
Common ways of creating new topological spaces from old are to restrict to a
subset, or take the Cartesian product of two or more, in much the same way as we
did with metric spaces.
Definition 2.18. Let (X, T) be a topological space, and let Z be a nonempty
subset of X. Then TZ = {U ∩ Z : U ∈ T} defines a topology on Z called the
subspace topology, or induced topology.
Obviously one should check that TZ does indeed define a topology on Z. Moreover if X is a metric space, then one should check that the topology induced on Z
by the metric restricted to Z is the same thing as the subspace topology coming
from the topology induced by the metric on X. That is, it doesn’t matter which
way we go round the following square:
(X, dX )
restrict
induce
/ (Z, dZ )
induce
(X, TdX )
restrict
$ / (Z, Td )
Z
This induced topology can be singled out as the only one that has certain nice
properties when it comes to composition of maps.
Proposition 2.19. Let Z be a subset of a topological space (X, T), equip Z with
the subspace topology TZ , and let i denote the inclusion map i : Z ,→ X that maps
x ∈ Z to x ∈ X. Then for any other topological space (Y, S)
(a) If f : X → Y is continuous, so is f ◦ i : Z → Y .
(b) A map g : Y → Z is continuous if and only if i ◦ g : Y → X is continuous.
Moreover, TZ is the unique topology such that property (b) holds for all choices of
Y and g.
24
Proof. For any subset A ⊂ X we have i−1 (A) = A ∩ Z, and so if U ∈ T then
i−1 (U ) = U ∩ Z ∈ TZ . In particular i is continuous.
If f : X → Y is a continuous map then f ◦i is the composition of two continuous
functions, and hence also continuous. Similarly if g : Y → Z is continuous. So
suppose now that g : Y → Z is a map such that i ◦ g is continuous. Then for all
U ∈ T we have
g −1 (U ∩ Z) = g −1 (i−1 (U )) = (i ◦ g)−1 (U ) ∈ S,
and hence g is continuous, since every open subset of Z is of the form U ∩ Z for
some U ∈ T.
Finally, suppose that T1 is a topology on Z such that (b) holds for all choices
of Y and g. Consider the following diagram:
(Y, S)
g
II
II
II
I
i◦g II$
/ (Z, T1 )
t
tt
tt
t
t
ztt i
(X, T)
This must commute whenever we choose (Y, S) and g such that either g is continuous or i◦g is continuous. We do this in two different ways, first with Y = Z, S = T1
and g = id. In this case we have that g is continuous, and so i ◦ g = i is continuous
from (Z, T1 ) to (X, T). Hence for any U ∈ T we have that i−1 (U ) = U ∩ Z ∈ T1 .
But this says that TZ ⊂ T1 , that is TZ is weaker than T1 .
Now take Y = Z and g = id again, but instead take S = TZ . This time
i ◦ g = i : (Z, TZ ) → (X, T), which we have already shown is continuous, hence we
must have that g is continuous. So for each U ∈ T1 , g −1 (U ) = U ∈ TZ , that is,
T1 ⊂ TZ . Thus T1 = TZ as required.
We shall now turn to a consideration of product spaces. Consider a curve in
R2 , which can be thought of as a map f : R → R2 (f might possibly be defined
on some subinterval of R). Our work on metric spaces has given us a number of
metrics on R2 that all lead to the same topology, and in this way we can specify
whether or not the curve is continuous. Alternatively we can always decompose
the map into components:
f (t) = x(t), y(t) ,
where x and y are now functions R → R. Naively we expect the curve to be
continuous if and only if these component functions are continuous.
If we now turn to abstract sets X1 and X2 , and write X = X1 × X2 , then
we can define the coordinate projections pi : X → Xi by pi (x1 , x2 ) = xi . Then,
given any other set Y , any map f : Y → X determines two maps pi ◦ f : Y → Xi ,
and conversely any two maps fi : Y → Xi give a map f : Y → X through
f (y) = f1 (y), f2 (y) . If we have topologies on X1 and X2 , we would like to put
a topology on the product X such that for any other topological space (Y, S), a
map f : Y → X is continuous if and only if both the component maps pi ◦ f are
continuous.
25
Definition 2.20. Let (X1 , T1 ) and (X2 , T2 ) be topological spaces, and let X =
X1 × X2 . The product topology T on X is the following collection of sets:
T = {U ⊂ X : for each (x1 , x2 ) ∈ U, there Ui ∈ Ti such that
x1 ∈ U1 , x2 ∈ U2 and U1 × U2 ⊂ U }
In particular T contains all sets of the form U1 × U2 for Ui ∈ Ti , but also more
— see the remark below. To see that the above definition makes sense, we must
do the following:
Lemma 2.21. The collection T defined above is indeed a topology on X.
Proof. Now X = X1 × X2 by definition, and Xi ∈ Ti for i = 1, 2, from which it
follows that X ∈ T. Similarly, ∅ ∈ T, since it satisfies conditions in the definition
vacuously.
S
Let {Uλ }λ∈Λ be any family of subsets from T and let (x1 , x2 ) ∈ λ UΛ . Then
(x1 , x2 ) ∈ Uλ0 for some λ0 ∈ I. Hence there are Vi ∈ Ti such that
S
x1 ∈ V1 , x2 ∈ V2 and V1 × V2 ⊂ Uλ0 ⊂ λ Uλ .
and so this union lies in T.
Finally, let U, V ∈ T and let (x1 , x2 ) ∈ U ∩ V . Then we can find Ui , Vi ∈ Ti
(i = 1, 2) such that
xi ∈ Ui , xi ∈ Vi , and U1 × U2 ⊂ U, V1 × V2 ⊂ V.
But Ui ∩ Vi ∈ Ti and xi ∈ Ui ∩ Vi for i = 1, 2, and moreover
(x1 , x2 ) ∈ (U1 ∩ V1 ) × (U2 ∩ V2 ) = (U1 × U2 ) ∩ (V1 × V2 ) ⊂ U ∩ V.
Thus U ∩ V ∈ T.
Remark. It is not hard to show that a subset of R is open (with respect to the
usual topology) if and only if it can be written as a union of open intervals. It
follows that a subset U ⊂ R2 is open with respect to the product topology if and
only if for each (x, y) ∈ U we can find a, b, c, d ∈ R such that
a < x < b, c < y < d and (a, b) × (c, d) ⊂ U.
This is illustrated by:
(a, d)
(b, d)
(x, y)
(a, c)
(b, c)
U
26
Sets of the form (a, b) × (c, d) are known as open rectangles (or open boxes,
especially in higher dimension). If we set
ε = min{x − a, b − x, y − c, d − y} > 0
then B∞ (x, y), ε ⊂ U , where B∞ denotes the open ball defined with respect
to the d∞ metric, which is Lipschitz equivalent to the metrics d1 and d2 from
Example 1.3. It follows that these three metrics all induce the same topology on
R2 , which coincides with the product topology.
It is important to note that there are open sets in R2 that are not open rectangles, or indeed of the form U1 × U2 . The sets of this form are only a basis for the
topology on R2 . For instance it is not hard to show that the open ball B(0, 1) of
radius 1 centred on the origin cannot be written as U1 × U2 for open sets Ui ⊂ R
Although the sets in T include things other than (generalised) open boxes, we
can usefully characterise the open sets in terms of these simple products:
Lemma 2.22. Let X1 , X2 and X be as above, then any U ∈ T is a union of sets
of the form U1 × U2 for Ui ∈ Ti .
Proof. Let U ∈ T, then for each x = (x1 , x2 ) ∈ U we can find Ux ∈ T1 and Vx ∈ T2
such that
x1 ∈ Ux , x2 ∈ Vx and Ux × Vx ⊂ U.
In now follows that
U=
S
x∈U (Ux
× Vx )
as required.
Corollary 2.23. Let X1 , X2 and X be as above, let (Y, S) be another topological
space, and consider a map f : Y → X. Then f is continuous if and only if
f −1 (U1 × U2 ) ∈ S for all Ui ∈ Ti .
S
Proof. If U ⊂ X is open then it can be written as U = λ Uλ × Vλ for families
{Uλ } ⊂ T1 and {Vλ } ⊂ T2 . The result now follows since
f −1
S
λ Uλ
S
× Vλ = λ f −1 (Uλ × Vλ )
The product topology is set up in such a way that the projections pi are
continuous, and also so that we use them to characterise continuity of maps from
any other topological space into X.
Lemma 2.24. Let X1 , X2 and X be as above. The projections pi : X → Xi are
continuous.
Proof. This follows since if U1 ∈ T1 then p−1
1 (U1 ) = U1 × X2 ∈ T.
Proposition 2.25. Let X1 , X2 and X be as above, let (Y, S) be another topological
space and let f : Y → X be a map. Then f is continuous if and only if the maps
pi ◦ f : Y → Xi are continuous.
27
Proof. If f is continuous then the maps pi ◦f are compositions of continuous maps,
hence continuous. So suppose instead that f is a map for which the component
maps pi ◦ f are continuous. To show that f is continuous it is enough to show that
f −1 (U1 × U2 ) ∈ S for any Ui ∈ Ti by Corollary 2.23. But
y ∈ f −1 (U1 × U2 ) ⇔ f (y) ∈ U1 × U2
⇔ (p1 ◦ f )(y) ∈ U1 and (p2 ◦ f )(y) ∈ U2
⇔ y ∈ (p1 ◦ f )−1 (U1 ) ∩ (p2 ◦ f )−1 (U2 ).
That is,
f −1 (U1 × U2 ) = (p1 ◦ f )−1 (U1 ) ∩ (p2 ◦ f )−1 (U2 ).
Continuity of the component functions pi ◦ f implies that (p1 ◦ f )−1 (U1 ) ∈ S and
(p2 ◦ f )−1 (U2 ) ∈ S, and so we are done.
A commutative diagram for the above would be the following:
2 X1
O
p1 ◦f
p1
Y
f
/ X = X1 × X2
p2 ◦f
p2
,X
2
As with the subspace topology, it can be shown that the product topology is
the only one for which Proposition 2.25 holds for all choices of space (Y, S) and
map f . Perhaps more important to mention at this juncture is that the above
procedure clearly extends products of any finite number of topological spaces. In
fact itQextends further: let {(Xλ , Tλ )}i∈I be any family of topological
S spaces, then
X = λ∈Λ Xλ as a set is defined to be the set of maps f : Λ → λ Xλ such that
f (λ) ∈ Xλ for each λ, and for each λ ∈ Λ the map pλ : X → Xλ is defined by
setting pλ (f ) = f (λ). The space X is then equipped with the weakest topology T
such that all of the pλ are continuous. [Note, if X is given the discrete topology
then all of the pλ are automatically continuous, hence the set of topologies for
which the projections are continuous is nonempty. To get T take the intersection
of all the topologies that have this property.] This definition coincides with the
one given above when |Λ| = 2.
Quotient spaces; homeomorphisms
Consider a sheet of paper. If it is square we can think of it as the subset [0, 1]×[0, 1]
of the plane. If it is rectangular then we should first scale one of the sides to make
it the same length as the other. If we now stick one edge to its opposite edge,
28
preserving the orientation, then we get a cylinder:
What we are doing is “identifying” the point (0, y) with the point (1, y) for all
y ∈ [0, 1], and this is denoted by having the arrows on the opposite edges pointing
in the same direction. Thus our square with these arrows on it is “essentially the
same as” a cylinder. What happens if we change the orientation arrows? That is,
we stick (0, y) to (1, 1 − y) for each y ∈ [0, 1]?
The resulting shape is the Möbius band, which is a surface that has only one
edge and one side, unlike the the cylinder which has two edges and two sides. An
alternative way to proceed from the cylinder is rather than unsticking the edge
that we had already glued together, we could identify the other two sides, again
doing so in the same direction, and we end up with the torus (=doughnut).
Here the two different types of arrows are used to indicate the different gluing
processes.
So we should now make two things more precise: what it means to glue or
identify points of a topological space, and what it means to say that one topological
space is essentially the same as another.
In the gluing procedure we are saying that we shall treat certain points as the
same, e.g. (0, y) and (1, y) in the case of the cylinder. Thus we are partitioning our
set, or, equivalently, defining an equivalence relation on the set. For the cylinder
two of the sets making up the partition are {(1/2, 1/2)} and {(0, 3/4), (1, 3/4)}.
29
Recall that the partitions of a set X are in one-to-one correspondence with
the equivalence relations on X. More importantly they are in one-to-one correspondence with the onto functions from X (up to bijective equivalence of their
ranges) as discussed in Section 0. In particular if {Aλ }λ∈Λ is a partition of X then
f : X → Λ defined by f (x) = λ if x ∈ Aλ is both well-defined and onto. So rather
than deal with partitions we shall use onto functions below.
Definition 2.26. Let (X, T) be a topological space, X0 a set, and q : X → X0 an
onto map. Define Tq to be the following collection of subsets of X0 :
Tq = {U ⊂ X0 : q −1 (U ) ∈ T}.
Then Tq is a topology on X0 , called the quotient or identification topology. The
map q is called the quotient or identification map.
Obviously one should check that Tq really is a topology on X0 . It is clear from
the definition that the quotient map q is (T, Tq )-continuous, which gives one half
of the next result. Also, note that Tq is the strongest topology on X0 for which q
is continuous.
Proposition 2.27. Let (X, T), X0 , q and Tq be as in the definition above, let
(Y, S) be another topological space, and let f0 : X0 → Y be a map. Then f0 is
continuous if and only if f := f0 ◦ q is continuous.
This situation is given by the diagram
X@
@@
@@
@
f @@
q
Y
/ X0
}
}
}}
}} f0
}
~}
Proof. If f0 is continuous, then f is the composition of two continuous maps, and
hence continuous. So suppose that we are given that f is continuous, and let
U ∈ S. Then
T 3 f −1 (U ) = (f0 ◦ q)−1 (U ) = q −1 (f0−1 (U )),
and so by definition of the quotient topology we have that f0−1 (U ) ∈ Tq . This
holds for all U ∈ S, hence f0 is continuous.
Returning to our cylinder, suppose we wish to consider an open neighbourhood
of a point on the join. When we unwrap the cylinder we get two points about which
we must draw an open ball, which in this case correspond to two half discs.
30
The other thing we had to clarify was in what sense we can identify topological
spaces. This is a key issue in the whole theory.
Definition 2.28. Let X and Y be topological spaces. A map f : X → Y is a
homeomorphism if f is a bijection such that both f and f −1 are continuous. The
spaces X and Y are said to be homeomorphic or topologically equivalent if such
a map f exists.
Homeomorphisms play the role in topology that bijections do in set theory or
isomorphisms do in group or ring theory. In the topological context we are dealing
with sets where we have an extra structure, namely a notion of continuity, and so
we want to reflect that in our choice of maps.
Example 2.29.
(i) Any two bounded open intervals in R are homeomorphic: the map f :
(a, b) → (c, d) defined through a combination of scaling and translation by
f (x) = c+(d−c)(x−a)/(b−a) is a homeomorphism. There are many others
between these spaces.
(ii) We do not have to restrict ourselves to bounded intervals: (−1, 1) and R
are homeomorphic. For example consider the map f : (−1, 1) → R where
f (x) = x/(1 − |x|), or the map g : (−1, 1) → R where g(x) = tan πx
2 .
(iii) A doughnut and a teacup are homeomorphic — they are both solid 3D shapes
that have only one hole in them. Neither are homeomorphic to the sphere
in R3 , which has no holes.
