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CHAPTER 1
Topological spaces
1. Definition of a topological space
Definition 1.1. A topology T on a set X is a collection of subsets of X subject to
the following three rules, called the axioms of a topology:
1. The empty set ∅ and all of X belong to T .
2. If Ui , i ∈ I is a collection of elements from T (indexed by the set I which may
well be infinite), then ∪i∈I Ui must belong to T as well. Said differently, T is
closed under taking arbitrary unions.
3. For elements U1 , ..., Un ∈ T , the set U1 ∩ ... ∩ Un must also belong to T . Thus
we demand that T be closed under taking finite intersections.
A topological space is a pair (X, T ) consisting of a set X and a topology T on X. A
subset U of X is called an open set if U ∈ T and a subset V ⊂ X is called closed if
X − V is open. A neighborhood of a point x ∈ X is any open set U ⊂ X that contains
x.
This definition is central to the remainder of the book and so, before moving on
to consider examples, we first pause to elucidate its various aspects. The choice of
the three axioms of a topology should not be too surprising given the results from
chapter ??. Specifically, they are modeled on the three properties of open subsets of
Euclidean space proved in theorem ??. Keeping in mind that open subsets of Rn were
used to recast the definition of continuity (see theorem ??), the attentive reader will
have little difficulty in guessing what the definition of a continuous function between
two topological spaces should be (for an answer, see definition ??).
Given a topological space (X, T ) we shall often simply write X when the topology
T is understood from context, and refer to X as a topological space. On the other
hand, when several topological spaces are involved in a discussion, we may label the
topology T by TX to indicate that it belongs with X. For example, we shall write
(X, TX ) or (Y, TY ) to label topological spaces.
The notion of open and closed subset of a topological space X shall be crucial
to all subsequent chapters. Whether or not a given subset A ⊂ X is open or not,
depends on the choice of a topology T on X. As we shall see in the examples below,
a set X almost always admits many different topologies and a subset A ⊂ X may
well be open with respect to some but not with respect to other topologies. The most
common misconception about open and closed sets among novices, is the notion that
these properties form a dichotomy:
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2
1. TOPOLOGICAL SPACES
Remark 1.2. Let X be a topological space and let A ⊂ X be any of its subsets.
The failure of A to be open does typically not imply that A is closed, and vice versa,
if A is not closed, typically this does not mean that A is open. In fact, there are many
examples of topological spaces X that have subsets A that are neither open nor closed
as well as subsets B that are both open and closed. Certainly B = ∅ and B = X are
examples of the latter.
Definition 1.3. Given a set X and two topologies T1 and T2 on X, we say that T1
is finer than T2 or, equivalently, that T2 is coarser than T1 , if T2 ⊂ T1 . We shall write
T2 ≤ T1 to denote this relation between the two topologies. As usual, we shall write
T2 < T1 to indicate that T2 ≤ T1 and T2 6= T1 .
The relation “≤”gives the set of all topologies on X a partial ordering in that
T1 ≤ T2 and T2 ≤ T1 implies that T1 = T2 . Likewise, T1 ≤ T2 and T2 ≤ T3 implies
T1 ≤ T3 . However, given two topologies T1 and T2 on X, neither of T1 ≤ T2 or T2 ≤ T1
has to hold.
Before exploring properties of open and closed subsets of a topological space X,
we turn to examine several examples of topological spaces. We recommend that the
reader not skip this next section, it will be used as a testing ground for many of the
concepts touched upon in later chapters.
2. Examples of topological spaces
This section is devoted to exploring some of the many examples of topological
spaces. Example 2.4 below shows that every set X can be equipped with a topology,
exhibiting that topological spaces are indeed very common animals in the jungle of
mathematics. The examples 2.1 – 2.3 discussing the Euclidean, the subspace and the
metric topology, are particularly relevant as they make frequent appearances throughout the text.
Example 2.1. The Euclidean topology The work we did in chapter ?? shows
that the Euclidean space Rn becomes a topological space if we equip it with the topology
TEu , henceforth referred to as the Euclidean topology:
TEu = {U ⊂ Rn | ∀x ∈ U ∃r > 0 such that Bx (r) ⊂ U }
Recall that Bx (r) denotes the open ball {y ∈ Rn | d(x, y) < r} from definition ??.
With this definition of TEu , the verification of the three axioms of topology is provided
courtesy of theorem ??.
Example 2.2. The relative or subspace topology. Let (X, T ) be a given
topological space and let A be a subset of X. Then A automatically inherits the
structure of a topological space from X by equipping it with the relative topology or
subspace topology TA defined as:
TA = {U ∩ A | U ∈ T }
2. EXAMPLES OF TOPOLOGICAL SPACES
3
We call the topological space (A, TA ) a subspace of X. Saying that “A is a subspace of
X”means that we have given the subset A of X the subspace topology.
It is quite straightforward to verify that TA satisfies the axioms of a topology on A:
1. Since ∅, X ∈ T then ∅ ∩ A = ∅ and X ∩ A = A belong to TA .
2. Let Ui ∈ TA , i ∈ I be given and set U = ∪i∈I Ui . For each Ui there exists a
Vi ∈ T such that Ui = Vi ∩ A. But then U = V ∩ A where V = ∪i∈I Vi ∈ T
showing that U ∈ TA .
