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Chapter 2
Topological Spaces
This chapter contains a very bare summary of some basic facts from topology.
2.1
Definition of Topology
A topology O on a set X is a collection of subsets of X satisfying the following
conditions:
(T1) ∅ ∈ O
(T2) X ∈ O
(T3) O is closed under finite intersections, i.e. if A, B ∈ O then A ∩ B ∈ O
(T4) O is closed under unions, i.e. if S ⊂ O then ∪S ∈ O
We have used the set-theoretic notation for unions: ∪S is the union of all
the sets in S.
Note also that by the usual induction argument, condition (T2) implies
that if A1 , ..., An are a finite number of open sets then A1 ∩ · · · ∩ An is also
open.
A topological space (X, OX ) is a set X along with a topology OX on it.
Usuallly, we just say “X is a topological space.”
If x ∈ X and U is an open set with x ∈ U then we say that U is a
neighborhood of x.
There are two extreme topologies on any set X:
• the indiscrete topology {∅, X}
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CHAPTER 2.
TOPOLOGICAL SPACES
• the discrete topology P(X) consisting of all subsets of X
A set A in O is said to be open in the topology O. The complement of an
open set is called a closed set. Taking complements of (T1)-(T4) it follows
that:
(C1) X is closed
(C2) ∅ is closed
(C3) the union of a finite number of closed sets is closed
(C4) the intersection of any family of closed sets is closed.
Let F ⊂ X. The interior F 0 of F is the union of all open sets contained
in F (F 0 could be empty, in case the only open set which is a subset of F is
the empty set). It is of course an open set and is the largest open set which is
a subset of F . The closure F of F is the intersection of all closed sets which
contain F as a subset. Thus F is a closed set, the smallest closed set which
contains F as a subset.
The intersection of any set of topologies on X is clearly also a topology.
Let S be any non-empty collection of subsets of X. Consider the collection
TS of all topologies which contain S, i.e. for which the sets of S are open.
Note that P(X) ∈ TS . Then
∩TS
is the smallest topology on X containing all the sets of S. It is called the
topology on X generated by S.
A topological space X is Hausdorff if any two points have disjoint neighborhoods: i.e. for any x, y ∈ X with x 6= y, there exist open sets U and V ,
with x ∈ U , y ∈ V , and U ∩ V = ∅.
2.2
Continuous maps
Let (X, OX ) and (Y, OY ) be topological spaces, and
f :X →Y
a mapping. We say that f is continuous if
f −1 (OY ) ⊂ OX ,
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2.3. COMPACT SETS
i.e. if for every open set B ⊂ Y the inverse image f −1 (A) is an open subset
of X.
It is clear that the identity map X → X is continuous, when X is given
any particular topology.
The other simple fact is that if f : X → Y is continuous and g : Y → Z
is continuous then the composite
g◦f :X →Z
is also continuous.
A mapping f : X → Y between topological spaces is a homeomorphism
of f is a bijection and both f and f −1 are continuous.
2.3
Compact Sets
In this section X is a topological space.
Let A ⊂ X. An open cover of A is a collection S of open sets such that
A ⊂ ∪S
It is usually convenient to index the sets of S: i.e we talk of an open cover
{Uα }α∈I of A, for some indexing set I.
If S is an open cover of A, then a subcover of S is a subset S 0 ⊂ S which
also covers A.
A set K ⊂ X is compact if every open cover has a finite subcover.
Observe that the union of a finite number of compact sets is compact.
Lemma 1 A closed subset of a compact set is closed.
Proof. Let C be closed and C ⊂ K where K is compact. If S is an open
cover of C then we obtain an open cover of K by throwing in the open set
C c along with S. Then S ∪ {C c } is an open cover of the compact set K and
so has a finite subcover S 0 . The S 0 is also automatically a finite cover of C,
and remains so if we discard C c from S 0 in case S 0 does contain C c . This
yields a finite subset of S which still covers C. QED
Lemma 2 In a Hausdorff space X:
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CHAPTER 2.