(iv) The letter ‘L’ is homeomorphic to the letter ‘V’ — one is got from the other
by bending the arms closer together or further apart. However, neither are
homeomorphic to ‘T’. This follows since if X and Y are homeomorphic, via
a map f : X → Y , then so are A and f (A) for any subset A of X. So now
note that if we remove the junction point of ‘T’ then we get three separate
parts, but if we remove any point from ‘L’ we get at most two parts.
It is straightforward to show that homeomorphism defines an equivalence relation on the set of topological spaces: if X and Y are homeomorphic, and if
Y and Z are homeomorphic, then so are X and Z — just compose the relevant
homeomorphisms. Thus R is homeomorphic to (a, b) for any choice of a < b from
R.
A given property that a space may or may not possess is called a topological
property if it is preserved by homeomorphisms. For instance, if X and Y are
homeomorphic and Y is Hausdorff, then it follows that X must also be a Hausdorff
space. This is actually a consequence of the following weaker result:
Proposition 2.30. Let f : X → Y be a continuous injective map between topological spaces, and suppose that Y is Hausdorff. Then X is also Hausdorff.
Proof. Let x1 , x2 ∈ X be distinct points. Since f is injective we have that f (x1 ) 6=
f (x2 ), and since Y is Hausdorff we know that there are some open sets V1 , V2 ⊂ Y
such that f (xi ) ∈ Vi , but V1 ∩ V2 = ∅. But f is also continuous, hence f −1 (V1 )
31
and f −1 (V2 ) are open subsets of X. Moreover we have xi ∈ f −1 (Vi ) and f −1 (V1 ) ∩
f −1 (V2 ) = ∅. Thus X is Hausdorff as required.
As a final word on homeomorphisms, consider what happens when we cut cylinders and Möbius strips in half. Cutting a cylinder in half produces two cylinders
which is clear from our usual picture of them:
This can also be shown by considering quotient topologies in the appropriate way:
two
cylinders
Here the change in arrows show which points are still being identified after the
cut. However, if we apply the same reasoning to the Möbius band we find the
following:
one cylinder
Thus it follows that if we cut the Möbius band in half we only get one piece!
Moreover that piece appears to be a cylinder. If you do this in practice you get
32
a cylinder with a double twist, which is a homeomorphic to the usual cylinder.
However no amount of deformation in R3 will allow you to move from one surface
to the other — the definition of homeomorphism applies to the surface/space itself,
and not to any ambient space we may viewing it in.
Exercises
1. List all possible topologies on {a, b, c}. Consequently find a two topologies T1
and T2 on a set that are not comparable. That is, we have neither T1 ⊂ T2 nor
T2 ⊂ T1 .
2. Give an example of subsets A and B of R for which A ∩ B, A ∩ B, A ∩ B and
A ∩ B are all different.
3. Let X be a topological space and A ⊂ X a subset. The boundary of A is defined
to be b(A) = A ∩ X \ A. Calculate b(A) when A = (0, 1] and
(i) X = R, with the usual topology
(ii) X = C, with the usual topology
(iii) X = R, with the discrete topology (iv) X = C, with the indiscrete topology
T
4. Let {Tλ }λ∈Λ be S
a family of topologies on a set X. Show that λ∈Λ Tλ is a topology
on X, but that λ∈Λ Tλ need not be.
5. Given a topological space (Y, S) and a map f : X → Y , show that the collection
Tf = {f −1 (U ) : U ∈ S} is a topology on X. Moreover, show that f is continuous with respect to this topology, and that Tf is the weakest topology with this
property.
Describe this topology when we take X = Y = R, S to be the usual topology, and
take f to be (i) a constant function, (ii) the function that maps (−∞, 0] to 0 and
(0, ∞) to 1, and (iii) f (x) = x.
6. Prove that any map f : X → Y is continuous if either X is equipped with the
discrete topology or Y is equipped with the indiscrete topology.
7. Show that f : X → Y is continuous if and only if it is continuous as a map onto
the subspace f (X).
8. Let (X, T) be a topological space and let U be an open subset of X. Show that if
V is a subset of U that is open in the subspace topology TU , then V is open as a
subset of X. Show that this can fail if we do not assume that U is open.
9. Let X1 be a topological space, and X2 a subset of X1 equipped with the subspace
topology. Let A be a subset of X2 , and denote by Ai the closure of A in Xi , for
i = 1, 2. Prove that (i) A2 = X2 ∩ A1 (ii) if X2 is closed in X1 then A1 = A2 .
10. Let X1 , X2 , Y1 and Y2 be topological spaces, and let f1 : X1 → Y1 and f2 : X2 → Y2
be maps. Define a map f : X1 × X2 → Y1 × Y2 by
f (x1 , x2 ) = f1 (x1 ), f2 (x2 ) .
Show that f is continuous if and only if f1 and f2 are continuous, where X1 × X2
and Y1 × Y2 are given their respective product topologies.
33
11.
(a) Let X1 and X2 be topological spaces, and let W be an open subset of X1 ×X2 .
Show that pi (W ) is an open subset of Xi for i = 1, 2, where pi is the projection
map X1 × X2 → Xi .
(b) Give an example of a closed subset W ⊂ R × R such that p1 (W ) is not closed
in R.
12. Let X and Y be topological spaces and suppose that E ⊂ X and F ⊂ Y are closed.
Show that E × F is closed in the topological product X × Y .
13. Let X = [−1, 1] equipped with the usual topology.
(a) Let f : X → [0, 1] be the function f (x) = |x|. Show that quotient topology
induced on [0, 1] by f coincides with the usual topology.
(b) Find a surjection g : X → [0, 1] for which the quotient topology induced by
g is not Hausdorff.
14. Prove that homeomorphism defines an equivalence relation on the class of all
topological spaces.
15. Show that if f : X → Y is a homeomorphism and A ⊂ X, then the restrictions
f |A : A → f (A) and f |X\A : X \ A → Y \ f (A) are both homeomorphisms.
16. Let X1 and X2 be topological spaces, and pick a2 ∈ X2 . Show that X1 is homeomorphic to the subspace X1 × {a2 } of X1 × X2 respectively. Show that X1 is
homeomorphic to the subset {(x, x) : x ∈ X1 } of X1 × X1 .
17. Let X be an infinite set, and let T be the Zariski topology, that is,
T = {∅} ∪ {U ⊂ X : X \ U is finite}.
Show that T is indeed a topology, but that it is not Hausdorff.
18. Prove the following:
(i) Any subspace of a Hausdorff space is Hausdorff.
(ii) The product of two Hausdorff spaces is Hausdorff.
(iii) If f : X → Y is continuous and injective, and Y Hausdorff, then so is X.
(iv) Being Hausdorff is a topological property, that is, it is preserved by homeomorphisms.
19. Show that in a Hausdorff space X, the set {x} is
(i) closed, and
(ii) the intersection of all open sets containing x.
20. Let f, g : X → Y be continuous maps between topological spaces, with Y Hausdorff. Show that W = {x ∈ X : f (x) = g(x)} is closed in X. Deduce that if
f : X → X is a continuous map and X is Hausdorff then the fixed point set
{x ∈ X : f (x) = x} is closed.
21. A T1 -space is a topological space (X, T) that satisfies the following: for any pair
34
of distinct points x, y ∈ X there are Ux , Uy ∈ T such that
x ∈ Ux , y ∈ Uy , but x ∈
/ Uy , y ∈
/ Ux .
Show that every Hausdorff space is a T1 -space, but that there are T1 -spaces that
are not Hausdorff.
35
3
Compactness
Compactness for topological spaces; the Heine-Borel Theorem
Several results in analysis concerning real-valued functions are trivial if the function f is defined on a set X that is finite. For example, if we have X = {x1 , . . . , xn }
then f (X) ⊂ R is bounded and achieves its bounds, namely
f (xi1 ) = min f (xi ) ≤ f (x) ≤ max f (xi ) = f (xi2 )
i
i
for some 1 ≤ i1 , i2 ≤ n. This result remains true if we replace X by a closed and
bounded subinterval of R, and we stipulate that f be continuous. That is, we are
dealing with a continuous function f : [a, b] → R for some a < b. This is often
given as part of the Intermediate Value Theorem. The particular property of [a, b]
that makes this result true is the subject of this chapter.
Definition 3.1. Let X be a set and Y a subset
of X. A cover for Y is any
S
collection U of subsets of X such that Y ⊂ U ∈U U . A subcover of the given
cover U is any subcollection VSof U such that V is again a cover for Y . That is, if
U ∈ V then U ∈ U, and Y ⊂ U ∈V U .
A cover U is finite if there are only finitely many sets in U. If X is a topological
space then U is an open cover if each set in U is an open set.
Example 3.2. The singleton sets {x} x∈R form a cover of R. Even though each
subset {x} is finite, there are infinitely many sets in the cover, so the cover is infinite. On the other hand {(−∞, 1), (−1, 1), (−1, ∞)} is a cover of R comprising only
three subsets, and hence is a finite cover. The subcollection {(−∞, 1), (−1, ∞)} is
a subcover, but the cover {(−∞, 1), (0, ∞)} is not a subcover since (0, ∞) is not
one of the original subsets.
Unsurprisingly, since we are dealing with topological spaces, it will be open
covers that are of interest in this course.
Definition 3.3. A topological space X is compact if every open cover of X has
a finite subcover.
Note what this definition says: if we have any family U of open sets such that
every element x ∈ X appears in at least one U ∈ U, then we actually only need a
finite number of sets to capture all of the points. This is not the same as saying
that there is at least one finite open cover of the space — this is always true since
we can take U = {X}.
Example 3.4.
(a) Any topological space with a finite number of points is compact — given
any open cover just pick out one set for each point in the space to produce
a finite subcover.
(b) A discrete topological space X is compact if and only if it is finite. To show
this, note that if X is infinite then U = {x} : x ∈ X is an open cover of
X containing infinitely many sets, and if we leave any of them out then the
corresponding points will not lie in the union of the remaining sets.
36
(c) The subset (0, 1) of R is not compact: consider
the family of open subsets
S
1
1
U = {( n , 1) : n ≥ 2}. It is clear that n≥2 ( n , 1) = (0, 1), so this is an
open cover. Any finite subfamily is of the form {( n11 , 1), . . . , ( n1k , 1)} for
some n1 , n2 , . . . , nk ∈ N, and so their union is ( N1 , 1) $ (0, 1) where N =
max{n1 , . . . , nk }. Similarly, (0, 1] is not compact.
As ever there is an equivalent definition of compactness in terms of closed
sets rather than open sets, and it is obtained from the above definition by taking
complements. This process, applied to a cover, will not yield another cover. Hence
we need the following:
Definition 3.5. A family {Aλ }Tλ∈Λ of nonempty subsets of a set X has the finite
intersection property (FIP) if γ∈Γ Aγ 6= ∅ for every finite set Γ ⊂ Λ.
Proposition
3.6. Let X be a topological space. Then X is compact if and only if
T
F
=
6
∅
for
every family {Fλ }λ∈Λ of closed sets that has the FIP.
λ
λ∈Λ
Proof. Suppose X is compact and {Fλ }λ∈Λ is a family of closed setsSsuch that
T
c
λ∈Λ Fλ = ∅. Taking complements and using De Morgan’s Laws gives
λ∈Λ Fλ =
Γ⊂Λ
X, and so S{Fλc }λ∈Λ is an open cover of X. Hence there is a finite subset
T
such that γ∈Γ Fγc = X, and taking complements once more gives γ∈Γ Fγ = ∅.
Thus {Fλ }λ∈Λ does not have the FIP. Thus any family of closed sets having the
FIP must have nonempty intersection.
Suppose the converse is true, that is, X is a topological space in which every
family of closed sets with the FIP has nonempty intersection. Let {Uλ }λ∈Λ be an
open cover of X, then
S
λ∈Λ Uλ
=X ⇒
T
c
λ∈Λ Uλ
= ∅.
Thus
{Uλc }λ∈Λ cannot have the FIP, hence
S there is a finite set Γ ⊂ Λ such that
T
c
γ∈Γ Uγ = X, and so we have a finite
γ∈Γ Uγ = ∅. This in turn implies that
subcover.
It is possible to consider compactness for subspaces of topological spaces, although we must take a little care to ensure that our definitions do not lead to
confusion.
Definition 3.7. Let (X, T) be a topological space and Y a nonempty subset of
X. Then Y is compact as a subspace if for any cover for Y taken from T, there is
a finite subcover.
Note that in the above we are choosing the sets that make up the cover from
T, that is they are open subsets of the larger space X rather than open sets of Y .
But in fact it turns out that Y is compact as a subspace of X if and only if it is
compact with respect to the subspace topology TY induced on Y by the topology
on X. To see this, suppose that Y is compact according to Definition 3.7 and that
U is any collection from TY that covers Y . Now each element of U is of the form
U ∩ Y for some U ∈ T, and if we select a U ∈ T in this way for each set in the
cover U, then we obtain a subcollection V of T that covers Y . Compactness of
37
Y in the sense of Definition 3.7 ensures that there is a finite subcover of V, say
{U1 , . . . , Un }, and then {U1 ∩ Y, . . . , Un ∩ Y } is a finite subcover of U. The reverse
implication is given by a very similar argument.
The above discussion thus frees us from having to be too pedantic about specifying what we mean when we say a subspace is compact.
Perhaps the most important example of a compact space is the following:
Theorem 3.8 (Heine-Borel). Any closed bounded interval [a, b] in R is compact.
Proof. Note that we may assume that a < b otherwise our set contains only one
point, and thus is clearly compact. So let U be any collection of open sets of R
that covers the interval [a, b], and set
G = {x ∈ [a, b] : [a, x] can be covered by a finite subcover of U}.
The elements of G are to be thought of as the ‘good’ points, and we are aiming to
show that b ∈ G.
Now a ∈ G, since there is some U ∈ U such that a ∈ U , and so {U } is a
finite subcover of U that covers [a, a] = {a}. Thus G 6= ∅ and G ⊂ [a, b], so if we
define g = sup G then g exists and also g ∈ [a, b]. Hence there is some U0 ∈ U
such that g ∈ U0 , and since U0 is open in R there must be some δ > 0 such that
(g − δ, g + δ) ⊂ U0 . However, by definition of g, there is some c ∈ G satisfying
g − δ < c ≤ g, and thus there is a finite subfamily {U1 , . . . , Un } of U that covers
[a, c]. Hence {U0 , U1 , . . . , Un } covers the subset [a, g + δ) of R, and we see that if
g < b then we would have contradicted the definition of g, since the points in the
nonempty set (g, g + δ) ∩ [a, b] would also lie in G. Thus g = b, and the set [a, b] is
covered by the finite subfamily {U0 , U1 , . . . , Un }. That is, b ∈ G as required.
The importance of the compactness of [a, b] lies in the following basic properties
of compactness. In particular we will shortly be able to characterise all compact
subsets of Rn .
Proposition 3.9. Any closed subspace of a compact space is compact.
Proof. Let (X, T) be a compact space and F ⊂ X a nonempty subset that is
closed.S Let U ⊂ T be an open cover of F consisting
of open subsets of X. Thus
S
c
c
F ⊂ U ∈U U , and so X = F ∪ F = F ∪ U ∈U U . But F c = X \ F is open
in X, hence {F c } ∪ U is an open cover of X. So we can choose a finite subcover
{U1 , . . . , Un , F c } of the cover for X, and it follows that the family {U1 , . . . , Un }
covers F as required.
Definition 3.10. A subset Y of a metric space (X, d) is bounded if there is some
constant K > 0 such that d(x, y) ≤ K for all x, y ∈ Y .