3. Let U1 , ..., Un ∈ TA and find sets V1 , ..., Vn ∈ T such that Ui = Vi ∩ A. Then the
set U = U1 ∩ ... ∩ Un equals V ∩ A where V = V1 ∩ ... ∩ Vn ∈ T and is therefore
contained in TA .
The subspace topology gives us immediately a myriad of examples of topological
spaces by applying it to various subsets of (Rn , TEu ) from the previous example. For
instance, each of
The n-sphere
The graph of f : Rn → Rm
The 2-dimensional torus
S n = {x ∈ Rn+1 | d(x, 0) = 1} ⊂ Rn+1
Γf = {(x, f (x)) ∈ Rn × Rm | x ∈ Rn } ⊂ Rn+m
T 2 = {x ∈ R4 | x21 + x22 = 1 and x23 + x24 = 1} ⊂ R4
becomes a topological space with the relative Euclidean topology. In the definition of
T 2 , the symbol x was used to denote the ordered quadruple (x1 , x2 , x3 , x4 ).
Example 2.3. Metric spaces. A metric space is a pair (X, d) consisting of a
non-empty set X and a function d : X × X → [0, ∞i, referred to as the metric on X,
that is subject to the next three axioms of a metric:
1. d(x, y) = 0 if and only if x = y.
2. Symmetry: d(x, y) = d(y, x) for all x, y ∈ X.
3. Triangle inequality: d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X.
When the metric d is understood from context, we will call X itself a metric space. In
analogy to the case of the Euclidean metric d on Rn , here too we can define what we
shall again call the open ball with center x ∈ U and radius r > 0 as
Bx (r) = {y ∈ X | d(x, y) < r}
Every metric space (X, d) comes equipped with a natural choice of topology Td , called
the metric topology, defined by
Td = {U ⊂ X | ∀p ∈ U ∃r > 0 such that Bp (r) ⊂ U }
The reader will, no doubt, have noticed that this definition agrees with the definition
of the Euclidean topology from example 2.1. Indeed, the premier examples of metric
spaces are (Rn , dp ) with p ∈ [1, ∞] (with dp as given in definition ??). The verification
of the axioms of a metric for dp is deferred to the exercises ?? and ??. The fact that
Td is indeed a topology on X, follows along the same lines as the proof of theorem ??
for the Euclidean space. Indeed, the proof of the later never uses the fact that it deals
with (Rn , d) explicitly.
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1. TOPOLOGICAL SPACES
We shall encounter more about metric spaces in later chapter (e.g. chapter ??) and
so for now we limit ourselves to only one additional example. Namely, let a, b ∈ R be
two arbitrary numbers with a < b and consider the set X = C 0 ([a, b], R) of continuous
functions f : [a, b] → R. Define d : X × X → [0, ∞i as
Z b
d(f, g) =
|f (t) − g(t)| dt
a
Then (X, d) is a metric space and thus becomes a topological space. The second and
third axiom of a metric are readily verified:
Z b
Z b
d(f, g) =
|f (t) − g(t)| dt =
|g(t) − f (t)| dt = d(g, f )
a
a
b
Z
|f (t) − h(t)| dt
d(f, h) =
a
b
Z
|(f (t) − g(t)) + (g(t) − h(t))| dt
=
a
Z
b
≤
|f (t) − g(t)| + |g(t) − h(t)| dt
Z b
Z b
=
|f (t) − g(t)| dt +
|g(t) − h(t)| dt
a
a
a
= d(f, g) + d(g, h)
The demonstration of axiom 1 of a metric is discussed in exercise ??.
Example 2.4. The discrete and indiscrete topologies Every set X always
admits topologies. Namely, every set can be made into a topological space by choosing
either the indiscrete topology Tindis (also referred to as the trivial topology) or the
discrete topology Tdis define as
Tindis = {∅, X}
Tdis = {A | A ⊂ X}
Thus Tindis only contains the empty set and all of X and is therefore the smallest
possible topology on X (according to the first axiom of a topology) while Tdis equals
the entire power set of X and is consequently the largest topology on X. In the notation
of definition 1.3, any topology T on X satisfies the double inequality Tindis ≤ T ≤ Tdis .
The axioms of a topology are trivially true for Tindis and Tdis and we omit their explicit
verification.
These two extreme topologies on X do not lead to interesting topological properties.
For example, as we shall see in section ??, every function on (X, Tdis ) is continuous.
Fertile ground for topological exploration lies with those topologies that live between
these two extreme cases.
