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Chapter 1
Metric spaces
Definition 1. A metric on X is a function d : X × X → R satisfying the
following conditions for all x, y, z ∈ X:
1. d(x, y) ≥ 0 and = 0 if and only if x = y.
2. d(x, y) = d(y, x).
3. d(x, y) ≤ d(x, z) + d(z, y).
A metric space is a pair (X, d) where X is a set and d is a metric on X.
The Euclidean metric on Rn refers to the function
v
u n
uX
d(x, y) = t (xi − yi )2 .
i=1
The fact that this is a metric is a consequence of the Cauchy-Schwarz inequality.
1.1
Limit points
Definition 2. A sequence (xn )n∈Z+ in a metric space (X, d) is said to converge
if there is a point x in X such that d(xn , x) → 0 as n → ∞. In other words,
for every ε > 0 there is an N such that n > N implies d(xn , x) < ε. When this
happens, we refer to x as the limit of the sequence.
Definition 3. Let E be a subset of a metric space (X, d). An isolated point of
E is a point x in E for which there is a δ > 0 such that for all y in E different
from x, d(x, y) ≥ δ.
Definition 4. Let E be a subset of a metric space (X, d). A limit point of E
is a point x in X that is the limit of a sequence in E but not an isolated point
of E. The set of limit points of E will be denoted by E 0 .
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CHAPTER 1. METRIC SPACES
1.2
Metric balls
Definition 5. The open ball centered at x of radius r, denoted by B(x, r), is
defined as the set of points y in X whose distance from x is less than r.
Note that a sequence converges to x if and only if every open ball about x
contains a tail of the sequence; also, x is an isolated point of E if and only if x
belongs to E and there is an open ball about x that contains no other points of
E. The following lemma characterizes limit points in terms of open balls.
Lemma 1. In a metric space, x is a limit point of E if and only if every open
ball about x intersects E in a point different from x.
Proof. (⇐) For each n ∈ Z+ , the set B(x, n1 ) ∩ E is nonempty, so it contains
some point xn . Since d(xn , x) < n1 , the sequence (xn ) converges to x. It remains
to check that x is not an isolated point of E, but that is essentially given by the
hypothesis.
(⇒) Suppose x is a limit point of E and let B be an open ball about x.
If x ∈ E, then B must intersect E in a point different than x, for otherwise,
x would be an isolated point of E. If x 6∈ E, then there is a sequence in E
converging to x and since B contains a tail of this sequence, it intersects E in
a point different from x. In either case, we have shown that B intersects E in
a point different from x.
Lemma 2. For any y in B(x, r) there is a δ > 0 such that B(y, δ) ⊂ B(x, r).
Proof. Take δ = r − d(x, y).
1.3
Closed sets
Definition 6. In a metric space, a set E is closed if it contains all its limit
points, i.e. E 0 ⊂ E. Its closure is defined by Ē = E ∪ E 0 .
Theorem 1. Let E be a subset of a metric space (X, d).
1. Ē = E if and only if E is closed.
2. Ē is closed.
Proof. The first statement is straightforward, so we only prove 2. Let x be a
limit point of Ē. Lemma 1 implies every open ball about x intersects Ē in
a point different from x. Hence, it suffices to show that any open ball that
intersects E 0 must also intersect E. Given an open ball B that intersects E 0 ,
choose any y ∈ B ∩ E 0 . By Lemma 2, there is an open ball B 0 contained in B
that contains y. Since y is a limit point of E, the open ball B 0 intersects E.
And since B contains B 0 , it intersects E as well.
Theorem 2. Let E be a subset of a metric space. Any closed set that contains
E must also contain Ē.
1.4. OPEN SETS
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Proof. Let F be a closed set containing E and suppose x is a limit point of E.
Then every open ball about x intersects F in a point different from x. Hence
x is a limit point of F and therefore contained in F , since F is closed. Since
x ∈ E 0 was arbitrary, it follows that Ē ⊂ F .
Theorem 2 says Ē is the smallest closed set containing E.
The next lemma should be compared with Lemma 1.
Lemma 3. In a metric space, x ∈ Ē if and only if every open ball about x
intersects E.
Proof. (⇒) If x ∈ E we are done, while if x ∈ E 0 , we apply Lemma 1.
(⇐) If x ∈ E we are done, while if x 6∈ E, we apply Lemma 1.
1.4
Open sets
Definition 7. Let U be a subset of a metric space. If U contains a ball about
x, we say x is an interior point of U . A set is open if all of its points are
interior.
Note that by Lemma 2, every open ball is an open set.
Theorem 3. In a metric space the following hold.