An alternative definition would say that there is some preferred point a ∈ X
and constant K 0 ≥ 0 such that d(a, x) ≤ K 0 for all x ∈ Y . That these definitions
actually single out the same class of subsets of X is an easy consequence of the
triangle inequality for metric spaces.
Definition 3.11. Let Y be a bounded subset of a metric space X. The diameter
of Y , denoted diam Y , is sup{d(x, y) : x, y ∈ Y }.
38
It follows from the previous two results that any closed and bounded subset
A ⊂ R is compact. Boundedness of A implies that A ⊂ [−R, R] for some R > 0,
which is compact by Theorem 3.8, and so A is a closed subset of a compact space,
hence compact by Proposition 3.9. In fact the converse is true, which follows from
the next two results which will be proved in far more generality than just dealing
with subsets of R.
Proposition 3.12. Let X be a compact metric space. Then X is bounded.
Proof. Pick any point x ∈ X. Then the collection of open balls {B(x, n)}n≥1 is a
family of open subsets of X that are nested, in the sense that
B(x, 1) ⊂ B(x, 2) ⊂ · · · ⊂ B(x, n) ⊂ B(x, n + 1) ⊂ · · ·
S
Moreover for any y ∈ X we have y ∈ B(x, n) for all n > d(x, y), and so n≥1 B(x, n) =
X. But X is compact, so there must be a finite subcover, from which it follows
that X ⊂ B(x, N ) for some N ≥ 1. Hence X is bounded, with diameter no greater
than 2N .
Proposition 3.13. Any compact subset of a Hausdorff space is closed.
Proof. Let C be a compact subspace of a Hausdorff space X, and pick x ∈ C c which
we keep fixed for now. For each y ∈ C we can find open sets U (x, y) and V (x, y)
such that x ∈ U (x, y), y ∈ V (x, y) and U (x, y) ∩ V (x, y) = ∅, since X is Hausdorff.
Thus {V (x, y) : y ∈ C} is an open cover of C, and so there must be an finite
subcover, {V (x1 , y1 ), . . . , V (xn , yn )} say. Let U (x) = U (x1 , y1 ) ∩ · · · ∩ U (xn , yn ),
then U (x) is open and xS
∈ U (x). Moreover U (x) has empty intersection with each
V (xi , yi ) = ∅, hence U (x) ∩ C = ∅. Now varying our
V (xi , yi ) and so U (x) ∩ ni=1 S
point x, it follows that C c = x∈C c U (x), hence is open, and so C is closed.
Before we deal with products of compact spaces we shall take a brief diversion
and consider their images under continuous maps.
Proposition 3.14. Let X be a compact space, Y a topological space, and f : X →
Y and continuous map. Then f (X) is a compact subset of Y .
Proof. Let U be an open cover of
S f (X), that is U is a collection {Uλ }λ∈Λ of open
subsets of Y such that f (X) ⊂ λ∈Λ Uλ . Now, by continuity of f , {f −1 (Uλ )}λ∈Λ
is a family of open subsets of X, and moreover
S
S
f (X) ⊂ λ∈Λ Uλ ⇒ X ⊂ λ∈Λ f −1 (Uλ ) (⊂ X).
Thus {f −1 (Uλ )}λ∈Λ is an open cover of the compact space X. Hence there is a
finite subcover, so for some finite set Γ ⊂ Λ
S
S
S
X = γ∈Γ f −1 (Uγ ) ⇒ f (X) = γ∈Γ f f −1 (Uγ ) ⊂ γ∈Γ Uγ ,
and so {Uγ }γ∈Γ is a finite subcover of f (X).
This very general result tells us that compactness is a topological property,
and leads immediately to easy corollaries that deal with familiar situations.
39
Corollary 3.15. Let X and Y be homeomorphic spaces. Then X is compact if
and only if Y is compact.
Proof. Let f : X → Y be a homeomorphism, and suppose that X is compact,
then Y = f (X) is compact. If, on the other hand, we are given that Y is compact,
then X = f −1 (Y ) is compact (where here we are, for once, really dealing with an
inverse function f −1 !).
Corollary 3.16. Let X be a compact space, Y a metric space and f : X → Y a
continuous map. Then f (X) is closed and bounded.
If Y = R then there are points x1 , x2 ∈ X such that f (x1 ) = inf f (X) and
f (x2 ) = sup f (X). That is, f attains its bounds.
Proof. By Proposition 3.14 we know that f (X) is compact, and hence it is closed
and bounded by Propositions 3.13 and 3.12. If Y = R, then the numbers inf f (X)
and sup f (X) exist. Moreover, for any bounded set A ⊂ R, either inf A ∈ A or inf A
is a limit of points from A. Since f (X) is closed we must have inf f (X) ∈ f (X),
hence the required point x1 must exist, and similarly for x2 .
Thus this corollary, together with the Heine-Borel Theorem, tells us a lot
about the behaviour of continuous maps f : [a, b] → R. More importantly the
above results apply in far more general situations. To get to such situations it is
useful to have the following:
Proposition 3.17. Let X1 and X2 be topological spaces and set X = X1 × X2 .
Then X is compact if and only if X1 and X2 are compact.
Proof. We do the easy half of this proposition first — suppose that X is compact.
Then Xi = pi (X) where pi is the coordinate projection. That is, Xi is the image
of a compact space under a continuous map, and hence is compact.
So now suppose instead that the components are both compact and let W be an
open cover of X. We call any subset A ⊂ X1 good if there is a finite subcollection
of W that covers A × X2 . We are aiming to show that X1 is a good set, and do
this in three steps.
Step 1: Let A
, . . . , Am ⊂ X1 be good subsets, then A = A1 ∪ · · · ∪ Am is good.
S1m
For A×X2 = i=1 (Ai ×X2 ), and so if Wi is a finite subcollection of W that covers
Ai × X2 , then W1 ∪ · · · ∪ Wn is a finite subcollection of W that covers A × X2 .
Step 2: Each point x ∈ X1 is locally good in the sense that there is an neighbourhood Ux of x that is good.
To show this, pick x ∈ X. Then for each y ∈ X2 there is some W (y) ∈ W
such that (x, y) ∈ W (y). But, by definition of the product topology, we can find
U (y) and V (y), open in X1 and X2 respectively, such that (x, y) ∈ U (y) × V (y) ⊂
W (y). Now {V (y) : y ∈ Y } is an open cover of the compact space X2 , since
y ∈ V (y) for each y, hence there is a finite subcover, {V (y1 ), . . . , V (yn )} say. Set
Ux = U (y1 ) ∩ · · · ∩ U (y1 ), then Ux is open in X and contains x. Moreover
S
S
Ux × X2 = Ux × ni=1 V (yi ) = ni=1 (Ux × V (yi ))
S
S
⊂ ni=1 (U (yi ) × V (yi )) ⊂ ni=1 W (yi ),
40
and so Ux is a good set, as required.
Step 3: X1 is a good set. This final step is achieved by noting that for each
x ∈ X1 we have found an open set Ux that is both good and contains x. Thus
in particular {Ux : x ∈ X} is an open cover of X1 . But then compactness
of X1
Sk
says that there is a finite subcover, {Ux1 , . . . , Uxk } say. So X1 = i=1 Uxi , a finite
union of good sets, and thus X1 is good by step 1, which is what we set out to
show.
The above now clearly extends to any finite product of compact spaces by an
induction argument. What is perhaps surprising is that this is also true for an
arbitrary product of compact spaces. This general form is Tychonoff’s Theorem,
and is the basis for many important results in analysis. An account of this is given
in Simmons’ book. It turns out that Tychonoff only proved the result in a special
case from which he then produced the Stone-Čech compactification. Meanwhile
Čech proved the general case of Tychonoff’s Theorem. . .
Putting a number of the above results together we finally arrive at a characterisation of the compact subsets of Rn , and hence also of Cm (since C is homeomorphic
to R2 , hence Cm is homeomorphic to R2m ).
Theorem 3.18 (Heine-Borel). A subset of Rn is compact if and only if it is
closed and bounded.
Proof. Let C be a compact subset of Rn . Then it is closed by Proposition 3.13
and bounded by Proposition 3.12.
Suppose instead that C is a closed and bounded subset of Rn . Then C ⊂
[−R, R]n for some suitably large R. Now [−R, R] ⊂ R is compact by Theorem 3.8,
and hence so is [−R, R]n by Proposition 3.17. But if C is a closed subset of Rn ,
then it is a closed subset of [−R, R]n , and hence compact by Proposition 3.9.
One way to think of compactness is that a topology that is compact cannot
have ‘too many’ open sets, in the sense that we cannot build up any open cover
that is badly behaved. On the other hand a topology that is Hausdorff has a
plentiful supply of open sets since we can use them to separate any two distinct
points. This two conditions thus work against each other — one can show that if
X is a set and T and S are topologies on X that make X Hausdorff and compact
respectively, and if T ⊂ S, then we must have T = S.
Compactness for metric spaces: sequential compactness
When dealing with metric spaces it often makes sense to use sequences, and indeed
when it comes to studying compactness there is no change to this rule of thumb. In
this section we show that there is another equivalent formulation of compactness
that is given in terms of sequences and their possible subsequences.
Definition 3.19. A subset C of a metric space X is sequentially compact in
itself (respectively in X) if every sequence in C has a subsequence that converges
to a point of C (resp. to a point of X).
41
It is clear that if C is sequentially compact in itself then it will be sequentially
compact in X. On the other hand by Proposition 1.23, if C is sequentially compact
in X then it is sequentially compact in itself if and only if it is closed.
The main result in this direction allows us to pass freely from sequential compactness, which is a natural concept for metric spaces, to the general idea of
compactness as explained above.
Theorem 3.20. Let X be a metric space. Then X is compact if and only if it is
sequentially compact.
Proof. Suppose first that X is compact, and let (xn )n≥1 be any sequence in X.
For each m ≥ 0 set Am = {xn : n > m} and Fm = Am . Since Am ⊃ Am+1 , we
have Fm ⊃ Fm+1 . Moreover, each Fm is closed and nonempty by construction,
and so
it follows that {Fm }m≥0 has the FIP. Hence, by Proposition 3.6, we must
T∞
have m=0 Fm 6= ∅.
T
1
So now let x ∈ ∞
m=0 Fm . Then x ∈ F0 = A0 , and so we must have B(x, 2 ) ∩
A0 6= ∅. So we can choose some n1 ≥ 1 such that d(xn1 , x) < 12 . Since x ∈
Fn1 = An1 , there is some n2 > n1 such that d(xn2 , x) < 14 . Repeating this, we find
n1 < n2 < n3 < · · · such that d(xnr , x) < 2−r , and hence xnr → x. Thus X is
sequentially compact.
Suppose instead that X is a sequentially compact metric space, and let U be
any open cover of X. Our first step is to show that
∃ε > 0 s.t. ∀x ∈ X ∃Ux ∈ U s.t. B(x, ε) ⊂ Ux .
(†)
So suppose that (†) does not hold, then
∀ε > 0 ∃x ∈ X s.t. ∀U ∈ U, B(x, ε) 6⊂ U.
For each n ≥ 1 we can put ε = 2−n in the above and find some xn ∈ X such
that B(xn , 2−n ) 6⊂ U for all U ∈ U. Since X is sequentially compact, there is a
subsequence (xnr )r≥0 of this sequence that converges to some x ∈ X, and since
U is a cover of X, there is some U ∈ U such that x ∈ U . But U is open so, for
some m ≥ 1, B(x, 2−m ) ⊂ U and, given this m, we can find some R ≥ 1 such that
d(xnr , x) < 2−m−1 for all r ≥ R.
So now choose any r ≥ R for which nr > m, then
B(xnr , 2−nr ) ⊂ B(xnr , 2−m−1 ) ⊂ B(x, 2−m ) ⊂ U,
using the triangle inequality for the second inclusion, and hence contradicting the
choice of xnr above. Hence (†) must hold after all.
Now suppose that U has no finite subcover. Pick any point x1 ∈ X, and
consider the corresponding Ux1 ∈ U for which B(x1 , ε) ⊂ Ux1 , whose existence is
guaranteed by (†). Since there is no finite subcover of U, Ux1 6= X, and so we
can choose x2 ∈
/ Ux1 . Then d(x1 , x2 ) ≥ ε, and if we consider the corresponding
Ux2 ∈ U, then Ux1 ∪ Ux2 6= X, hence B(x1 , ε) ∪ B(x2 , ε) 6= X, and so we can choose
x3 ∈ X such that d(x1 , x3 ) ≥ ε and d(x2 , x3 ) ≥ ε. Iterating this procedure we
42
obtain a sequence (xn )n≥1 such that d(xm , xn ) ≥ ε for all m < n. This sequence
cannot have a convergent subsequence, for if xnr → x then we would have
ε ≤ d(xnr , xnr+1 ) ≤ d(xnr , x) + d(x, xnr+1 ) → 0
as r → ∞. Hence the existence of this sequence contradicts our assumption of
sequential compactness, and we’re done.
This alternative characterisation of compactness for metric spaces leads to an
alternative proof of one half of the Heine-Borel Theorem, namely that any closed
and bounded subset of Rm is compact. It is based on the Bolzano-Weierstrass
Theorem:
Theorem 3.21 (Bolzano-Weierstrass). Every bounded sequence of real numbers has a convergent subsequence.
So let A ⊂ Rm be a closed and bounded subset, let (x(n) )n≥1 be any sequence
(n) (n)
(n)
(n)
in A, where x(n) = (x1 , x2 , . . . , xm ) for xi ∈ R. So we have
(1)
(1)
(1)
(2)
(2)
(2)
(3)
(3)
(3)
x(1) = (x1 , x2 , x3 , . . .)
x(2) = (x1 , x2 , x3 , . . .)
x(3) = (x1 , x2 , x3 , . . .)
etc.
(n)
The sequence of first coordinates (x1 )n≥1 is a bounded sequence in R, and so has
a convergent subsequence. So for some infinite subset σ1 ⊂ N the (sub)sequence
(n)
(n)
(x1 )n∈σ1 is convergent. But then (x2 )n∈σ1 is a bounded sequence in R, hence
(n)
there is an infinite subset σ2 ⊂ σ1 such that (x2 )n∈σ2 is convergent, and note also
(n)
that the subsequence (x1 )n∈σ2 of our first subsequence will still be convergent.
Continuing this way we will eventually produce an infinite subset σm ⊂ N for which
(n)
each sequence (xi )n∈σm of real numbers is convergent, and hence the sequence
(x(n) )n∈σm of points is also convergent to some x ∈ Rm . Finally, since A is closed,
it contains the limits of all sequences taken from A, so that x ∈ A. Thus A is
sequentially compact, hence compact.
Exercises
1. Show that every finite topological space is compact. Show that a discrete topological space is compact if and only if it is finite.
2. Let Y be a subspace of a topological space X and let Z be a nonempty subset of
Y . Show that Z is compact as a subspace of Y if and only if it is compact as a
subspace of X.
3. Let Y and Z be compact subspaces of a topological space X. Show that Y ∪ Z is
a compact subspace of X.
T
Let {Yλ }λ∈Λ be a family of compact and closed subsets of X such that λ Yλ 6= ∅.
Show that this intersection is compact. [Hint: use question 2.]
43
4. Let T1 and T2 be topologies on a set X, with T1 ⊂ T2 . Show that if X is compact
with respect to T2 then it is also compact with respect to T1 .