2. EXAMPLES OF TOPOLOGICAL SPACES
5
Example 2.5. Topologies on finite sets. On finite sets, topologies are by necessity also finite and can be listed by simply listing all of their elements. For instance,
let us take X = {1, 2, 3, 4, 5}. Then each of the following is a topology on X (exercise
??):
(a) T1 = {∅, X, {1, 2}, {3, 4, 5}}
(b) T2 = {∅, X, {1}, {2}, {1, 2}}
(c) T3 = {∅, X, {1, 2, 3}, {2, 3, 4}, {2, 3}, {1, 2, 3, 4}}
(d) T4 = {∅, X, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}
Example 2.6. Included point topology Let X be any non-empty set and let
p ∈ X be an arbitrary point. We define the included point topology on X as
Tp = {U ⊂ X | U is the empty set or p ∈ U }
To see that (X, Tp ) is a topological space, we need to verify the axioms of topology
for Tp (as given in definition 1.1):
1. Clearly ∅ ∈ Tp be definition. Also, X ∈ Tp since p ∈ X.
2. Let Ui ∈ Tp with i ∈ I where I is any indexing set and let U = ∪i∈I Ui . If each
Ui is the empty set then so is U and is therefore contained in Tp . If at least one
set Ui is not empty, then p ∈ Ui and thus p ∈ U showing again that U ∈ Tp . So
in either case U must belong to Tp .
3. Let U1 , ..., Un ∈ Tp and let V = ∩ni=1 Ui . If even one of U1 , ..., Un is empty then
V is empty as well and thus a member of Tp . If none of U1 , ..., Un is empty then
they each must contain p and therefore V must contain p as well. So in this
case V is also in Tp .
Example 2.7. The excluded point topology Let X again be any non-empty
set and, as in the previous example, pick an arbitrary point p ∈ X. The excluded point
topology T p on X is then define to be
T p = {U ⊂ X | U equals X or p ∈
/ U}
Let’s verify the axioms of a topology:
1. X belongs to T p be definition and ∅ belongs to T p since p ∈
/ ∅.
2. Let Ui ∈ T p , i ∈ I and set U = ∪i∈I Ui . If at least one Ui equals X then U = X
and so U ∈ T p . If none of the Ui equals X then no Ui can contain p and so p
cannot be contained in U either. Thus, in this case too, we get U ∈ T p .
3. Take U1 , ..., Un ∈ T p and let V = U1 ∩ ... ∩ Un . If all of the set Ui happen to
equal X then so does V and is thus automatically contained in T p . Conversely,
if there is at least one Ui not equal to X then that particular Ui cannot contain
p and consequently neither can V . Thus, in this case too, V is again an element
of T p .
Definition 2.8. Let X be any non-empty set. A partition P on X is a collection
of subsets of X such that
1. If A, B ∈ P are two distinct elements of P then A ∩ B = ∅.
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1. TOPOLOGICAL SPACES
2. The elements of P cover all of X: ∪A∈P A = X.
Thus a partition P is a way of dividing all of X into mutually disjoint set. The partition
P = {X} consisting of only X shall be referred to as the trivial partition.
Examples of partitions abound. For instance, if we take X = R, then
P1 = {[a, a + 1i | a ∈ Z}
P2 = {Q, R − Q}
(1)
P3 = {x + Z | x ∈ hπ, π + 1]}
are all examples of partitions.
Example 2.9. Partition topology Let X be a non-empty set and let P = {Ui | i ∈
I} be a partition on X. We then define the partition topology TP as
TP = {∪j∈J Uj | J ⊂ I}
Thus elements of TP are obtained by taking unions of set from P. To see that this is
a topology we again check that the three axioms of a topology are satisfied.
1. Choosing J = ∅ and J = I renders the set ∪j∈J Uj equal to the empty set and
all of X respectively.
2. Let Jk , k ∈ K be a family of subsets of I giving rise to the sets Vk = ∪j∈Jk Uj
from TP and let V = ∪k∈K Vk . Rewriting this definition of V we see that
V = ∪j∈L Uj
with
L = ∪k∈K Jk ⊂ I
Thus, of course, V belongs to TP .
3. This case proceeds in complete analogy with the previous point. Let V1 =
∪j∈J1 Uj , ..., Vn = ∪j∈Jn Uj and set W = ∩nk=1 Vk . But then
W = ∩j∈L Uj
with
L = ∩nk=1 Jk ⊂ I
we see again that W lies in P.
For instance, let us pick the partion P1 from (1) on X = R. Examples of open sets
in the associated partition topology TP1 are intervals of the form [a, bi as well as [a, ∞i
and h∞, bi with a, b ∈ Z.
Example 2.10. Finite complement topology. On a non-empty set X we define
the finite complement topology Tf c as
Tf c = {U ⊂ X | U = ∅
or
X − U is a finite set }
If X is itself a finite set then the finite complement topology agrees with the discrete
topology (from example 2.4) but if X is infinite, Tf c and Tdis are in fact quite different.
We defer the verification of the axioms of topology for Tf c to exercise ??.
Example 2.11. The countable complement topology. The countable complement topology Tcc on a set X is define as
Tcc = {U ⊂ X | U = ∅
or
X − U is a countable set }
2. EXAMPLES OF TOPOLOGICAL SPACES
7
Note that if X is itself countable then Tcc agrees with the discrete topology Tdis on X.
However, for example on X = R, the two topologies are different. Exercise ?? asks you
to verify the axioms of a topology for this example.
Example 2.12. The Fort topology. This example is obtained by combining the
excluded point topology with the finite complement topology from above. Assume
that X is an infinite set and let p ∈ X be an arbitrary point. We then define the Fort
topology TF ,p as
TF ,p = {U ⊂ X | Either X − U is finite or p ∈
/ U}
The verification of the axioms of a topology is left as an exercise
Example 2.13. The order topology In this example we suppose that the set X
is equipped with an ordering “≤”, a notion which we briefly review before defining the
associated order topology T≤ .