1. An arbitrary union of open sets is open.
2. A finite intersection of open sets is open.
S
Proof. 1. Let {Uα }α∈J be a collection of open sets and set U = α∈J Uα . Given
x ∈ U , there is some α such that x ∈ Uα . Since Uα is open, it contains some
open ball B about x. Since U contains Uα , it also contains B. Hence x is an
interior point of U and since x ∈ U was arbitrary, this shows U is open.
2. It suffices to show the intersection of two open sets is open. Given
x ∈ U1 ∩ U2 , let δi > 0 be such that B(x, δi ) ⊂ Ui . If δ = min(δ1 , δ2 ) we have
B(x, δ) ⊂ U1 ∩ U2 . Thus, x is an interior point of U1 ∩ U2 and since x ∈ U1 ∩ U2
was arbitrary, this shows U1 ∩ U2 is open.
Theorem 4. A subset of a metric space is closed if and only if its complement
is open.
Proof. (⇒) Let E be closed and x 6∈ E. Then x is not a limit point of E.
Since x 6∈ E, Lemma 1 implies there is an open ball B about x disjoint from E.
Thus x is an interior point of E c , the complement of E. And since x ∈ E c was
arbitrary, this shows E c is open.
(⇐) Suppose E c is open and let x be a limit point of E. We need to show
x ∈ E. We do this by contradiction. Suppose x 6∈ E. Then x is an interior
point of E c . Hence, there is an open ball about x disjoint from E. But then x
cannot be a limit point of E, by Lemma 2.
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CHAPTER 1. METRIC SPACES
Chapter 2
Topological spaces
Definition 8. A topology on a set X is a collection T of subsets of X satisfying
1. ∅ and X are elements in T .
2. An arbitrary union of elements in T is an element in T .
3. A finite intersection of elements in T is an element in T .
A topological space is a pair (X, T ) where X is a set and T a topology on
X. An open subset of X refers to an element of T . A nonempty open set is
called a neighborhood and a ”neighborhood of x” simply refers to an open set
containing x.
The trivial topology on X refers to the topology {∅, X} consisting of
exactly two elements, while the discrete topology on X refers to the set of all
subsets of X, a.k.a. the power set of X. By Theorem 3, the collection of open
subsets of a metric space (X, d) is a topology on X, called the metric topology
induced by d. The standard topology on Rn is the metric topology induced
by the Euclidean metric. A metrizable space refers to a topological space
whose topology can be induced from a metric. Euclidean spaces are metrizable,
by definition.
Definition 9. In a topological space a closed set is one whose complement is
open. The closure of a set is the intersection of all closed sets containing it.
By De Morgan’s Laws, an arbitrary intersection of closed sets is closed and
a finite union of closed sets is closed. In particular, Ē is closed because it
is an intersection of closed sets, by definition. Note that the statements in
Theorems 1-4 hold in setting of topological spaces because they are essentially
immediate consequences of the definitions.
Definition 10. In a topological space, an isolated point of a set E is a point
in E that has a neighborhood containing no other points of E.
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CHAPTER 2. TOPOLOGICAL SPACES
Definition 11. In a topological space, a limit point of a set E is a point x in
X such that every neighborhood of x intersects E in a point different from x.
Definition 12. In a topological space, a point x is an interior point of a set
E if E contains some neighborhood of x.
These notions reduce to the ones given earlier in the context of metric spaces.
The next two lemmas verify that the same is true for open and closed sets.
Lemma 4. A set is open if and only if all its points are interior points.
Proof. (⇒) If a set is open, then it is a neighborhood of all its points so that
all its points are interior points. (⇐) Conversely, if all points of a set E are
interior, then for each x ∈ E there is a neighborhood Ux of x such that Ux ⊂ E.
It follows that E is the union of the open sets Ux as x ranges over the elements
of E. Hence, E is open.
Lemma 5. A set E is closed if and only if it contains all its limit points.
Proof. (⇒) Suppose E is closed and let x be a limit point of E. To show that
x belongs to E, let F be an arbitrary closed set containing E. If x is not in
E, then the complement of F is a neighborhood of x that is disjoint from E,
contradicting the fact that x is a limit point of E. Hence, x belongs to E. Since
x was an arbitrary limit point of E, this shows that E contains all its limit
points.
(⇐) Let x be a point in the complement of E. Then x is not a limit point
of E, so there is a neighborhood of x that is disjoint from E. Thus, x is an
interior point of the complement of E. It follows that the complement of E is
open. Hence, E is closed.
2.1
Hausdorff and T1 Axiom
Definition 13. In a topological space, a sequence (xn )n∈Z+ is said to converge
to a point x if every neighborhood of x contains a tail of the sequence; that is,
for every neighborhood U of x there is an N such that n > N implies xn ∈ U .