5. Which of the following subsets of R and R2 are compact?
(ii) [0, ∞)
(i) [0, 1)
2
2
2
(iv) {(x, y) ∈ R : x + y = 1}
(vi) {(x, y) ∈ R2 : x2 + y 2 < 1}
(iii) Q ∩ [0, 1]
(v) {(x, y) ∈ R2 : x2 + y 2 ≤ 1}
(vii) {(x, y) ∈ R2 : x ≥ 1, 0 ≤ y ≤ x1 }
6. Let (Fn )n≥1 be a sequence of nonempty closed subsets
T of a compact topological
space X such that Fn ⊃ Fn+1 for all n. Show that n≥1 Fn 6= ∅.
7. Let C1 and C2 be compact subspaces of a Hausdorff space. Show that C1 ∩ C2 is
compact.
8. Suppose that X is a compact topological space, Y is a Hausdorff topological space
and f : X → Y a continuous bijection. Show that X and Y are homeomorphic.
[Hint: consider the closed subsets viewpoint of continuity.]
9. Show that R equipped with the Zariski topology is compact.
10. Let X be a compact metric space and suppose that f : X → X a continuous
map for which f (x) 6= x for all x ∈ X. Show that there is some ε > 0 such that
d(f (x), x) ≥ ε for all x ∈ X.
[Hint: consider the map g : X → R given by g(x) = d(f (x), x).]
11. Let X be a compact metric space and f : X → X a continuous map. Show that
there is a nonempty closed set F ⊂ X such that f (F ) = F .
[Hint: consider the sequence F1 = f (X), Fn+1 = f (Fn ), and question 6.]
12. Show that a compact metric space is separable.
13. Let (X, T) be a topological space. Set X 0 = X ∪ {∞}, where ∞ is any object not
in X, and
T0 = T ∪ {V ∪ {∞} : V ⊂ X, X \ V is compact and closed in X}.
Prove that (X 0 , T0 ) is a compact space containing X as a subspace.
[This is the one-point or Alexandroff compactification — see Simmons p.163]
44
4
Connectedness
Equivalent definitions; subintervals of R
This section of our course deals with the idea of how many ‘chunks’ or ‘pieces’ a
given topological space splits up into. Intuitively a connected space is one that does
not fall apart, for instance the subspace [0, 1] of R would be considered connected,
but [0, 1] ∪ [2, 3] is not. However, what about more complicated spaces such as
Q or the graph of the function y = sin x1 ? We shall discuss two different ways of
formulating this idea more precisely.
Definition 4.1. A topological space X is connected if it cannot be written as
X = U ∪ V where U and V are disjoint nonempty open subsets of X.
Every author has different terminology for the representation of a space X as
U ∪ V in terms of disjoint open sets U and V . One route is to call the pair (U, V )
a partition (into open sets), with the idea being that any set theoretic partition
of a topological space should be done using open sets, and hoping to thus avoid
confusion. Simmons is a little more inventive:
Definition 4.2. A space X is disconnected if it can be written as X = U ∪ V
for disjoint nonempty sets U and V . If X is disconnected then any representation
of this form is called a disconnection of X, noting that there may be many such
representations. Indeed, the space X is called totally disconnected if for every
pair of distinct points x, y ∈ X there is a disconnection (U, V ) such that x ∈ U
and y ∈ V .
Note. There are other definitions of totally disconnected in the literature, which
are similar in spirit to the one above but not necessarily equivalent.
There are several important things to spot arising from these definitions. One
is that a one point space, X = {x}, is both connected and totally disconnected!
Also note that if (U, V ) is a disconnection for a space X, then
X = U ∪ V and U ∩ V = ∅ ⇒ U = V c and V = U c ,
and so the sets that make up the disconnection are both open and closed. It follows
that a topological space X is connected if and only if the only sets that are both
open and closed are ∅ and X. Some authors used the word ‘clopen’ to describe a
set that is both open and closed. Finally, note that any totally disconnected space
is Hausdorff.
We now want to explore this new concept in the context of subspaces of R. It
will turn out that the connected subspaces of R are precisely the intervals, and in
showing this we make use of the following easily proved lemma:
Lemma 4.3. A subset X ⊂ R is an interval if and only if whenever x, y, z ∈ R
are three points satisfying x < z < y and x, y ∈ X, we must have z ∈ X.
Theorem 4.4. A subspace X ⊂ R is connected if and only if it is an interval.
45
Proof. Suppose X is not an interval, then we can find x < z < y in R such that
x, y ∈ X, but z ∈
/ X. Now (−∞, z) and (z, ∞) are open subsets of R, hence
U = X ∩ (−∞, z) and V = X ∩ (z, ∞) are open subsets of X such that x ∈ U and
y ∈ V . But X = U ∪ V since z ∈
/ X, and U ∩ V = ∅ by construction. Thus (U, V )
is a disconnection of X, and so X is not connected.
For the converse we first show that for any a < b there is no disconnection
(U, V ) of [a, b] such that a ∈ U and b ∈ V . To see this, suppose that such a
pair (U, V ) of open sets does exist. Then U and V are also closed subsets of [a, b]
and hence closed and bounded subsets of R. Moreover b ∈ V , so V 6= ∅. Setting
c = inf V , then V ⊂ [a, b] implies that c ∈ [a, b], and V being closed implies that
c ∈ V . In particular c 6= a since a ∈ U . But c is a lower bound for V and a < c,
hence c − n1 ∈ U for all n ≥ (c − a)−1 . Then c = limn (c − n1 ) ∈ U , since U is closed,
giving U ∩ V 6= ∅, which contradicts the fact that (U, V ) is a disconnection.
Now suppose that X is an interval that is disconnected. So X = U 0 ∪ V 0 for
nonempty disjoint open subsets U 0 and V 0 of X. Without loss of generality we
can pick a ∈ U 0 and b ∈ V 0 such that a < b, but then putting U = U 0 ∩ [a, b]
and V = V 0 ∩ [a, b] gives a disconnection (U, V ) of [a, b] with a ∈ U and b ∈ V .
So, by contradiction once again, X cannot be disconnected, hence every interval
is connected.
It follows from this characterisation that Q is not connected, since it is certainly
not an interval of R. In fact the rational numbers form a totally disconnected
subspace of R: given any p, q ∈ Q with p < q we can
find some irrational r such
that p < r < q, and then Q ∩ (−∞, r), Q ∩ (r, ∞) is a disconnection of Q that
separates p and q.
An alternative to talking about disconnections is the idea that if a space falls
apart into more than one component then we should be able to define a continuous
function on the space such that the values taken by the function on one component
bear no relation to those taken on the other component. This forms the basis for
an equivalent definition of connectedness. Since a disconnection is only made up
of two subsets, it is sufficient to consider functions that take only two values.
Proposition 4.5. A topological space X connected if and only if there is no continuous map from X onto a discrete two point space.
Remark. Since all discrete two point spaces are homeomorphic to one another, it
follows that we can focus on the existence or nonexistence of continuous maps from
X onto {0, 1}.
Proof. Suppose that X is a topological space and that f : X → {0, 1} is a
continuous surjection. Set U = f −1 ({0}) and V = f −1 ({1}), then U ∪ V =
f −1 ({0} ∪ {1}) = X and U ∩ V = f −1 ({0} ∩ {1}) = ∅. Moreover, since f is onto,
neither U nor V can be empty, and, since f is continuous, both of these sets are
open. Hence we have a disconnection, and so X is not connected.
Conversely, suppose that X is not connected, so then there is a disconnection
(U, V ) of X. Define a map f : X → {0, 1} by
(
0 if x ∈ U,
f (x) =
1 if x ∈ V.
46
This is possible since U ∩ V = ∅, and f is onto since U 6= ∅ =
6 V . Moreover, the
open subsets of {0, 1} are ∅, {0}, {1} and {0, 1}, and their preimages are ∅, U , V
and X respectively. Hence f is a continuous surjection X → {0, 1}.
An immediate corollary of the next result is that connectedness is a topological
property.
Corollary 4.6. Let f : X → Y be a continuous map between topological spaces,
with X connected. Then f (X) is a connected subspace of Y .
Proof. Suppose that f (X) is not connected, then there must be a continuous
surjection g : f (X) → {0, 1}. It follows that the composition g ◦ f : X → {0, 1} is
continuous and onto, and hence X cannot be connected.
Combining this with the characterisation of connected sets of R (Theorem 4.4)
gives the following:
Corollary 4.7. If X is a connected space and f : X → R is a continuous map,
then f (X) is an interval.
The obvious application of this result (together with our work on compactness)
to real analysis, is a proof by abstract means of the Intermediate Value Theorem:
if f : [a, b] → R is a continuous map then f ([a, b]) is a compact and connected
subset of R, hence must be of the form [c, d] for some c < d in R.
As with all of the properties we have considered, we would like to know how
connectedness behaves when we pass to subspaces, build products etc. Indeed, we
have already seen examples of spaces that are not connected, but are subspaces of
connected space (e.g. Q ⊂ R).
Proposition 4.8. Let X be a topological space, let {Aλ }λ∈Λ be a family of subspaces such that each Aλ is connected, and
S suppose that for some λ0 ∈ Λ we have
Aλ ∩ Aλ0 6= ∅ for all λ ∈ Λ. Then A = λ∈Λ Aλ is connected.
Proof. Let f : A → {0, 1} be a continuous map. We must show that f is not onto.
Since f is continuous, the restriction f |Aλ is continuous for each λ ∈ Λ, and so
must be constant. That is, for some yλ ∈ {0, 1} we have f (x) = yλ for all x ∈ Aλ .
But then for each λ we can choose xλ ∈ Aλ ∩ Aλ0 , so yλ = f (xλ ) = yλ0 , and hence
f (x) = yλ0 for all x ∈ A. Thus A is connected.
Corollary 4.9. Let X1 and X2 be connected spaces. Then X = X1 × X2 is
connected if and only if X1 and X2 are connected.
Proof. First note that Xi is the image of X under the continuous projection map
pi , and so each Xi must be connected if X is connected, by Proposition 4.6.
Suppose instead that X1 and X2 are connected. We shall apply Proposition 4.8
with Λ = {0}∪X2 . Pick x ∈ X1 and set A0 = {x}×X2 . Then A0 is homeomorphic
to X2 (exercise), and hence connected. Similarly, for each y ∈ X2 set Ay =
X1 × {y}, then each Ay is homeomorphic to X1 and hence connected. Moreover,
A0 ∩ Ay = {(x, y)} =
6 ∅, and
S
S
X = y∈X2 Ay ⊂ A0 ∪ y∈X2 Ay ⊂ X.
47
Path-connectedness
When considering whether or not a subset S of R2 or R3 is connected, our mental
picture is probably to check if every pair of points in S can be joined by some sort
of path that lies in S. This is the basis of a very important, but slightly different,
method of approaching this topic.
Definition 4.10. Given two points a, b in a topological space X, a path from a
to b is a continuous map f : [0, 1] → X such that f (0) = a and f (1) = b.
Definition 4.11. A topological space is path-connected if every two points in
the space can be connected by a path.
This definition turns out to be a specialisation of the one we have been using
up to now:
Proposition 4.12. Any path-connected space is connected.
Proof. Suppose that X is a path-connected space and let f : X → {0, 1} be a
continuous map. Let x, y ∈ X, then there a path from x to Y , i.e. a continuous
map g : [0, 1] → X with g(0) = x and g(1) = y. But then f ◦ g : [0, 1] → {0, 1} is
a continuous map on the connected space [0, 1], and hence must be constant. So
now we have
f (x) = f g(0) = f g(1) = f (y),
thus f must be constant, and so X is connected.
It turns out that the converse is not true — an example of a connected space
that is not path-connected is given on the exercise sheets. However for a large
class of spaces the converse is true, as shown in Proposition 4.14 below. First,
however, we need to consider the composition of paths.
Lemma 4.13. Let X be a topological space and suppose that x, y, z ∈ X are points
such that x and y can be connected by a path, and y and z can be connected by a
path. Then x and z can be connected by a path.
Proof. By hypothesis there are continuous maps f : [0, 1] → X and g : [0, 1] → X
such that f (0) = x, f (1) = g(0) = y and g(1) = z. Defining h : [0, 1] → X by
(
f (2t)
if t ∈ [0, 21 ],
h(t) =
g(2t − 1) if t ∈ [ 12 , 1]
gives a continuous map such that h(0) = x and h(1) = z.
Proposition 4.14. Any open connected subset U of Rn is path-connected.
Proof. Fix a point a ∈ X, and let H be the set of points that can be connected to
a by a path in U . We must show that K = U \ H is empty.
First we shall show that H is open in U (equivalently, open in Rn ). So let
x ∈ H, then for some ε > 0 we have B(x, ε) ⊂ U . Given any y ∈ B(x, ε) we
can connect it to x, the centre of this ball, by the radial path. Thus, there is a
48
path from a to x, and then a path from x to y, and these can be composed using
Lemma 4.13 to give a path from a to y. Hence B(x, ε) ⊂ H, and so H is open in
U (and Rn ). A similar argument shows that K must also be open. But now by
definition we have H ∪ K = U , and H ∩ K = ∅. Furthermore, U is connected, and
H 6= ∅, hence K = ∅ as required.
Exercises
1. Let T1 and T2 be topologies on a set X satisfying T1 ⊂ T2 . Is (X, T1 ) connected
if (X, T2 ) is connected? Is (X, T2 ) connected if (X, T1 ) is connected?
2. Let X be a connected space containing more than one point, and suppose that
{x} is closed for each x ∈ X. Show that the number of points in X is infinite.
3. Prove that any continuous function f : [a, b] → [a, b] has a fixed point.
[Hint: consider the function g : [a, b] → R given by g(x) = f (x) − x.]
4.
(a) Prove that if X is a connected space and there is a continuous map f : X → R
that is not constant, then X is uncountably infinite.
(b) Prove that if X is a connected metric space with more than one point, then
it has uncountably many points.
5. Prove that there is no continuous injective map f : R2 → R. Can the spaces R2
and R be homeomorphic? Is the line R homeomorphic to [−1, 1]?
6. Let f : [0, 1] → R be a continuous map such that f (0) = f (1). Let n ≥ 2.
Show that there is some x ∈ [0, 1] such that f (x) = f (x + n1 ). [Hint: let g(x) =
f (x) − f (x + n1 ) and show that g(0) + g( n1 ) + · · · + g( n−1
n ) = 0.]
7. Prove that any discrete space is totally disconnected.
2
8. Give an example
T of connected subsets Cn ⊂ R such that Cn ⊃ Cn+1 for all n ≥ 1,
but such that n≥1 Cn is not connected. Can you find an example for which all of
the Cn are closed?
9. Prove that there is no continuous map f : [0, 1] → R satisfying x ∈ Q ⇔ f (x) ∈
/ Q.
10. Consider the following subset of R2 : A = {(x, y) ∈ R : x 6= 0, y = sin x1 }. Is A
connected or disconnected? Answer the same question for the subsets A ∪ {(0, 0)}
and A ∪ ({0} × [−1, 1]).
[Hint: first prove the following result. If B and C are subsets of a topological
space X with B connected and B ⊂ C ⊂ B, then C must also be connected. To
show this note that any continuous map f : C → {0, 1} restricts to a continuous
map B → {0, 1}, which must be constant.]
49
5
Completeness and Uniformity
Uniform convergence and continuity
When dealing with a collection of functions defined on some given set X, we would
like some notion of whether or not two of these functions are close to each other,
or whether a sequence from the collection converges. A natural method is to look
at the behaviour at each point x in the space, but this does not always guarantee
respectable outcomes.
Definition 5.1. Let X be a set and (fn ) a sequence of functions from X into a
metric space (Y, d). The sequence converges pointwise to a function f : X → Y if
f (x) = limn fn (x) for each x ∈ X. Here we are using the definition of convergence
in a metric space as introduced in the first section. So fn converges pointwise to
f if
∀x ∈ X, ε > 0 ∃N ≥ 1 s.t. d f (x), fn (x) < ε ∀n ≥ N.