Recall than a relation r on X is simply a subset of X × X. It is customary to write
xry if (x, y) ∈ r. A typical choice for the name of a relation is a “relation symbol”,
such as ∼, ≡, ≤, ≺ etc. For example, if we called a relation ∼ then we would write
x ∼ y to indicate that (x, y) belongs to this relation.
A (total) ordering “≤”on X is a relation on X subject to the conditions
1. “≤”is reflexive: x ≤ x for any x ∈ X.
2. “≤”is antisymmetric: If x ≤ y and y ≤ x then x = y.
3. “≤”is transitive: If x ≤ y and y ≤ z then x ≤ z.
4. “≤”satisfies the trichotomy law: For all x, y ∈ X, either x ≤ y or y ≤ x.
We shall write x < y to mean that x ≤ y but x 6= y.
Given an ordering “≤”on a set X with at least two elements, let us consider the
following special subset of X
Lx = {y ∈ X | y < x}
and
Rx = {z ∈ X | x < z}
In terms of these we define the order topology T≤ associated to the ordering “≤”as
T≤ = {U ⊂ X | U is obtained by taking arbitrary unions of
finite intersections of the sets Lx and Ry with x, y ∈ X}
Examples of elements from T≤ are Lx ∩ Ry which we shall denote by hx, yi and refer
to as the open intervals of the order topology. Note that hx, yi is an empty set unless
there exists an element z ∈ X with x < z < y.
Since the set T≤ is closed under finite intersection and arbitrary unions (by definition
of T≤ ), axioms (2) and (3) of a topology (definition 1.1) are trivially true. To see that
the empty set and X belong to T≤ we proceed as follows. Given two distinct elements
x, y (recall that we assumed that X has at least two elements), we have either x < y
or y < x (according to trichotomy axiom above). Suppose that y < x, then hx, yi = ∅
showing that ∅ ∈ T≤ . On the other hand, for these same x, y ∈ X we obtain X = Lx ∪Ry
(exercise ??) and so X ∈ T≤ .
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1. TOPOLOGICAL SPACES
As an illustration of the order topology, we consider the lexicographic ordering on
R . Let ≤ denote the standard ordering of the real number R and extend it to an
ordering on Rn by the following rule: (x1 , ..., xn ) < (y1 , ..., yn ) if
n
x1 < y 1
x1 = y1 and x2 < y2
x1 = y1 , x2 = y2 and x3 < y3
..
.
or
or
or
x1 = y1 , x2 = y2 , ..., xn−1 = yn−1 and xn < yn
or
The thus obtained ordering (exercise ??) is called the lexicographic ordering on Rn .
When n = 1, this topology equals the Euclidean topology on R (from example 2.1).
However, when n ≥ 2 the resulting order topology on Rn is quite different from its
Euclidean counterpart. For example, when n = 2, consider the interval h(0, 0), (1, 0)i ⊆
R2 given by
h(0, 0), (1, 0)i = {(x, y) ∈ R2 | 0 < x < 1} ∪ {(0, y) ∈ R2 | y > 0} ∪ {(1, y) ∈ R2 | y < 0}
Figure 1 shows this interval drawn in the Euclidean plane. Note that this interval
h(0, 0), (1, 0)i does not belong to the Euclidean topology on R2 .
1
Figure 1. The shaded region in R2 represents the interval h(0, 0), (1, 0)i
in the order topology on R2 associated to the lexicographic ordering. The
dotted lines are not part of the region while the full lines are. The region
extends infinitely vertically in both directions.
Example 2.14. The lower and upper limit topologies on R The lower-limit
topology Tll on R is defined as
Tll = {U ⊂ X | U is obtained by taking unions of finite intersections of
sets [a, bi with a, b ∈ R}
3. PROPERTIES OF OPEN AND CLOSED SETS
9
The definition of Tll shows that it is closed under finite intersections and arbitrary
unions while the empty set and R belong to Tll because, for example, ∅ = [0, 1i ∩ [3, 5i
and R = ∪a∈Z [a, a + 1i. Thus, Tll is indeed a topology.
The upper limit topology Tul on R is defined analogously by replacing the sets [a, bi
in the definition of Tll by the sets ha, b].
Notice that the open intervals ha, bi belong both to Tll and to Tul since, for example,
∞ [
1
ha, bi =
a + , b ∈ Tll
n
n1
The starting value of n in the above union is chosen large enough so that a + 1/n < b.
¿From this observation it is not hard to show that TEu ⊂ Tll and TEu ⊂ Tul (exercise
??). However, the set [0, 1i belong to Tll but not to TEu showing that TEu 6= Tll . A
similar observation applies to the upper limit topology.
Example 2.15. The topologist’s sine curve. This example and the two subsequent ones, do not define new types of topologies, but rather look at subspaces of
(R2 , TEu ) (see examples 2.1 and 2.2) with certain special properties that shall be explored in later chapters.
We define the topologist’s sine curve to be the subspace X ⊂ R2 given by
X = {(x, sin(1/x)) | x ∈ h0, 1]} ∪ ({0} × [0, 1])
Thus, the topologist’s sine curve is the union of the graph of sin(1/x) over h0, 1] with
the closed segment [−1, 1] on the y-axis. This space is illustrated in figure 2a.