In this case, we call x a limit of the sequence (xn ).
Unlike in a metric space, the limit of a sequence need not be unique. Figure FORK illustrates how a convergent sequence can fail to have a unique limit.
The space, which we call the fork, is obtained by taking two copies of the real
line and ”glueing them together” along their negative real axes.1 Note that
the two origins are not identified, and each is a limit of the sequence ( −1
n )n∈Z+
approaching along the identified negative axes.
Definition 14. A topological space is Hausdorff if for any two points x and y
there exist disjoint neighborhoods about each of them. In other words, for some
neighborhood U of x and some neighborhood V of y we have U ∩ V = ∅.
1 The quotient topology is a general construction that formalizes the idea of glueing spaces
along various subsets that we shall discuss later.
2.2. FIRST COUNTABILITY
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It is not hard to see that metric spaces are Hausdorff. Indeed, if two points
are a distance δ > 0 apart, the balls of radius 2δ about each of them will be disjoint. The next lemma shows the Hausdorff property is responsible for uniqueness of the limits of convergent sequences.
Lemma 6. Convergent sequences in a Hausdorff space have unique limits.
Proof. Suppose x and y are limits of a convergent sequence (xn ). Suppose on
the contrary that x 6= y. Then we may choose disjoint neighborhoods U and V
of x and y, respectively. But then U and V both contain a tail of the sequence,
hence they cannot be disjoint. This contradiction establishes that x = y.
Observe that the fork is not Hausdorff, as the required condition fails for the
two origins. (Open sets in the glued space come from open sets in original space
before glueing.) In fact, these are the only two points for which the Hausdorff
condition fails. Note that since metric spaces are Hausdorff, the fork gives an
example of a space that is not metrizable.
The following is a slightly weaker condition that the fork does satisfy.
Definition 15. A topological space is said to satisfy the T1 Axiom if for any
pair of points x and y there is a neighborhood U about x that does not contain
y and a neighborhood V of y that does not contain x.
Lemma 7. A topological space is T1 if and only if finite sets are closed.
Proof. (⇐) Given a pair x and y, let U and V be the complement of y and x,
respectively. These are neighborhoods since singletons are closed.
(⇒) It suffices to prove that the T1 Axiom implies singletons are closed.
Given a singleton {x}, consider y in its complement, i.e. y 6= x. Then there is a
neighborhood V about y that does not contain x. Thus, y is an interior point of
the complement of {x}. Since y 6= x was arbitrary, this shows the complement
of {x} is open, hence the singleton {x} is closed.
2.2
First countability
In a metric space, a limit point of a set E is the limit of a sequence in E, by
definition. In a general topological space, this might not be true. The main
reason for why it holds in a metric space is the existence of the collection of
open balls B(x, n1 ), which was used in the proof of Lemma 1. The essential
properties of this collection are captured in the following.
Definition 16. A collection B of neighborhoods at x is a basis at x if every
neighborhood of x contains some element of B. A topological space is said to be
first countable if it has a countable basis at each of its points.
It is not hard to see that metric spaces are first countable. Indeed, every
neighborhood of a point x contains an open ball B(x, n1 ) for some sufficiently
large integer n by virtue of the fact that x is an interior point of the neighborhood. The converse does not hold, as we shall see in the next chapter.
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CHAPTER 2. TOPOLOGICAL SPACES
Lemma 8. If x is a limit point of E in a first countable space, then there is a
sequence in E that converges to x.
Proof. TLet {Un } be a countable basis at x. Note that the collection {Vn } where
n
Vn = i=1 Ui is a countable basis at x too. Hence, we may assume {Un } is
nested, i.e. for all n, Un+1 ⊂ Un .
Since x is a limit point, for each n, there is a point xn in Un ∩ E different
from x. It only remains to show that the sequence (xn ) converges to x. Indeed,
given a neighborhood U of x, there is an N such that UN ⊂ U . Moreover, for
any n > N , xn ∈ Un ⊂ UN ⊂ U . Hence, U contains a tail the of sequence.
Remark. The notion of a net allows for ”sequences” that aren’t necessarily
countable and such that if x is a limit point x of E then there is a net in E that
”converges” to x.
Chapter 3
Order topology
Definition 17. A total order on a set X is a relation ≺ on X satisfying the
following properties1
1. a ≺ a never holds.
2. If a 6= b then either a ≺ b or b ≺ a.
3. If a ≺ b and b ≺ c then a ≺ c.
A totally ordered set is a pair (X, ≺) where X is a set and ≺ is a total order
on X.
Definition 18. An open interval in a totally ordered set (X, ≺) refers to a
nonempty subset of the form {x ∈ X : a ≺ x ≺ b} where a, b ∈ X with a ≺ b.