Here the N depends on the choice of both x and ε.
The sequence converges uniformly to the function f if
∀ε > 0 ∃N ≥ 1 s.t. d f (x), fn (x) < ε ∀x ∈ X, n ≥ N.
That is, there is some N that does the job for all of the points x ∈ X simultaneously.
It is possible in certain circumstances to reinterpret uniform convergence as the
usual convergence with respect to a given metric. For instance let X = C[0, 1], the
space of continuous functions from [0, 1] to C, equipped with the following metric:
d∞ (f, g) = sup |f (x) − g(x)|.
x∈[0,1]
Now suppose that f ∈ X and (fn ) is a sequence in X such that fn → f with
respect to d∞ . This means that for any given ε > 0 we can find some N ≥ 1 such
that d∞ (f, fn ) < ε for every n ≥ N . But if this is so then
n ≥ N ⇒ |f (t) − g(t)| ≤ sup |f (x) − g(x)| = d∞ (f, g) < ε ∀t ∈ [0, 1]
x∈[0,1]
That is, the functions fn converge uniformly to the function f . For this reason d∞
is often known as the uniform metric.
Note that d∞ can be defined in the same way on the set B(X; R) of all bounded
functions from any given set X into R, and again a sequence in B(X; R) converges
to some other function in B(X; R) uniformly if and only if it converges with respect
to the uniform metric. Here we need to restrict ourselves to bounded functions in
order to make sense of d∞ , but the concept of uniform convergence applies more
widely: if we consider the functions fn and f defined on R by
fn (x) = x2 + n1 ,
f (x) = x2
then fn converges to uniformly to f , since fn (x) − f (x) =
x ∈ R, but none of the functions involved are bounded.
50
1
n
for all choices of
Example 5.2. Consider the functions fn (x) = xn defined for x ∈ [0, 1]. Then we
have fn (x) → f (x) pointwise, where
(
0 if x ∈ [0, 1),
f (x) =
1 if x = 1.
Thus this sequence of continuous functions converges pointwise to a discontinuous
function. It follows from the next result that fn does not converge uniformly to f .
Proposition 5.3. Let X and Y be metric spaces, and let (fn ) be a sequence of
functions from X to Y each of which is continuous at x0 ∈ X. If fn converges
uniformly to a function f : X → Y , then f is continuous at x0 .
Remark. Thus, if a family of continuous functions converges uniformly then the
limit must be a continuous function.
Proof. Fix ε > 0. Since fn → f uniformly there is some N ≥ 1 such that if
n ≥ N , then dY f (x), fn (x) < 3ε for every x ∈ X. In particular this is true for
n = N . But fN is continuous
at the point x0 , and so there is some δ > 0 such
ε
that dY fN (x), fN (x0 ) < 3 whenever x ∈ X satisfies dX (x, x0 ) < δ. So now if
dX (x, x0 ) < δ then
dY f (x), f (x0 ) ≤ dY f (x), fN (x) +dY fN (x), fN (x0 ) +dY fN (x0 ), f (x0 )
ε
< × 3 = ε,
3
and thus f is continuous at x0 as required.
The above proposition is one that is frequently invoked whenever we need to
change the order in which limits are taken. In effect what we have proved is that
if fn → f uniformly, and all of the fn are continuous at x0 then
lim lim fn (x) = lim lim fn (x) .
n→∞ x→x0
x→x0 n→∞
Example 5.4. As an example of a sequence of continuous functions that converge
pointwise but not uniformly to a continuous function (indeed, a constant function)
consider the following ‘moving bump’:
1
1
1
1
f1
1
2
f2
51
1
1
3
1
f3
1
We have that fn (x) → 0 for every x ∈ [0, 1], but fn ( 2n
) = 1 for each n. So given
any 0 < ε < 1 there can be no N ≥ 1 for which |fn (x) − 0| < ε for every x ∈ [0, 1]
and n ≥ N .
A counterpart to uniform convergence is uniform continuity, and this now requires that our domain X be a metric space, rather than merely a set.
Definition 5.5. Let X and Y be metric spaces. A function f : X → Y is
uniformly continuous if for each ε > 0 there is some δ > 0 such that
dX (x, x0 ) < δ ⇒ dY f (x), f (x0 ) < ε.
The distinction between continuity as discussed earlier and uniform continuity
is that for the given ε we can find a δ such that whenever we pick any two points
x and x0 that are within δ of each other, then their images are within ε of each
other. That is, the δ is independent of the points in question.
Example 5.6. The simplest example of a uniformly continuous map is constructed
as follows: given a pair of metric spaces X and Y , pick a point y0 ∈ Y and
define
0
f : X → Y by setting f (x) = y0 for every x ∈ X. Then dY f (x), f (x ) = 0 for
every pair of points x, x0 ∈ X, and so given ε > 0 we may take δ to be any positive
number.
Uniform continuity plays an important role in real and complex analysis, as it is
a significant strengthening of the usual notion of continuity. The next result shows
that we encounter this happy situation more frequently than one may imagine on
first seeing the definition.
Proposition 5.7. Let X and Y be metric spaces, with X compact. Then any
continuous function from X to Y is uniformly continuous.
Proof. Let ε > 0. For each x ∈ X we can find some δx > 0 such that
ε
dX (x, x0 ) < 2δx ⇒ dY f (x), f (x0 ) <
2
(∗)
and the resulting collection of open balls, {B(x, δx ) : x ∈ X} (note the halving of
the radii), is an open cover of X. Hence there is a finite subcover, {B(x1 , δx1 ), . . . , B(xn , δxn )}
say, since X is compact.
Set δ = min{δx1 , . . . , δxn }, and consider any pair of points x, x0 ∈ X such
that dX (x, x0 ) < δ. Now x ∈ B(xi , δxi ) for some 1 ≤ i ≤ n. So by (∗) we have
dY f (x), f (xi ) < 2ε . Also,
dX (x0 , xi ) ≤ dX (x0 , x) + dX (x, xi ) < δ + δxi < 2δxi .
But then dY f (x0 ), f (xi ) < 2ε by (∗) again, and putting these two inequalities
together by means of the triangle inequality gives
ε ε
dY f (x), f (x0 ) ≤ dY f (x), f (xi ) + dY f (xi ), f (x0 ) < + = ε
2 2
as required.
52
2δx
δx x
X
Thus whenever we are dealing with a continuous function defined on some
compact subset of R or C we know automatically that it is in fact uniformly continuous. Sometimes we may want to work with functions defined on noncompact
sets (for instance all of R and C) and would prefer to have global uniform continuity, but may have to make do with restricting our function to each of an increasing
sequence of compact subsets whose union is the whole space that we are interested
in (for instance ([−n, n])n≥1 in the case of R.)
It is not hard to show that Proposition 5.3 extends to the case when we are
dealing with uniformly continuous functions, by imitating the ‘3ε-proof’ of that
result.
Proposition 5.8. Let X and Y be metric space and let (fn ) be a sequence of
uniformly continuous functions from X to Y . Suppose that the sequence converges
uniformly to a map f : X → Y , then f is uniformly continuous.
Complete metric spaces
Recall that a sequence (xn )n≥1 in a metric space X converges to the point x ∈ X
if for each ε > 0 we can find some N ≥ 1 such that
n ≥ N ⇒ d(xn , x) < ε.
That is, all elements in the sequence must be reasonably close to/arbitrarily close
to x after a certain point. So if m, n ≥ N then, by the triangle inequality,
d(xm , xn ) ≤ d(xm , x) + d(x, xn ) < 2ε.
Thus, since ε > 0 was arbitrary, the individual elements of the sequence must be
as close to each other as we like after some point.
Definition 5.9. Let X be a metric space. A sequence (xn )n≥1 is a Cauchy sequence if for each ε > 0 we can find some N ≥ 1 such that
m, n ≥ N ⇒ d(xm , xn ) < ε.
This is written d(xm , xn ) → 0 as m, n → ∞.
If every Cauchy sequence in X is convergent then the space X is said to be
complete.
53
Remark. In particular we see that every convergent sequence in a metric space is
Cauchy. Whether or not the converse is true is the more interesting question.
Perhaps the most basic and fundamental example of a complete metric space is
R equipped with the usual metric. Indeed, completeness is built into this space in
terms of what is usually known as the ‘Completeness Axiom’ which states that any
nonempty set that is bounded above has a least upper bound. This can then be
used to show that every Cauchy sequence converges. Conversely, if we assume that
R obeys its usual algebraic and order properties, and that every Cauchy sequence
converges, then each nonempty set that is bounded above must have a least upper
bound.
The next space usually encountered after giving the definition of completeness
is then Q, the rational numbers. This space (with the usual metric/topology) is
not complete. In √
particular, since it is dense in R we can find a sequence in Q
that converges to 2 ∈ R \ Q. Hence this sequence will be Cauchy, but cannot
converge to a rational number since the limits of sequences in a metric space are
unique.
It is should also now be noted that this is a property that is more properly
associated to the metric space structure than the topological structure. Recall that
R and (0, 1) are homeomorphic as topological spaces, but the former is complete
whereas the latter is not. To see this we can take the sequence (1/(n + 1))n≥1
in (0, 1) which is Cauchy, but again cannot actually converge to a point in this
subspace of R, since 1/(n + 1) → 0 as n → ∞.
Proposition 5.10. Let Y be a subspace of a metric space X.
(a) If Y is complete then Y is a closed subset of X.
(b) If X is complete and Y is closed then Y is complete.
Proof. (a) Suppose that Y is complete, and let (yn ) be a sequence in Y that
converges to some x ∈ X. But then (yn ) is a Cauchy sequence in Y , and hence
converges to some y ∈ Y by completeness of Y . However the limits of sequences in
metric spaces are unique (Proposition 2.12), and so x = y ∈ Y . Thus Y is closed
by Corollary 1.24.
(b) Suppose that X is complete and that Y is closed in X. Now let (yn ) be
a Cauchy sequence in Y . It is also a Cauchy sequence in X, hence convergent
to some x ∈ X since X is complete. However since Y is closed we have that
x = limn yn ∈ Y by Corollary 1.24, and so Y is complete.
Note. We do not require X to be complete in part (a) above — [1, 2] is a complete
subspace of the incomplete space (0, 3).
The concept of completeness is also intimately related to compactness. To see
this we first prove the following lemma:
Lemma 5.11. Let (xn )n≥1 be a Cauchy sequence in a metric space X, and suppose
that some subsequence (xn(k) )k≥1 converges to some point x ∈ X. Then xn → x
as n → ∞.
54
Proof. Choose ε > 0. Then (xn )n≥1 is Cauchy and so there is some N ≥ 1 such
that that d(xm , xn ) < 2ε for all m, n ≥ N . Also, since xn(k) → x as k → ∞, we
can find some K ≥ 1 such that d(x, xn(k) ) < 2ε whenever k ≥ K.
So now pick k such that both k ≥ K and n(k) ≥ N . Then
d(x, xn ) ≤ d(x, xn(k) ) + d(xn(k) , xn ) <
ε ε
+ =ε
2 2
for all n ≥ N by the triangle inequality, and so xn → x as required.
Proposition 5.12. A compact metric space is complete.
Proof. Let X be a compact metric space and (xn )n≥1 a Cauchy sequence taken
from X. Since X is compact we can extract a convergent subsequence, (xn(k) )k≥1
say, with limit x. Applying the lemma above we get that xn → x as well, and so
X is complete.
The above shows that all compact metric spaces are also complete, but the
converse is not true — consider the space R.
Cantor’s Intersection Theorem gives an alternative characterisation of completeness, based on nested sequences of sets with decreasing diameters.
Lemma 5.13. Let X be a metric space, and A ⊂ X a bounded subset. Then
diam A = diam A.
Proof. Since A ⊂ A, it is clear that diam A ≤ diam A. On the other hand if
x, y ∈ A then we can pick sequences (xn ), (yn ) ⊂ A such that xn → x and yn → y
by Proposition 1.23. So then
d(x, y) ≤ d(x, xn ) + d(xn , yn ) + d(yn , y)
≤ d(x, xn ) + diam A + d(yn , y)
for all n ≥ 1. The right hand side converges to diam A as n → ∞, and so
d(x, y) ≤ diam A for all x, y ∈ A, which implies that diam A ≤ diam A.
Lemma 5.14. Let (xn )n≥1 be a sequence in a metric space X, and for each m ≥ 1
define Am = {xm , xm+1 , . . .}. Then (xn ) is a Cauchy sequence if and only if
diam Am → 0 as m → ∞.
Note. Any Cauchy sequence is automatically bounded, and so the diameters mentioned above do make sense.
Proof. This follows since (xn ) is Cauchy if and only if for each ε > 0 there is
some N ≥ 1 such that d(xm , xn ) < ε for all m, n ≥ N , which holds if and only if
diam An ≤ ε for all n ≥ N .
Theorem 5.15 (Cantor). A metric space X is complete if, and only if, for any
sequence T
(Fn ) of nonempty closed sets that satisfy Fn ⊃ Fn+1 and diam Fn → 0,
we have n≥1 Fn 6= ∅.
T
Remark. Since diam Fn → 0, it follows that diam n≥1 Fn = 0, and so the intersection is either the empty set or contains precisely one point.
55
Proof. Suppose that X is complete and that (Fn ) is a sequence as in the statement
of the theorem. For each n ≥ 1 choose some xn ∈ Fn . Then An = {xn , xn+1 , . . .} ⊂
Fn , by the nesting property, hence diam An ≤ diam Fn → 0, and so this sequence
is Cauchy by Lemma 5.14. Since X is complete, xn → x for some x ∈TX. For each
m we have x = limn→∞,n≥m xn ∈ Fm , since Fm is closed, hence x ∈ n≥1 Fn , and
so the intersection is nonempty.
Suppose conversely that any nested sequence of closed sets whose diameters
tend to 0 has nonempty intersection, and pick any Cauchy sequence (xn ) in X.
Setting An = {xn , xn+1 , . . .} as before, we have diam An = diam An → 0 by
Lemmas 5.13 and 5.14. Also AT
n ⊃ An+1 , and so An ⊃ An+1 . Thus, by our
assumption on X, we must have n≥1 An 6= ∅. Pick a point x in this intersection,
then for each n we have x, xn ∈ An and so d(x, xn ) ≤ diam An → 0, hence the
Cauchy sequence converges to x as required.
Applications of completeness: fixed points and category
One of the main uses of completeness is to make use of the guarantee of existence
of limits to Cauchy sequences to produce existence results results in a wide variety
of contexts, in particular when solving differential equations. Rather than use
the definition of completeness explicitly, these existence proofs traditionally make
use of one of a number of ‘fixed-point theorems’, whose proofs in turn rely on
completeness. These theorems can often also be used guarantee uniqueness of the
solution.
Definition 5.16. Let X be a set and f : X → X a map. A point x ∈ X is a fixed
point of f if f (x) = x. The set {x ∈ X : x = f (x)} is called the fixed point set
of the map f .
Definition 5.17. Let X be a metric space. A map f : X → X is a contraction if
there is some constant K < 1 such that d f (x), f (y) ≤ Kd(x, y) for all x, y ∈ X.
Remark. Note that if we take f to be a constant map then it is a contraction since
we can take K = 0.
Lemma 5.18. A contraction f : X → X is uniformly continuous.