Example 2.16. The infinite broom Let In ⊆ R2 be the closed straight-line
segment joining the origin (0, 0) to the point (1, 1/n) for n = 1, 2, 3, .... The infinite
broom is then the subspace X of (Rn , TEu ) defined by
X = (∪∞
n=1 In ) ∪ ([0, 1] × {0})
See figure 2b for an illustration.
Example 2.17. Hawaiian earrings For n = 1, 2, 3, ... let Cn ⊂ R2 be the circle
with center (1/n, 0) and with radius rn = 1/n. Note that all off these circles pass
through the origin (0, 0).The Hawaiian earrings is the subspace X of (R2 , TEu ) given
as the union of these circles, see figure 2c:
∞
2
2
2
2
X = ∪∞
n=1 Cn = ∪n=1 {(x, y) ∈ R | (x − 1/n) + y = (1/n) }
3. Properties of open and closed sets
This section discusses some general properties of open and closed sets of a topological space (X, T ). Recall that a subset U ⊂ X is called open if U ∈ T , a subset A ⊂ X
is called closed if X − A ∈ T .
While typically a random subset Y ⊂ X is neither open nor closed, as we shall see
it can be “approximated”by both kinds of sets.
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1. TOPOLOGICAL SPACES
1
1
−1
(a)
...
..
.
(b)
(c)
Figure 2. (a) The topologist’s sine curve. (b) The infinite broom. (c)
The Hawaiian earrings.
Definition 3.1. Let A ⊂ X be an arbitrary subset of X. Then we define the
interior Å (often also denoted by Int(A)) and the closure Ā of A (sometimes written
as Cl(A)) as
Å = union of all open sets contained in A
Ā = intersection of all closed sets containing A
Additionally, we define the boundary or frontier ∂A of A as ∂A = Ā − Å.
Lemma 3.2. Let A be a subset of the topological space X. Then
3. PROPERTIES OF OPEN AND CLOSED SETS
11
(a) The interior Å of A is an open set and it is the largest open set contained in
A. The equality A = Å holds if and only if A is open.
(b) The closure Ā of A is a closed set and it is the smallest closed set containing
A. The equality Ā = A holds if and only if A is closed.
(c) A point x ∈ X belongs to Ā if and only if every neighborhood of x intersects A.
Proof. These claims follow readily from the definition of interior and closure of a
set.
(a) Since Å is obtained as a union of open sets (those contained in A) it is an open
set. If U is any other open set containing A, then U is one of the sets from the union
which defines Å and is thus contained in Å, showing that the interior is the largest open
set contained in A. If Å = A then clearly A must be open since Å is open. Conversely,
if A is open then it is clearly the largest open set containing A and thus by necessity
equal to Å. (b) By definition, Ā is an intersection of closed set and must therefore be
closed. If B is any closed set containing A, then B occurs in the intersection of sets
defining Ā and hence Ā ⊂ B. This shows that Ā is the smallest closed set containing
A. If Ā = A then A is closed, since Ā is. If A = Ā then A itself is the smallest closed
set containing A and is thereby equal to Ā.
(c) Suppose firstly that x ∈ Ā and let U be a neighborhood of x. If we had A∩U = ∅
then Ā − U would be a closed set containing A and would properly be contained in Ā,
an impossibility according to part (b) of the lemma. Thus we must have A ∩ U = ∅.
Conversely, suppose that x ∈ X is a point such that U ∩ A = ∅ for every neighborhood U of x. If we had x ∈
/ Ā, we could take U = X − Ā which would give an
immediate contradiction. Thus we must have x ∈ Ā.
Definition 3.1 shows that the inclusions
Å ⊂ A ⊂ Ā
hold for every subset A of X while lemma 3.2 shows that Å is always an open set and
Ā is always a closed set. It is in this sense that A can be approximated by both an
open set (its interior) and a closed set (its closure). How good an approximation of A
is given by the sets Å and Ā depends heavily on the topology on X.
Example 3.3. Consider the set R and its subset A = h0, 1i. Find the interior
Å, the closure Ā and the boundary ∂A of A with respect to the following choices of
topologies on R:
1. The Euclidean topology (see example 2.1). With this toplogy A itself is open
and so, by lemma 3.2, the interior of A equals A itself. The closure of A is
a closed set cointaining A. We guess that Ā = [0, 1]. Indeed, [0, 1] is closed
and contains A and the only smaller subsets cotaining A are [0, 1i, h0, 1] and A
itself, neither of which is closed. Thus
Å = A = h0, 1i
Ā = [0, 1]
a result which confirms our Euclidean intuition.
∂A = {0, 1}
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1. TOPOLOGICAL SPACES
2. The included point topology with p = 0 (see example 2.6). Since p ∈
/ A we see
that A is in fact closed so that Ā = A. On the other hand, the only open set
from Tp that is contained in A is the empty set, thus Å = ∅. Therefore,
Å = ∅
Ā = A = h0, 1i
∂A = A = h0, 1i
3. The particular point topology with p = 1/2 (see again example 2.6). Since
p ∈ A we see that Å = A. However, since p ∈ A, the only closed set containing
A is all of R showing that Ā = R. We arrive at
Å = A = h0, 1i
Ā = R
∂A = h−∞, 0] ∪ [1, ∞i
4. The finite complement topology (see example 2.10). In this topology, closed set
are finite subsets of R and all of R. This makes is clear that Ā = R. No subset
of A has finite complement showing that Å = ∅. In summary
Å = ∅
Ā = R
∂A = R
The following lemma provides an alternative definition of the boundary ∂A.