An open ray is a nonempty subset either of the form {x ∈ X : x ≺ a} or of
the form {x ∈ X : a ≺ x}. Closed intervals and closed rays are similarly
defined with ≺ replaced by , where a b means a ≺ b or a = b.
Given a collection B of subsets of a set X one might ask for the existence of
a topology on X in which each element of B is open. Such a topology always
exists, since the discrete topology is one such. Now it is easy to see that if T1
and T2 are topologies on X, then so is T1 ∩ T2 ; and this holds more generally for
arbitrary intersections of topologies. The smallest topology containing B refers
to the intersection of all topologies on X that contain B.
Definition 19. The order topology on a totally ordered set X is the smallest
topology on X that contains all open intervals and all open rays.
3.1
Well-ordered sets
Definition 20. A well-ordered set is a totally ordered set that has the wellordering property: every nonempty subset has a smallest element, i.e. an element x such that for all y 6= x, x ≺ y.
1A
relation on X is a subset R of X × X. The notation ”xRy” means (x, y) ∈ R.
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CHAPTER 3. ORDER TOPOLOGY
The order topology on ({1, . . . , n}, <) coincides with the discrete topology,
but that on (Z+ , <) does not. (We shall see later that every infinite open set
contains an open ray.)
The minimal uncountable well-ordered set SΩ satisfies:
1. For each α ∈ §Ω , the open ray Sα = {β : β ≺ α} is countable.
2. Every countable subset Z of SΩ is bounded above, i.e. for some α ∈ §Ω ,
β ≺ α for all β ∈ Z.
The well-ordered set S̄Ω = SΩ ∪ {Ω} is obtained from SΩ by ”giving” it a
largest element Ω.
Lemma 9. Ω is a limit point of SΩ , but no sequence in SΩ converges to Ω.
Proof. Note that every neighborhood of Ω contains an open ray and that every
open ray intersects SΩ . Hence, every neighborhood of Ω intersects SΩ , proving
the first part of the lemma. Now let (αn )n∈Z+ be any sequence in SΩ . Being
countable, its image is bounded above by some α. The open ray {β ∈ S̄Ω : α ≺
β} is a neighborhood of Ω that does not contain any elements of the sequence.
Hence, Ω is not the limit of the sequence. Since the sequence in SΩ was arbitrary,
this proves the second part of the lemma.
By similar reasoning, one can show that S̄Ω does not have a countable basis
at Ω. Hence, it is not first countable. Since every metric space is first countable,
S̄Ω is not metrizable. It is not hard to see from this that SΩ is not metrizable.
Also, one can show (using the first property mentioned above) that SΩ is first
countable. Thus, SΩ provides an example of a topological space that is first
countable but not metrizable.
3.2
Bases for a topology
We have now seen two ways of constructing a topology by starting with a collection B of subsets of X then enlarging the collection to form a topology T
on X. In the setting of metric spaces, B was the set of open balls and T the
collection of sets with the property that all of their points are interior points.
In the context of the order topology, we took the smallest topology containing
all open intervals and open rays. In this section, we shall see how both of these
are part of the same general construction.
Definition 21. A collection B of subsets of X is a basis for a topology if
1. Every point of X is contained in some element of B.
2. If B1 and B2 are elements of B and x ∈ B1 ∩B2 then there exists a B3 ∈ B
such that x ∈ B3 and B3 ⊂ B1 ∩ B2 .
The topology generated by the basis B refers to the smallest topology containing B, i.e. the intersection of all topologies on X that contain B.
3.3. SECOND COUNTABILITY
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Theorem 5. Let T be the topology generated by a basis B on X. Then the
following are equivalent.
1. U ∈ T
2. U can be written as a union of elements from B.
3. For each x ∈ U there is a B ∈ B such that x ∈ B and B ⊂ U .
Proof. It is obvious that 3. implies 2. Since the collection of open subsets of a
topology is closed under arbitrary unions, any topology that contains B must
also contain arbitrary unions of elements of B. Hence, 2. implies 1. To see that
1. implies 3. it suffices to verify that the collection T3 of subsets of X that
satisfy 3. is a topology on X. The first two conditions in the definition of a
topology are straightforward, while the third condition follows easily form the
second condition in the definition of a basis.
Remark. The reader may have noticed that the first condition in the definition
of a basis was not used In the proof of Theorem 5. It is included for otherwise
there would be points whose only neighborhood is the entire space.
3.3
Second countability
Existence of a countable basis for standard topology on R.
Existence of a countable dense subset is sufficient in a metric space.
Rn has a countable basis.
Lower limit topology on R
3.4
Definition of a manifold
Dictionary order, non commutativity
Long line SΩ × (0, 1] does not have a countable basis.