Proof. Let K be a constant as in the definition above, and let ε > 0. Then, for
any x, y ∈ X satisfying d(x, y) < ε, we have
d f (x), f (y) ≤ Kd(x, y) ≤ d(x, y) < ε,
and hence f is uniformly continuous.
Theorem 5.19 (Banach). Let f be a contraction of a metric space X. Then f
has a unique fixed point.
Proof. We first show that there is at least one fixed point. Choose an arbitrary
point x0 ∈ X and define xn for n ≥ 1 iteratively by setting xn = f (xn−1 ). The
contraction condition and an easy induction argument gives
d(xj+1 , xj ) ≤ K j d(x1 , x0 )
56
for all j ≥ 0. Thus, for any n > m, we have
d(xn , xm ) ≤ d(xn , xn−1 ) + · · · + d(xm+1 , xm )
≤ K n−1 + · · · + K m d(x1 , 0)
Km − Kn
1
=
d(x1 , x0 ) ≤ K m ×
d(x1 , x0 ).
1−K
1−K
Now K m → 0 as m → ∞, since 0 ≤ K < 1, and the other term on the right is
fixed. It thus follows that (xn ) is a Cauchy sequence, and so must converge to
some limit x.
But f is a continuous
map from X to itself, and since x = limn xn we must
have f (x) = f limn xn = limn f (xn ). Also f (xn ) = xn+1 → x as n → ∞, and so
f (x) = x. Thus the limit of this sequence is indeed a fixed point.
Finally, to show uniqueness,
suppose that x and y are both fixed points for f .
Then d(x, y) = d f (x), f (y) ≤ Kd(x, y). For this inequality to hold we must have
d(x, y) = 0, otherwise we would have K ≥ 1. Hence we must have x = y, and so
there is only one fixed point.
As stated above, fixed point theorems can be used to prove the existence and
uniqueness of solutions to differential equations. The above theorem can also be
applied to a related sort of problem, namely an integral equation of the form
Z b
F (x) = λ
K(x, y)F (y) dy + G(x),
a
where a < b are fixed and K : [a, b] × [a, b] → R and G : [a, b] → R are known
continuous functions. In fact to apply Theorem 5.19 we must pick λ so that
|λ| < {(b − a)M }−1 , where M is any constant such that |K(x, y)| ≤ M for all
(x, y) ∈ [a, b] × [a, b]. That such an M exists is guaranteed by the fact that K is a
continuous map on the compact set [a, b] × [a, b].
This line of reasoning works most naturally if we equip the space C[a, b] of
continuous maps from [a, b] to C with the uniform metric d∞ , since it is then a
complete metric space (exercise!). On the other hand, it is not complete if given
the d1 or d2 metrics defined in terms of integrals over [a, b]. In such a situation
if we want to continue to use one of these metrics for some other reason and
use completeness arguments, then we would like to ‘extend’ or ‘enlarge’ the space
C[a, b] in some manner to make it complete.
Definition 5.20. Let X and Y be metric spaces. A map f : X → Y is an
isometry if d f (x), f (x0 ) = d(x, x0 ) for all x, x0 ∈ X.
Remark. An isometry is necessarily uniformly continuous (cf. Lemma 5.18). It
preserves distances, and hence the metric space structure of X.
Theorem 5.21. Let X be a metric space. Then there is another metric space X0
and an isometry ι : X → X0 such that
(i) X0 is complete.
(ii) ι(X) = X0 .
57
Remark. The space X0 is known as the completion of X. It is a complete metric
space that ‘contains a copy of X’ as a dense subspace. The most famous example
is the real numbers R, which are the completion of the rational numbers Q.
Note that if X is complete then it is not hard to show that ι(X) is a complete
subspace of the complete space X0 , and so must be closed by Proposition 5.10.
But then X0 = ι(X) = ι(X), and so X and X0 are, essentially, the same space.
Proof. (Sketch) Let X1 denote the set of all Cauchy sequences coming from X.
That is
X1 = {(xn )n≥1 ⊂ X : (xn ) Cauchy}.
We define a relation ∼ on X1 by
(xn ) ∼ (yn ) if d(xn , yn ) → 0 as n → ∞.
It is not hard to show that ∼ is an equivalence relation. Let X0 denote the set of
equivalence classes, then we equip X0 with the metric
d [(xn )], [(yn )] = sup d(xn , yn ),
n
where [(xn )] denotes the equivalence class of (xn ). We must check that this is welldefined (that is, it does not depend on the choice of representative from [(xn )]),
and is also a metric on X0 that makes X0 into a complete space.
Finally, we must construct the map ι. This is done by mapping x ∈ X to
ι(x) ∈ X0 , where ι(x) = [(x, x, x, . . .)]. It is not hard to see that ι is indeed an
isometry, and moreover that ι(X) is dense in X0 .
Finally, we turn to the concept of category for topological spaces.
Definition 5.22. Let A be a subset of a topological space X. The interior of A
is the largest open set contained in A, and is denoted A◦ .
Remark. That the set A◦ exists can be shown by taking the union of all open sets
that are contained in A.
Example 5.23. If X = R with the usual topology then (0, 1)◦ = [0, 1]◦ = [0, 1)◦ =
(0, 1). Indeed, the interior of any open set is equal to the set itself. Note that we
must also pay attention to the space we are in, since if we were to take X = [0, 1]
then [0, 1)◦ = [0, 1).
The connection between interiors and closures is given by:
Lemma 5.24. Let X be a topological space. Then X \A◦ = X \ A for any A ⊂ X.
Proof. By definition A◦ ⊂ A, hence X \ A◦ ⊃ X \ A. But X \ A◦ is closed, since
A◦ is open, and thus
X \ A◦ ⊃ X \ A ⊃ X \ A ⇒ A◦ ⊂ X \ X \ A ⊂ A.
Now note that X \ X \ A is an open subset of A that contains A◦ , and so we
must have A◦ = X \ X \ A .
58
Definition 5.25. A subset A of a topological space X is nowhere dense if its
closure has empty interior, that is (A)◦ = ∅.
Example 5.26. The one point set {0} ⊂ R is nowhere dense since {0} = {0} and
{0}◦ = ∅. Similarly the two point set {0, 1} is nowhere dense, and in fact it is clear
that any finite subset of R is nowhere dense.
The set { n1 : n ≥ 1}, which is a countable union of singleton sets, is also
nowhere dense — taking its closure only adds the point 0 to the set, but still we
will be unable to fit an open set/open ball inside. Note however that Q has empty
interior, since any open subset of R contains both rational and irrational numbers,
but Q is not nowhere dense since Q = R, and so (Q)◦ = R.
The second of the next two results gives an alternative characterisation of what
is means to be nowhere dense, which highlights the idea that nowhere dense subsets
should be somehow ‘small’.
Lemma 5.27. A subset B of a topological space X is dense if and only if U ∩B 6= ∅
for every nonempty open set U of X.
Proof. If B is dense then B = X, so if U is a nonempty set then for any x ∈ U
we have x ∈ B and so must have U ∩ B 6= ∅ by Proposition 2.14.
On the other hand suppose that U ∩B 6= ∅ for all nonempty open sets U . Then
for any x ∈ X and neighbourhood U of x we have U ∩ B 6= ∅, and so x ∈ B, giving
B = X.
Lemma 5.28. Let A be a subset of a topological space X. Then A is nowhere
dense if and only if every nonempty open set U contains a nonempty open subset
V that is disjoint from A. Furthermore, if X is a metric space it is enough to
consider open balls in place of the set V .
Proof. By Lemma 5.24 we have X \ (A)◦ = X \ A, and so A is nowhere dense if
and only if X \ A = X, that is if and only if X \ A is dense in X. Hence A is
nowhere dense if and only if for each nonempty subset U we have U ∩ (X \ A) 6= ∅
by Lemma 5.27.
So if A is nowhere dense then for each open U 6= ∅, the set U ∩ (X \ A) 6= ∅ is
an open subset that is disjoint from A, and hence disjoint from A ⊂ A.
Suppose conversely that each open U 6= ∅ contains a nonempty open subset V
such that V ∩ A = ∅. This is equivalent to X \ V ⊃ A, and so X \ V ⊃ A, since
X \ V is closed. But then V ∩ A = ∅, and thus
∅=
6 V = V ∩ (X \ A) ⊂ U ∩ (X \ A),
showing that A is nowhere dense.
It is a straightforward matter to specialise the result to the case of metric
spaces and open balls.
The following is the Baire Category Theorem, the full impact of which is far
from obvious at first, but which has very significant ramifications in functional
analysis such as the Uniform Boundedness Principle, which in turn is used to
establish the Open Mapping Theorem and the Closed Graph Theorem.
59
Theorem 5.29 (Baire). Let X be a complete metric space,
S and suppose that
(An )n≥1 is a sequence of nowhere
dense subsets. Then X \ n≥1 An is a dense
S
subset of X. In particular n≥1 An 6= X.
Proof. Let U be a nonempty open set. It follows S
from Lemma 5.27 that we must
show that it has nonempty intersection with X \ n≥1 An .
Now A1 is nowhere dense, so by Lemma 5.28 we can pick some x1 ∈ U and
ε1 > 0 such that B(x1 , 2ε1 ) ∩ A1 = ∅. Let F1 be the closed set Bc (x1 , ε1 ), where
Bc (y, δ) = {z ∈ X : d(y, z) ≤ δ}. Then F1 ⊂ B(x1 , 2ε1 ) and so F1 ∩ A1 = ∅.
Suppose now that we have chosen x1 , . . . , xn and ε1 , . . . , εn so that each Fi =
Bc (xi , εi ) satisfies Fi ∩ Ai = ∅, that F1 ⊃ F2 ⊃ · · · ⊃ Fn , and that εi+1 ≤
2−1 εi . But then B(xn , εn ) is nonempty and open, and so contains an open ball
B(xn+1 , 2εn+1 ) that is disjoint from the nowhere dense subset An+1 , and we can
thus set Fn+1 = Bc (xn+1 , εn+1 ), to ensure that Fn+1 ∩ An+1 = ∅, that Fn ⊃ Fn+1 ,
and that εn+1 ≤ 2−1 εn .
But now diam Fn ≤ 2εn ≤ T
22−n ε1 → 0, and so by Cantor’s Intersection Theorem (Theorem 5.15) we have n≥1 Fn = {x} forSsome x ∈ X. Since x ∈ Fn for
each n, we have x ∈
/ A for each n, and so x ∈ X \ n≥1 An . Moreover x ∈ F1 ⊂ U ,
S n and so U ∩ X \ n≥1 An 6= ∅ as required.
Consider the Euclidean plane X = R2 , equipped with any of the sensible
metrics considered in Example 1.3. It is not hard to show that X is complete, using
the completeness of R. Consider any straight line l in R, so for some constants
a, b, c ∈ R we have
l = {(x, y) ∈ R2 : ax + by = c} = f −1 ({c})
where f : R2 → R is the continuous map f (x, y) = ax + by. In particular l is
the preimage of a single point and hence closed. It is clear that any ball about a
point on the line must also contain points off of the line, so that l◦ = ∅, and since
l = l we have that l is nowhere dense. It follows from Baire’s Category Theorem
that R2 cannot be written as a union of a countable family of lines. This readily
generalises to show that for any n ≥ 2, the n-dimensional Euclidean space Rn
cannot be written as a countable union of subspaces of lower dimension.
The same is not true of the incomplete space Q2 however, since we have
[
Q = {(0, x) : x ∈ Q} ∪
{(x, qx) : x ∈ Q}
q∈Q
writing Q2 as the union of the line x = 0 together with each line of rational
slope through the origin, since if x1 6= 0 then (x1 , y1 ) lies on the line y = qx for
q = y1 /x1 ∈ Q.
The following terminology was devised by Baire: a subset A of a topological
space X is of the first category if it can be written as a countable union of nowhere
dense sets. The set is of the second category if it is not of the first category. This is
somewhat derided terminology, since many authors believe it does not describe at
all the concepts involved. There are alternatives such as meagre and nonmeagre.
In any case, what Baire’s (Category) Theorem shows is that a complete metric
space is of the second category in itself.
60
Exercises
1. Which of the following sequences of functions [0, 1] → R converge uniformly?
(i) fn (x) =
x
1 + nx
2
(ii) fn (x) = nxe−nx
(iii) fn (x) =
xn
1 + xn
2. Let X be a set, (fn ) a sequence of functions X → R and f : X → R a function
such that fn → f pointwise. Define Mn = supx∈X |fn (x) − f (x)| whenever this
supremum exists. Show that fn → f uniformly if and only if Mn → 0 as n → ∞.
3. Let X be a set and Y a metric space. Suppose (fn ) is a sequence of functions
X → Y that converges uniformly to a map f : X → Y . Suppose that each fn
is bounded (that is, fn (X) is a bounded subspace of the metric space Y ). Show
that
(i) f is bounded.
(ii) There is a uniform bound
for the functions. That is, there is some K > 0
such that d fn (x), fn (x0 ) < K for all n ≥ 1 and x, x0 ∈ X.
4. Construct a sequence of functions fn : R → R none of which is continuous at
x = 0, but such that (fn ) converges uniformly on R to a continuous function.
5. Let X, Y and Z be metric spaces and let f : X → Y and g : Y → Z be
uniformly continuous maps. Show that the composition g ◦f : X → Z is uniformly
continuous.
6. State the negation of the definition of uniform continuity and use this to prove
that the map f : (0, 1) → R given by f (x) = x1 is not uniformly continuous.
On the other hand, show that a function g : (0, 1) → R that is continuous, monotonic and bounded must be uniformly continuous on (0, 1).
7. Let X be a set and Y a complete metric space. A sequence (fn ) of functions
X → Y is uniformly
Cauchy if for each given ε > 0 there is some N ≥ 1 such
that d fm (x), fn (x) < ε for all m, n ≥ N and x ∈ X. Show that (fn ) converges
uniformly to some map f : X → Y if and only if the sequence if uniformly Cauchy.
8. Prove that if a sequence of uniformly continuous functions converges uniformly,
then the limit function is also uniformly continuous.
9. Prove that any nonempty set X is made into a complete metric space if it is given
the discrete metric.
10. Which of the following spaces are complete?
n1
o
n
1o
(i)
: n ≥ 1 ∪ {0}
(ii) Q ∩ [0, 1]
(iii) (x, y) ∈ R2 : x > 0, y ≥
n
x
11. Show that the uniformly continuous image of a complete metric space need not be
complete.
12. Let X and Y be metric spaces and f : X → Y a bijection such that f is uniformly
continuous and f −1 is continuous.
61
(i) Show that if (xn ) is a Cauchy sequence in X, then f (xn ) is a Cauchy
sequence in Y .
(ii) Show that if (yn ) is a convergent sequence in Y then f −1 (yn ) is a convergent sequence in X.
(iii) Deduce that if Y is complete then X is complete.
13. Let l∞ denote
numbers. Show that setting
the set of all bounded sequences of real
∞
d (xn ), (yn ) = supn |xn − yn | defines a metric on l that turns it into a complete
metric space.
Let X = {(xn ) ∈ l∞ : supn |xn | ≤ 1}. Show that X is a proper, closed and
bounded subset of l∞ . Deduce that X is complete but, by considering sequences
taken from X, show that X is not compact.
14.
(a) Let f : (0, 1) → (0, 1) be given by f (x) =
without a fixed point.
x2
3 .
Show that f (x) is a contraction
(b) Let f : [1, ∞) → [1, ∞) be given by f (x) = x+x−1 . Show that |f (x)−f (y)| <
|x − y| for any distinct x, y, that [1, ∞) is complete, but that f has no fixed point.