Lemma 3.4. Let X be a topological space and let A be a subset of X. Then
(a) X − A = X − Å.
(b) Int(X − A) = X − Ā.
(c) ∂A = Ā ∩ X − A.
Proof. Parts (a) and (b) are left as an exercise.
(c) This portion of the theorem follows now easily from part (a):
∂A = Ā − Å = Ā ∩ (X − Å) = Ā ∩ X − A
Definition 3.5. Let X be a topological space. A subset A of X is called dense in
X if Ā = X. The space X is called separable if there exists a countable dense subset
A of X.
As many topological spaces X are uncountable as sets, the notion of separability
provides a measure of how big X is as a topological space. A dense subset A ⊂ X has
the property that it intersects every open set of X (corollary 3.7 below). Thus, if we
can find a countable dense subset of X (i.e. if X is separable), we should think of X
as being “relatively small”as a topological space.
Example 3.6. The subset A = h0, 1i of R is dense with respect to either the
particular point topology Tp with p = 1/2 or with respect to the finite complement
topology Tf c (see example 3.3).
Corollary 3.7. A subset A of the topological space (X, T ) is dense if and only if
A ∩ U 6= ∅ for all U ∈ T other than U = ∅.
Proof. This is a direct consequence of part (c) of lemma 3.1.
4. BASES AND SUBBASES OF A TOPOLOGY
13
Example 3.8. The set Q of rational number is dense in (R, TEu ) since it intersects
every open interval ha, bi (as follows from Archimides’ axiom for the real numbers, cf.
[]). More generally, by the same principle, Qn is dense in (Rn , TEu ). Consequently, since
Qn is a countable set for each n ≥ 0, we find that (Rn , TEu ) is a separable topological
space for all n ≥ 0.
4. Bases and subbases of a topology
The reader familiar with linear algebra will recall that a basis for a finite dimensional
vector space V is a set of vectors {e1 , ..., en } ⊂ V such that every vector v ∈ V can
be written uniquely as a linear combination v = λ1 e1 + .. + λn en . Thus, while V is
typically an infinite set, we can capture the totality of its vectors with the finite set
{e1 , ..., en } by relying on the vector space operations of “vector addition”and “scalar
multiplication”.
As a topology T on a set X is a collection of subsets of X, it comes equipped
with two operations among its elements, namely those of taking unions and taking
intersections. It is thus conceivable, in analogy with the vector space basis, that there
are subsets of T which “generate”all of T by means of taking unions of their elements.
This is indeed that case. In fact, we will consider both subsets B ⊂ T which “generate”all of T by means of only taking unions of elements from B, and subsets S ⊂ T
which will generate T by relying on both unions and intersection. The first of these
cases will lead the notion of a basis for (X, T ), the closest analogy to a vector space
basis. The second will lead to the notion of a subbasis, one that is without analogue
in the vector space world.
Definition 4.1. Let (X, T ) be a topological space and let B and S be subsets of
T such that
(a) Every set U ∈ T is a union of sets from B.
(b) Every set U ∈ T is a union of finite intersections of sets from S.
Then B is called a basis for the topology T while S is called a subbasis for the topology
T . We also say that T is generated by B or that T is generated by S.
Thus, if for example B = {Ui ∈ T | i ∈ I} and S = {Vj ∈ T | j ∈ J } (where
as usual, I and J are two abstract indexing sets) then properties (a) and (b) from
definition 4.1 mean that every set U ∈ T can we written as
(a) U = ∪i∈I1 Ui
for some subset I1 of I.
(b) U = ∪`∈L (∩j∈J` Vj )
for some family of finite subsets J` of J with ` ∈ L.
The indexing set L from (b) above, is of course allowed to be infinite. While we will
typically start out with a topological space (X, T ) and then seek bases and subbases
for T , it is meaningful to ask about going the other way. That is, given a set X and
two collections B, S of subsets of X, under what conditions are B and S a basis and a
subbasis for some topology T . The next two lemmas address this question.
14
1. TOPOLOGICAL SPACES
Lemma 4.2. A collection B of subsets of a set X is a basis for some topology T if
and only if B has the properties
1. X = ∪B∈B B
2. For every two B1 , B2 ∈ B and every point x ∈ B1 ∩ B2 , there exists an element
B3 ∈ B such that x ∈ B3 ⊂ B1 ∩ B2 .
If these two conditions are met, the topology T generated by B consists of all possible
unions of elements from B:
T = {U ⊂ X | U is a union of sets from B}
Proof. One implication of the lemma is immediate. Namely, if B is a basis for the
topology T then, since X ∈ T , it must be that X = ∪B∈B B. Likewise, if B1 , B2 ∈ B,
then B1 ∩ B2 is an open set and therefore a union of elements from B.