15. Let X = C[a, b] be the space of continuous maps from [a, b] to C, together with
the metric d∞ . Prove that (X, d∞ ) is a complete metric space.
Let K : [a, b] × [a, b] → C and G : [a, b] → C be continuous maps, and choose
λ ∈ C. For each F ∈ X define a map Ψ(F ) : [a, b] → C by
b
Z
Ψ(F )(x) = λ
K(x, y)F (y) dy + G(x).
a
Explain why there is a constant M such that |K(x, y)| ≤ M for all x, y ∈ [a, b].
Given that Ψ(F ) ∈ X for all F ∈ X (this is most easily proved using results from
the measure theory course), show that Ψ : X → X is a contraction whenever
−1
|λ| < (b − a)M
. Show that there is a unique solution to the integral equation
Z
F (x) = λ
b
K(x, y)F (y) dy + G(x).
a
[Unique here means that there is only one continuous map that satisfies the above.]
16. Let X be a complete metric space, and (Fn )n≥1 a sequence of closed subsets of
X such that
T Fn ⊃ Fn+1 for each n ≥ 1, and such that diam(Fn ) → 0 as n → ∞.
Show that n≥1 Fn contains exactly one point.
[Hint: Pick an xn from each Fn , and use the condition on diameters to show that
the sequence (xn ) is Cauchy. Then show that the limit is the unique point in the
intersection. This is (one half of) Cantor’s Intersection Theorem.]
62
6
Hints to the Exercises
Metric spaces
R1
1. (b) Solution: If f (t) = 0 for all t then clearly 0 f (t) dt = 0, so suppose that f is
a nonzero function. Hence f (t0 ) > 0 for some point t0 ∈ (0, 1). By continuity of f
there is some δ > 0 such that (t0 − δ, t0 + δ) ⊂ [0, 1], and that
|f (t) − f (t0 )| < 21 |f (t0 )|
∀t ∈ (t0 − δ, t0 + δ)
Hence, for all such t we have f (t0 )−f (t) < 21 f (t0 ) which implies that f (t) > 21 f (t0 ),
and so
Z 1
Z t0 +δ
Z t0 −δ
Z 1
f (t) dt
f (t) dt +
f (t) dt +
f (t) dt =
t0 −δ
t0 +δ
0
0
Z
t0 +δ
Z
f (t) dt ≥
≥
t0 −δ
t0 −δ
t0 +δ
1
2 f (t0 ) dt
= δf (t0 ) > 0,
where the first inequality holds since f (t) ≥ 0, so that the first and third integrals
on the right hand side of the first line must be nonnegative.
So now if f, g ∈ C[0, 1], then the function t 7→ |f (t) − g(t)| is nonnegative and
continuous, and so
d1 (f, g) ≥ 0, with d1 (f, g) = 0 ⇔ |f (t) − g(t)| = 0 ∀t ⇔ f = g,
by the above. Since |f (t) − g(t)| = |g(t) − f (t)| for all t, d1 (f, g) = d1 (g, f ), and,
if h ∈ C[0, 1] then
Z 1
d1 (f, g) =
|f (t) − h(t) + h(t) − g(t)| dt
0
Z 1
|f (t) − h(t)| + |h(t) − g(t)| dt = d1 (f, h) + d1 (h, g).
≤
0
3. Hint: The tricky part is to show that d0 satisfies the triangle inequality. This will
a
b
c
follow if we can show that if a, b, c ∈ [0, ∞) satisfy a ≤ b + c, then 1+a
≤ 1+b
+ 1+c
.
This latter inequality is equivalent to
b(1 + a)(1 + c) + c(1 + a)(1 + b) − a(1 + b)(1 + c) ≥ 0,
which is easily shown to hold on multiplying out the terms.
5. Solution: First note that the two conditions together give
0 = d(x, x) ≤ d(x, z) + d(x, z) = 2d(x, z) ⇒ d(x, z) ≥ 0
∀x, z ∈ X.
Thus d(x, y) ≥ 0 for all x, y ∈ X, with equality if and only if x = y by M10 . Also,
by M20 ,
d(x, y) ≤ d(x, x) + d(y, x) = d(y, x) ∀x, y ∈ X.
Swapping the roles of x and y gives d(y, x) ≤ d(x, y), and hence d(x, y) = d(y, x)
for all x, y ∈ X. Finally, the triangle inequality follows easily from the given
inequality using the symmetry we have just proved.
63
7. Hint: The only one that requires the traditional ε-δ proof is the function f + g.
The function tf is a special case of the function f g, with g taken to be a constant
(hence continuous) function. Then note that |f | is the composition of f with
| · | : R → R, hence continuous. The function f 2 is similarly continuous, and the
remaining functions can now be written as
min{f, g} = 21 (f +g−|f −g|), max{f, g} = 21 (f +g−|f −g|), f g =
1
2
(f +g)2 −f 2 −g 2 .
and so can all be made up from functions that we know are continuous from the
above.
9. Hint: An easy solution to the first part is as follows: let U be an open subset of R
and set Λ = {(p, q) ∈ Q × Q : (p, q) ⊂ U }. Since Q × Q is countable,Sthe index set
Λ must also be countable. Furthermore it is easy to see that U = (p,q)∈Λ (p, q),
so that U is a union of a countable number of open intervals.
To make sure that the union is taken over a family of pairwise disjoint intervals
requires more work — see Theorem E on page 62 of Simmons.
14. Hint: Pick a positive integer n such that
some m ∈ Z such that m
n ∈ (x, x + ε).
1
n
< ε and explain why there must be
15. Hint: Build up a sequence (xn )n≥1 ⊂ A iteratively as follows: since x ∈ A we must
have B(x, ε) ∩ A 6= ∅ for any ε > 0, and so we can choose x1 ∈ B(x, ε) ∩ A. But
/
x∈
A, so x1 6= x, hence d(x, x1 ) > 0. Now we can choose some x2 ∈ B x, d(x, x1 ) ∩A,
and the three points x, x1 and x2 are all distinct. Continuing we get a sequence
that in A ∩ B(x, ε) that converges to x and all of whose terms are distinct from
one another.
17. Solution: Clearly d∞ (x1 , x2 ), (y1 , y2 ) ≤ d1 (x1 , x2 ), (y1 , y2 ) . The CauchySchwarz inequality ((†) from question 1) gives
√
d1 (x1 , x2 ), (y1 , y2 ) = 1 × |x1 − y1 | + 1 × |x2 − y2 | ≤ 2d2 (x1 , x2 ), (y1 , y2 ) .
Finally,
2
2 1/2
d2 (x1 , x2 ), (y1 , y2 ) ≤ d∞ (x1 , x2 ), (y1 , y2 ) + d∞ (x1 , x2 ), (y1 , y2 )
√
= 2d∞ (x1 , x2 ), (y1 , y2 ) .
Putting these together gives
√
d∞ (x1 , x2 ), (y1 , y2 ) ≤ d1 (x1 , x2 ), (y1 , y2 ) ≤ 2d2 (x1 , x2 ), (y1 , y2 )
≤ 2d∞ (x1 , x2 ), (y1 , y2 ) ,
from which the Lipschitz equivalences follow.
Consider the sequence ( n1 , 0) n≥1 in R2 . It is easy to see that ( n1 , 0) → (0, 0)
with respect to each of the metrics d1 , d2 and d∞ . For d1 , say, to be Lipschitz
equivalent to the discrete metric dd we would need constants 0 < a ≤ b such that
a d1 (x1 , x2 ), (y1 , y2 ) ≤ dd (x1 , x2 ), (y1 , y2 ) ≤ b d1 (x1 , x2 ), (y1 , y2 ) .
64
1
But then b d1 ( n1 , 0), (0, 0) = nb → 0, which
would
then
force
d
(
,
0),
(0,
0)
→ 0,
d
n
1
whereas in fact we have dd ( n , 0), (0, 0) = 1 for all n. Thus no such b can exist,
hence the metrics cannot be Lipschitz equivalent.
Topological spaces
2. Solution: Let A = [0, 1)∪[2, 3) and B = [1, 2). We shall show that A = [0, 1]∪[2, 3]
and B = [1, 2], from which we get
A∩B = {2},
A∩B = {1},
A∩B = {1, 2} and A ∩ B = ∅
(since A∩B = ∅).
Recall that for any subset C of a metric space X, C ⊂ C, and x ∈ C if and
only if B(x, ε) ∩ C 6= ∅ for all ε > 0. So to begin with A ⊂ A. Let x < 0 then
ε := −x > 0 and B(x, ε) = (2x, 0), hence B(x, ε)∩A = ∅, so that x ∈
/ A. Similarly,
if x > 3 then ε := x − 3 > 0 and B(x, ε) = (3, 2x − 3), so that B(x, ε) ∩ A = ∅,
and hence x ∈
/ A. Also, if x ∈ (1, 2) then ε := min{x − 1, 2 − x} > 0 and
B(x, ε) = (x − ε, x + ε) ⊂ (1, 2), so that x ∈
/ A. Finally, we must consider the
cases x = 1 and x = 3. Now B(1, ε) ∩ A = (ε, 1) for all 0 < ε < 1, and so
1 ∈ A. Similarly B(3, ε) ∩ A = (2 − ε, 3) for all 0 < ε < 1, and so 3 ∈ A. Hence
A = [0, 1] ∪ [2, 3] as claimed.
The proof that B = [1, 2] is essentially the same, and so omitted.
[Alternative solution: A = Q, B = R \ Q.]
T
4. Solution: Let T = λ∈Λ Tλ . Now ∅ ∈ Tλ and X ∈ Tλ for all λ ∈ Λ, hence
∅ ∈ T and
Uγ ∈ Tλ for all λ ∈ Λ,
S X ∈ T. Let {Uγ }γ∈Γ ⊂ T then, for each
S λ ∈ Λ, T
hence γ Uγ ∈ T, since Tλ is a topology, and so γ Uγ ∈ λ Tλ = T.
T Similarly,
let {V1 T
, . . . , Vn } ⊂TT. Then, for each λ ∈ Λ, V1 , . . . , Vn ∈ Tλ , and so ni=1 Vi ∈ T.
Hence ni=1 Vi ∈ λ Tλ = T as required, and thus T is a topology on X.
To see that a union of topologies need not be a topology, consider the following
example: let X = {1, 2, 3} and S = {∅, {1}, X}, T = {∅, {2}, X}. Since ∅ ⊂ {1} ⊂
X it follows that S is a topology on X, and similarly for T. However their union
is {∅, {1}, {2}, X}. We have {1}, {2} ∈ S ∪ T, but {1, 2} = {1} ∪ {2} ∈
/ S ∪ T, so
S ∪ T is not a topology.
7. Solution: For any A ⊂ Y and x ∈ X we have f (x) ∈ A if and only
if f (x) ∈
A ∩ f (X), since certainly x ∈ f (X), and so f −1 (A) = f −1 A ∩ f (X) .
Suppose f : X → Y is continuous, and let U ⊂ f (X) be an open subset of
f (X). Then U = V ∩ f (X) for some open subset V ⊂ Y of Y . But then f −1 (U ) =
f −1 V ∩ f (X) = f −1 (V ), which is open in X since V is open in Y and f : X → Y
is continuous.
If, on the other hand, f : X → f (X) is continuous, then for any open set V ⊂ Y
of Y we have f −1 (V ) = f −1 V ∩ f (X) , which is open in X since V ∩ f (X) is
open in f (X).
8. Solution: We have U ∈ T, and V ∈ TU = {U ∩ W : W ∈ T}. Thus there is some
W ∈ T such that V = U ∩ W . So V is the intersection of two open subsets of X,
hence V is an open subset of X.
65
To show that this can fail if U is not open, let X = R, U = (0, 1] and V = (0, 1].
Then U ∈ TU (since TU is a topology on U ), and so V = U ∈ TU . But V is not
open in R — there is no ε > 0 such that B(1, ε) = (1 − ε, 1 + ε) ⊂ V .
9. Solution: (i) Let x ∈ X2 . Note that any X2 -neighbourhood R of x can be
written as X2 ∩ S for an X1 -neighbourhood S of x, and conversely given any X1 neighbourhood S of x, R = X2 ∩ S is an X2 -neighbourhood of x. Moreover, if
R = X2 ∩ S then R ∩ A = (X2 ∩ S) ∩ A = S ∩ A, since A ⊂ X2 . So now
x ∈ A2 ⇔ x ∈ X2 and R ∩ A 6= ∅ ∀X2 -nhds. R of x
⇔ x ∈ X2 and S ∩ A 6= ∅ ∀X1 -nhds. S of x ⇔ x ∈ X2 ∩ A1
Thus A2 = X2 ∩ A1 .
(ii) By part (i) we have A2 ⊂ A1 . However, if X2 is closed in X1 then so is the
intersection X2 ∩ A1 = A2 . That is, A2 is closed in X1 , and contains A, hence
A1 ⊂ A2 . The two inclusions give A1 = A2 . [Further exercise: give an example
where A2 is a proper subset of A1 — this will require X2 to be a non-closed subset
of X1 .]
10. Solution: We know that a map g : Z → Y1 × Y2 is continuous if and only if
the coordinate maps q1 ◦ g : Z → Y1 and q2 ◦ g : Z → Y2 are continuous, where
qi (y1 , y2 ) = yi . Thus f is continuous if and only if the maps f i = qi ◦ f are
continuous, where
f 1 : X1 ×X2 → Y1 ,
f 2 : X1 ×X2 → Y2 ,
(x1 , x2 ) 7→ f1 (x1 ),
(x1 , x2 ) 7→ f2 (x2 ).
Suppose first that f1 and f2 are continuous, then f 1 = f1 ◦ p1 and f 2 = f2 ◦ p2
where pi are the projections pi (x1 , x2 ) = xi . By definition of the product topology
the pi are continuous, hence the compositions f 1 and f 2 are continuous, and thus
so is f .
Conversely, suppose that the map f is continuous, hence f 1 and f 2 are continuous.
Fix points a1 ∈ X1 and a2 ∈ X2 , then f1 = f 1 ◦ ι1 and f2 = f 2 ◦ ι2 where ι1 : X1 →
X1 ×X2 and ι2 : X2 → X1 ×X2 are given by ι1 (x1 ) = (x1 , a2 ) and ι2 (x2 ) = (a1 , x2 ).
But the component maps of ι1 are the identity map (p1 ◦ι1 )(x1 ) = x1 on X1 , and the
constant map (p2 ◦ι1 )(x1 ) = a2 from X1 to X2 , both of which are continuous, hence
ι1 is continuous. Similarly ι2 is continuous, and so each fi must be continuous.
12. Hint: This follows almost immediately as soon as you establish the identity
(X × Y ) \ (E × F ) = (X \ E) × Y ∪ X × (Y \ F ) .
13. (b) Hint: Suppose that (X, T) and (Y, S) are topological spaces, that f : X → Y
is a surjective map, and let Sf denote the quotient topology induced on Y by f .
If f is continuous with respect to the topologies T and S, then S ⊂ Sf . In our
example [0, 1] has a natural Hausdorff topology coming from R, so if we want a
map f : [−1, 1] → [0, 1] such that the quotient topology is not Hausdorff, it follows
that f must be discontinuous with respect to the usual topologies involved.
15. Hint: This requires similar pedantry as that from the solution of question 7.
66
16. Hint: For the first part consider the map ι1 from the answer to question 10, and
argue similarly.
19. Solution: (i) For each y ∈ X \{x} there are open sets Uy and Vy suchSthat y ∈ Uy ,
x ∈ Vy and Uy ∩Vy = ∅. In particular x ∈
/ Uy . Thus we have X\{x} = y∈X\{x} Uy ,
a union of open sets, hence open, and so {x} is closed.