We thus turn to the other implication. We assume that B is a collection of subsets
of X subject to conditions 1. and 2. from the lemma and let T be the set T = {U ⊂
X | U is a union of elements from B}. It is then clear that X ∈ T (this uses condition
1.) and ∅ ∈ T (the empty set is the empty union of any sets). By definition of T ,
it is obviously closed under unions since unions of unions are again just unions. To
see that T is closed under finite intersection it suffices to show that it is closed under
2f-old intersections. Thus, let U1 , U2 ∈ T , we need to show that U1 ∩ U2 is a union of
elements from B. Let x ∈ U1 ∩ U2 be any point. By definition of T , there must exist
elements B1 , B2 ∈ B with x ∈ Bi ⊂ Ui , i = 1, 2. But then by property 2., there is an
element Bx ∈ B such that x ∈ Bx ⊂ B1 ∩ B2 . But then clearly
U1 ∩ U2 = ∩x∈U1 ∩U2 Bx
This shows that T is indeed a topology on X. Moreover, the definition of T shows
that B is a basis for T , as claimed.
Lemma 4.3. A collection S of subsets of a set X is a subbasis for a topology T on
X, if and only if X = ∪S∈S S. In this case, the topology T is obtained as
T = {U ⊂ X | U is a union of finite intersections of sets from S}
Proof. If indeed S is a subbasis for a topology on X, then the condition X =
∪S∈S S needs to be satisfied.
Conversely, suppose that X = ∪S∈S S and let T be defined as in the statement
of the lemma. Then clearly ∅ ∈ T while X ∈ T by the stated condition on S. The
fact that T is closed under finite intersections and arbitrary unions, follows from the
definition of T , showing that T is a topology. Likewise, the fact S is a subbasis of T
also follows from the definition.
Lemmas 4.2 and 4.3 give us ways to define topologies on a set X by either picking
a basis or a subbasis first and letting them generate the topology. Before proceeding,
we pause to consider several examples.
4. BASES AND SUBBASES OF A TOPOLOGY
15
Example 4.4. The sets
B1 = {ha, bi | a, b ∈ R, a < b}
S1 = {h−∞, bi, ha, ∞i | a, b ∈ R}
are a basis and subbasis for (R, TEu ) (recall that TEu denotes the Euclidean topology
on R). Similarly, by relying on the density of the rational numbers Q in R, one can
show that
B2 = {ha, bi | a, b ∈ Q, a < b}
S2 = {h−∞, bi, ha, ∞i | a, b ∈ Q}
are also a basis and a subbasis for (R, TEu ). The key difference between the two
examples is that the sets B2 and S2 are countable sets while both of B1 and S1 are
uncountable.
Example 4.5. Consider the included point topology Tp on R (see example 2.6).
The sets {p} and {x, p}, for any x ∈ R, are all open sets. However, neither {p} nor
{x, p} can be obtained as a union of other nonempty open sets and must therefore
be part of every basis B. Therefore, every basis B for (R, Tp ) has uncountably many
elements. The smallest possible basis for (R, Tp ) is
B = {{p}, {x, p} | x ∈ R − {p}}
Example 4.6. Let (X, ≤) be an ordered set and consider the order topology T≤
on X (from example 2.13). Then a basis and a subbasis for (X, T≤ ) are given by
B = {ha, bi | a, b ∈ X, a < b}
and
S = {La , Rb | a, b ∈ X}
In the case of X = R and with ≤ being the usual ordering of real numbers, these two
sets agree with B1 and S1 from example 4.4.
Example 4.7. Consider the set R and let S be the collection of subsets of R given
by
S = {x + Q+ , y + Q− | x, y ∈ R}
where Q+ and Q− are the sets of the positive and of the negative rational numbers
respectively. According to lemma 4.3, S is a subbasis for a topology TS on R. An
example of an open set in this topology is h−1, 1i ∩ Q.
Just as vector spaces are divided into finite dimensional and infinite dimensional
examples according to whether or not they possess a finite basis or not, so too topological spaces can be group into two distinct categories. The first attempt to define the
analogue of a finite dimensional vector space for topological spaces, might be to demand the existence of finite basis for the topology. However, since a topology generated
by a finite basis is by necessity finite, this definition would exclude most interesting
examples (for instance, no Euclidean space (Rn , TEu ) with n ≥ 1 admits a finite basis).
Instead, we will divide topological spaces into two categories according to whether or
not they possess a countable basis or not.
16
1. TOPOLOGICAL SPACES
Definition 4.8. A topological space (X, T ) is called second countable if it possesses
a countable basis B.
We should think of a second countable topological space as being “small”and of
one that isn’t second countable, as being “large”. Another measure of “size”for a
topological space the we encountered previously was that of being separable (definition
3.5). As the next lemma shows, these two concepts are related.
Lemma 4.9. If the topological space (X, T ) is second countable then it is also separable. The converse is false in general (see example 4.10 below).
Proof. Let B = {Ui | i = 1, 2, 3, ...} be a countable basis for (X, T ). Without
loss of generality we can assume that Ui 6= ∅. Let a∈ Ui be any element and let A =
{a1 , a2 , ...} ⊂ X. Then A is a countable set and we claim that Ā = X. For if not, then
there would exist an element x ∈ X − Ā. Since X − Ā is an open set, there must exist
a Uj ∈ B such that Uj ⊂ X − Ā. But then aj ∈ (X − Ā) ∩ A, a contradiction. Thus
Ā = X and hence X is separable.
Example 4.4 shows that the Euclidean space (R, TEu ) is second countable while
example 4.5 shows that (R, Tp ) is not second countable. While it is not hard to show
that all Euclidean spaces (Rn , TEu ) are second countable, we shall defer this to chapter
?? where we will show that products of second countable spaces are again second
countable.
Example 4.10. As we already saw (example 4.5) that the set of real numbers R
with the included point topology Tp isn’t a second countable space. On the other hand,
let A = {p} ⊂ R and notice that Ā = R (since closed sets in this topology are sets
either not containing p or else all of R). Thus (R, Tp ) is separable.
We finish this section by considering a local version of the notion of second countability.
Definition 4.11. Let (X, T ) be a topological space and let x ∈ X be a point in
X. A neighborhood basis around x is a collection Bx of neighborhoods of x such that
for every neighborhood U of x there is an element V ∈ Bx with x ∈ V ⊂ U . We say
that (X, T ) is first countable if every point x ∈ X possesses a countable neighborhood
basis.
It should be clear that a second countable space is automatically first countable.
The converse is of course false as the next example demonstrates.
Example 4.12. Consider the space (R, Tp ) where Tp is the include point topology
(example 2.6) and let x ∈ R be any point. A neighborhood basis Bx for x is given by

; if x = p
 {{p}}
Bx =
 {{x, p}}
; if x 6= p
Thus (R, Tp ) is first countable but it isn’t second countable according to example 4.5.
4. BASES AND SUBBASES OF A TOPOLOGY
17
Example 4.13. A metric space (X, d) equipped with the metric topology Td (see
example 2.3) is always first countable. Namely, given a point x ∈ X we can define Bx
as
Bx = {Bx (r) | r ∈ Q+ }
where, as before, Q+ is the set of positive rational number. Clearly Bx is a countable
set and if U is any neighborhood of x, then there must exist some real number r > 0
such that Bx (r) ⊂ U . Taking any r0 ∈ h0, ri ∩ Q gives an element Bx (r0 ) ∈ Bx with
x ∈ Bx (r0 ) ⊂ Bx (r).
We saw that second countability and separability are generally two distinct measures for the size of a topology (lemma 4.9). However, for metric spaces, these two
notions actually agree. The reason for this lies in the first countability.
Lemma 4.14. A metric space (X, d) equipped with the metric topology Td is separable
precisely when it is second countable.
Proof. Let A = {a1 , a2 , a3 , ...} be a countable dense subset of X, for each ai ∈ A
let Bi = {Bai (r) | r ∈ Q+ } be a countable neighborhood basis for ai and set B = ∪∞
i=1 Bi .
Since B is a countable union of countable sets, it must itself also be countable set. To see
that B is a basis for (X, T ), let U ⊂ X be an open set and let x ∈ U be any point. There
has to exist a real number r > 0 such that Bx (r) ⊂ U (be definition of Td , see example
2.3). Consider the open set Bx (r/2). By corollary 3.7 the intersection A ∩ Bx (r/2) has
to be nonempty so, without loss of generality, suppose that a1 ∈ A ∩ Bx (r/2). But
then Ba1 (r/2) ∈ B1 ⊂ B, clearly x ∈ Ba1 (r/2) and, moreover, Ba1 (r/2) ⊂ Bx (r) for if
y ∈ Ba1 (r/2) then d(y, x) ≤ d(y, a1 ) + d(a1 , x) < r/2 + r/2 = r. Thus, we have shown
that for every point x ∈ U there is a set Ux ∈ B with x ∈ Ux ⊂ U . But then clearly
U = ∪x∈U Ux showing that B is a basis.
Remark 4.15. It is not true in general that a separable and first countable space
is second countable. We refer the interested reader to [] for examples.
Theorem 4.16. Let (X, TX ) be a topological space and let Y ⊂ X be a subspace of
X. Let A be chosen from the following set of properties of topological spaces


Separabiltiy


Second countability
A∈
 First countability 
Then, if X has property A, so does Y .
Proof. • A is “Separability”
Let A ⊂ X be a countable dense subset of X and set AY = A ∩ Y . Then AY is a
countable and dense subset of Y (denseness follows from corollary 3.7 and the definition
of the relative topology on Y ).
• A is “Second countability”
If B = {U1 , U2 , U3 , ...} is a countable basis for the topology on X, then BY = {U1 ∩
Y, U2 ∩ Y, U3 ∩ Y, ...} is a countable basis for the relative topology on Y .
18
1. TOPOLOGICAL SPACES
• A is “First countability”
If By = {U1 , U2 , U3 , ...} is a countable neighborhood basis for the point y ∈ X, then
By,Y = {U1 ∩ Y, U2 ∩ Y, U3 ∩ Y, ...} is a countable neighborhood basis for y ∈ Y with
respect to the relative topology on Y .
5. Exercises
5.1. For p ∈ [1, ∞], verify that dp : Rn × Rn → [0, ∞i satisfies the axioms of a
metric.