T
(ii) Consider the set F = U ∈F U , where F = {U ⊂ X : U open, x ∈ U }. Now
X ∈ F, so F is nonempty, and x ∈ F since x ∈ U for each U ∈ F. Thus {x} ⊂ F .
Let y ∈ X with y 6= x. Then there are open sets U1 and U2 such that y ∈ U1 ,
x ∈ U2 and U1 ∩ U2 = ∅. So then U2 ∈ F, but y ∈
/ U2 , hence y ∈
/ F . Since y was
arbitrary we have F = {x} as required.
21. Hint: Consider question 17 for a suitable choice of X.
Compact spaces
2. Solution: Let T denote the topology on X and TY the subspace topology on Y .
Suppose that Z is compact as a subspace of YS, and let {Uλ }λ∈Λ ⊂ T be a collectionSof open subsetsSof X such that Z ⊂ λ∈Λ Uλ . Since Z ⊂ Y we have
Z⊂
λ∈Λ Uλ ∩ Y =
λ∈Λ (Uλ ∩ Y ), and so {Uλ ∩ Y }λ∈Λ ⊂ TY is a cover of Z
made up of open subsets of YS. Thus, since Z is compact in
S Y , there is a finite
subset Γ ⊂ Λ such that Z ⊂ γ∈Γ (Uγ ∩ Y ). But then Z ⊂ γ∈Γ Uγ , and so Z is
compact as a subspace of X.
Conversely, suppose that Z is compact as a subspace of S
X, and let {Vλ }λ∈Λ ⊂ TY
be a collection of open subsets of Y such that Z ⊂ λ∈Λ Vλ . Then for each
λ ∈ ΛS there is some Uλ ∈ T such that Vλ = Uλ ∩ Y ⊂ Uλ , and it follows that
Z ⊂ λ∈Λ Uλ . Thus {Uλ }λ∈Λ is an open cover for Z from T, hence there must
be a finite
{UλS
1 , . . . , Uλn }. But then, once again since Z ⊂ Y , we have
Sn subcover
n
U
∩
Y
=
Z⊂
λ
k
k=1 Vλk , and so {Vλ1 , . . . , Vλn } ⊂ TY is a finite subcover
k=1
of the cover {Vλ }λ∈Λ from TY . Thus Z is compact as a subspace of Y .
5. Hint: A subset of R or R2 is compact if and only if it is closed and bounded (the
Heine-Borel Theorem), so (iv) and (v) are compact, but the other four are not
compact.
6. Solution: Let {Fn1 , . . . , Fnk } be any finite subcollection taken from the sets making up the sequence (Fn ), and suppose T
without loss of generality that n1 < · · · <
nk . Then, since the sequence is nested, ki=1 Fni = Fnk 6= ∅. That T
is the collection
of closed sets has the FIP, and since X is compact we must have n≥1 Fn 6= ∅.
9. Solution: Let TZ denote the Zariski topology on R, that is
TZ = {∅} ∪ {U ⊂ R : R \ U is finite}.
Let {Uλ }λ∈Λ ⊂ TZ be any open cover of R. At least one set in the cover must
be nonempty, say Uλ0 . Hence R \ Uλ0 is finite and can be written as {x1 , . . . , xn }.
Then, since {Uλ }λ∈Λ is a cover of R, there
S are λ1 , . . . , λn ∈ Λ such that xj ∈ Uλj
for all 1 ≤ j ≤ n. Hence {x1 , . . . , xn } ⊂ nj=1 Uλj , and
S
R = R \ {x1 , . . . , xn } ∪ {x1 , . . . , xn } ⊂ Uλ0 ∪ nj=1 Uλj ⊂ R.
67
Thus {Uλ0 , . . . , Uλn } is a finite subcover and so R is compact in this topology.
10. Hint: The map g is the composition of the continuous maps
X 3 x 7→ (f (x), x) ∈ X × X and X × X 3 (x1 , x2 ) 7→ d(x1 , x2 ) ∈ R,
and so is continuous. Thus g(X) is a compact and hence closed subset of R, and
0∈
/ g(X) since this would imply that there is some x ∈ X such that x = f (x).
Thus 0 does not lie in the closed set g(X), which leads to the existence of the
required ε.
11. Solution: Since X is compact, so is F1 = f (X), and hence so is each Fn by
induction. In particular each Fn is closed since X is Hausdorff. Moreover, let
x ∈ F2 , then x = f (y) for some y ∈ F1 . But y ∈ X as well, hence x ∈ F1 , and so
F2 ⊂ F1 . Applying f to both sides
T gives F3 = f (F2 ) ⊂ f (F1 ) = F2 ; iterating gives
Fn+1 ⊂ Fn for all n. Thus F := n≥1 Fn 6= ∅ by question 6.
T
T
T
T
T
Now f (F ) = f n≥1 Fn ⊂ n≥1 f (Fn ) = n≥1 Fn+1 = m≥2 Fm = m≥1 Fm = F ,
where we have used the fact that the sequence is nested to get the penultimate equality. Finally pick an x ∈ F . We must show that x ∈ f (F ). But
x ∈ Fn+1 for each n ≥ 1, so for each such n we can pick some xn ∈ Fn such
that x = f (xn ). Since X is compact it is also sequentially compact, and so there
is some subsequence (xnk )k≥1 such that xnk → x0 for some x0 ∈ X. In fact,
since Fn ⊃ Fn+1 for each n, it follows that {xnk : k ≥ K} ⊂ FnK for each
K ≥ 1, and so x0 = limk→∞,k≥K xnk ∈ FnK , from which we get x0 ∈ F , and also
x = f (xnk ) → f (x0 ), so that x ∈ f (F ) as required.
Connected spaces
15. Solution: Suppose that X is finite, then we can write it as X = {x1 , . . . , xn } where
n ≥ 2. Since the singleton {x1 } is closed it follows that {x2 , . . . , xn } = {x1 }c is
open. Moreover, {x2 , . . . , xn } = {x2 } ∪ · · · ∪ {xn } is a finite union of closed sets,
hence closed, and so {x1 } = {x2 , . . . , xn }c is also open. Hence ({x1 }, {x2 , . . . , xn })
is a disconnection for X, which contradicts our assumption that X be connected.
Thus X must contain infinitely many points after all.
17. Hint: (a) f (X) is a connected subspace of R, hence it is an interval, and since f
is not constant this interval must contain an uncountable number of points.
(b) Fix a point x0 ∈ X and consider the map X 3 x 7→ d(x, x0 ) ∈ R.
18. Hint: Suppose f : R2 → R is continuous and injective. Since it is not constant,
f (R2 ) is an interval containing more than one point. Denote this interval by I,
and choose a ∈ I such that I ∩ (−∞, a) and I ∩ (a, ∞) are both nonempty. Let
x ∈ R2 be the unique point in R2 such that f (x) = a, and consider the restriction
of f to R2 \ {x}. The identity f (R2 \ {x}) = I \ {a} now gives a contradiction.
21. Solution: Consider the subsets Cn = {(x, y) ∈ R2 : y ≤ 0 or y ≥ ex − n1 }. We
have Cn = A ∪ Bn where
A = {(x, y) ∈ R2 : y ≤ 0} and Bn = {(x, y) ∈ R2 : y ≥ ex − n1 }
68
2
Now A = F −1 (−∞, 0] and Bn = G−1
n (−∞, 0] , where F, Gn : R → R are the
functions F (x, y) = y and Gn (x, y) = ex − n1 − y. Since F and Gn are clearly
continuous, the preimages A and Bn of the closed set (−∞, 0] ⊂ R are closed, as
is their union Cn .
Note also that both sets are convex (given any two points in A (or Bn ), the line
segment joining them lies entirely inside A (resp. Bn )), and so they are certainly
path-connected. Moreover the point (− log n − 1, 0) lies in A ∩ Bn , and it follows
that the union Cn = A ∪ Bn is path-connected, hence a connected subset of R2 .
SinceTBn ⊃ Bn+1 for all n, we have Cn = A ∪ Bn ⊃ A ∪ Bn+1 = Cn+1 . Let
C = n≥1 Cn . Now if (x, y) ∈ R2 with y ≤ 0 then (x, y) ∈ Cn for all n, and so
(x, y) ∈ C. If, on the other hand, (x, y) ∈ R2 with y > 0, then (x, y) ∈ C if and
only if y ≥ ex − n1 for all n, that is, if and only if y ≥ ex . Hence we have
C = {(x, y) ∈ R2 : y ≤ 0} ∪ {(x, y) ∈ R2 : y ≥ ex } = A ∪ B
where B = {(x, y) ∈ R2 : y ≥ ex }. But now U = {(x, y) ∈ R2 : y < 13 ex } and
V = {(x, y) ∈ R2 : y > 23 ex } are open disjoint subsets of R2 , and moreover A ⊂ U
and B ⊂ V . Thus A = A ∩ U and B = B ∩ V are open subsets of C that are
clearly nonempty and disjoint. Hence (A, B) is a disconnection for C, and so C is
disconnected as required.
22. Hint:
We
need
a
continuous
map
f
such
that
f
[0,
1]
∩
Q
⊂ R \ Q and f [0, 1] \
Q ⊂ Q. Clearly f cannot be constant, and so f [0, 1] is an interval containing
more than one point. How many irrational numbers lie in this interval?
Uniformity and completeness
1. Hint: (i) Converges uniformly to the constant function x 7→ 0
(ii) There is a sequence (xn )n≥1 ⊂ [0, 1] such that fn (xn ) → ∞.
(
0 if x ∈ [0, 1),
(iii) The sequence converges pointwise to f (x) = 1
if x = 1.
2
3. Solution: (i) Since fn → f uniformly there is some N ≥ 1 such that
d fn (x), f (x) < 1 ∀x ∈ X whenever n ≥ N.
(∗)
But fN (X) is a bounded subset of Y , and so there must be some constant KN
such that d fN (x), fN (x0 ) < KN for all x, x0 ∈ X. Hence
d f (x), f (x0 ) ≤ d f (x), fN (x) + d fN (x), fN (x0 ) + d fN (x0 ), f (x0 ) < KN + 2
(†)
for all x, x0 ∈ X, and thus f is also a bounded function (i.e. f (X) is bounded in
Y ).
(ii) For each n ≥ N we have by (∗) and (†) that for any choice of x, x0 ∈ X
d fn (x), fn (x0 ) ≤ d fn (x), f (x) + d f (x), f (x0 ) + d f (x0 ), fn (x0 ) < KN + 4.
69
Also, f1 , . . . , fN −1are all bounded, so there are constants K1 , . . . , KN −1 that satisfy d fi (x), fi (x0 ) < Ki for each 1 ≤ i ≤ N − 1 and all x, x0 ∈ X. Thus
d fn (x), fn (x0 ) < K ∀n ≥ 1 and x, x0 ∈ X
where K = max{K1 , . . . , KN −1 , KN + 4}.
4. Hint: One possible example is the following sequence: fn (x) =
(
0
1
n
if x ≤ 0
if x > 0
6. Solution: A function g : X → Y between metric spaces is not uniformly continuous if there is some ε > 0 such that for every choice of δ > 0 we can find a pair
of points x1 , x2 ∈ X such that d(x1 , x2 ) < δ and d g(x1 ), g(x2 ) ≥ ε.
So now consider f : (0, 1) → R given by f (x) = x1 , where we are using the usual
metric on (0, 1) and R. Let ε = 1 and choose any δ > 0. Pick an n > 1 such that
1
1
1
n < δ, and set x1 = n , x2 = n+1 . Then
1
1 1
1
1
< < δ,
d(x1 , x2 ) = −
= −
n n+1
n n+1
n
but
d f (x1 ), f (x2 ) = |n − (n + 1)| = 1,
and so f is not uniformly continuous.
10. Hint: Recall that a subspace of a complete metric space is itself complete if and
only if it is closed. Thus (i) and (iii) are complete but (ii) is not complete.
13. Solution: We have l∞ = {(xn )n≥1 ⊂ R : supn |xn | < ∞} and d(x, y) = supn |xn −
yn |. First note that if x = (xn ) ∈ l∞ and y = (yn ) ∈ l∞ then (xn − yn )n≥1 is also
bounded, and so d(x, y) is well-defined. Moreover clearly d(x, y) = d(y, x) ≥ 0,
and if d(x, y) = 0 then |xn − yn | = 0 for all n, so that x = y. Finally, let
x, y, z ∈ l∞ , then for each m ≥ 1
|xm − ym | ≤ |xm − zm | + |zm − ym | ≤ supn |xn − zn | + supn |zn − yn |
and so taking the supremum over all m of the left hand side gives d(x, y) ≤
d(x, z) + d(z, y) as required. Thus d really is a metric on l∞ .
Now let (xk )k≥1 be a Cauchy sequence in l∞ . So each xk = (xkn )n≥1 is a bounded
sequence in R, and for any given ε > 0 there is some K ≥ 1 such that d(xk , xj ) < ε
for all k, j ≥ K. Thus for each m ≥ 1 we have
|xkm − xjm | ≤ supn |xkn − xjn | = d(xk , xj ) < ε
(†)
whenever k, j ≥ K. That is, (xkm )k≥1 is a Cauchy sequence in R for each m. But
R is a complete space, and so there is some xm ∈ R such that xkm → xm as k → ∞.
We must now show that the sequence x := (xm )m≥1 lies in l∞ , and that xk → x
with respect to the metric d. Going back to (†) we see that |xkm − xK
m | < ε for all
K , and so by
k ≥ K and m ≥ 1. But since xkm → xm , we have xkm − xK
→
x
−
x
m
m
m
the continuity of the modulus function we must have |xm − xK
m | ≤ ε for all m ≥ 1.
70
Moreover xK ∈ l∞ , and so there is a constant M > 0 such that |xK
m | ≤ M for all
m ≥ 1. Hence
K
|xm | ≤ |xm − xK
m | + |xm | ≤ ε + M
for all m ≥ 1, and so x ∈ l∞ . Again from (†) we have, letting k → ∞, |xm −xjm | ≤ ε
for all j ≥ K and m ≥ 1. Taking the supremum over m of the left hand side gives
d(x, xj ) ≤ ε for all j ≥ K, and since ε was arbitrary we get that xj → x as
required.
Note that (2, 0, 0, 0, . . .) ∈ l∞ \ X, and so X is a proper subset of X. To show that
it is closed, let (xk )k≥1 ⊂ X be a sequence such that xk → x ∈ X with repsect to
d∞ . Since each xk ∈ X we have |xkm | ≤ supn |xkn | ≤ 1 for all k, m ≥ 1. But
0 ≤ |xkm − xm | ≤ supn |xkn − xn | = d∞ (xk , x) → 0
as k → ∞, from which we get xkm → xm as k → ∞. Hence |xm | = limk |xkm | ≤ 1,
since it is a limit of numbers lying in [0, 1]. Thus supm |xm | ≤ 1, hence the limit
x ∈ X, so X is a closed subspace of the complete space l∞ , and so must be
complete.
If x, y ∈ X, then supn |xn | ≤ 1 and supn |yn | ≤ 1, and so d(x, y) = supn |xn −yn | ≤
supn |xn | + supn |yn | ≤ 2. That is, X is a bounded subset of l∞ .
Finally, consider the sequence (em )m≥1 ⊂ X where em = (0, 0, . . . , 0, 1, 0, . . .),
with the only nonzero entry being 1 which occurs in the mth position. Then
d(em , en ) = 1 whenever m 6= n, from which it follows that no subsequence can be
a Cauchy sequence, and so no subsequence can converge.
71