TOPOLOGICAL SPACES
(i) if K is a compact subset of X and y ∈ X a point outside K then y and
K have disjoint neighborhoods, i.e. there is an open neighborhood W y
of y and an open set Vy ⊃ K with Wy ∩ Vy = ∅.
(ii) every compact subset of X is closed.
(iii) any two disjoint compact subsets of X have disjoint open neighborhoods,
i.e. if C and D are compact subsets of X then there exist open sets
U ⊃ C and V ⊃ D with U ∩ V = ∅.
The finiteness argument used in the proof here is typical.
Proof. Let y be a point outside the compact set K. For each x ∈ K,
Hausdorffness gives us disjoint open sets Ux and Fx , with x ∈ Ux and y ∈ Fx .
The open sets Ux , with x running over K, form an open cover of K. By
compactness there exist x1 , .., xN ∈ K such that Ux1 , ..., UxN cover K, i.e.
K ⊂ V y = U x1 ∪ · · · ∪ U xN
On the other hand we have the open set
Wy = F y 1 ∩ · · · ∩ F y N
which contains y. Observe that Vy and Wy are disjoint. This proves (i).
From (i) we see that each point outside the compact set K has an open
neighborhood contained entirely inside the complement K c , which implies
K c is open, i.e. K is closed. This proves (ii).
The finiteness argument used to prove (i) also proves (iii). By (i), for any
y ∈ D, there exist disjoint open sets Vy and Wy with y ∈ Wy and C ⊂ Vy . The
sets Wy form an open covering of D which has a finite subcover Wy1 , ..., Wym .
Let U = Vy1 ∩ · · · ∩ Vym and V = Wy1 ∪ · · · ∪ Wym . Then U and V are open
sets, they are disjoint, and C ⊂ U and D ⊂ V . QED
2.4
Locally Compact Hausdorff Spaces
Let X be a topological space. We say that X is locally compact if for each
point x ∈ X there is an open set U and a compact set K such that
x∈U ⊂K
2.4. LOCALLY COMPACT HAUSDORFF SPACES
5
If X is Hausdorff, then K would be closed and so for a Hausdorff space, being
locally compact means that each point has a neighborhood U whose closure
U is compact.
The following separation result is useful:
Lemma 3 Suppose X is a locally compact Hausdorff space. If C is a closed
subset of X and x a point outside C then x and C have disjoint open neighborhoods, the neighborhood of x having compact closure. Equivalently, if W
is an open neigborhood of a point x ∈ X then there is an open neighborhood
V of x such that the closure V is compact and V ⊂ W .
Proof We prove the second formulation.
Because X is locally compact Hausdorff, there is an open neighborhood
U of x and compact K with U ⊂ K. Since W c and K are closed, W c ∩ K is
closed. Being closed and a subset of the compact set K, it is also compact. So
there is an open neighborhood F of x with closure F lying in the complement
of W c ∩ K. Let V = F ∩ U . Then V is an open neighborhood of x, and
V ⊂U ⊂K
So V is compact. Moreover, since V is a subset of F the set V lies in the
complement of W c ∩ K, i.e.
V ⊂ W ∪ Kc
But we have already seen that V is a subset of K, so
V ⊂W
which is what we wanted. QED
The following stronger separation result follows easily from the preceding
result:
Lemma 4 Suppose X is a locally compact Hausdorff space. If C is a closed
subset of X and D a compact set disjoint from C then C and D have disjoint
open neighborhoods, the neighborhood of D being also compact. Equivalently,
if K is a compact set and U and open set with K ⊂ U then there is an open
set V , with compact closure V , such that
K⊂V ⊂V ⊂U
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2.5
CHAPTER 2.
TOPOLOGICAL SPACES
Urysohn’s Lemma
The main theorem of use for us is Urysohn’s Lemma:
Theorem 1 Suppose X is a locally compact Hausdorff space. Let K be a
compact subset of X and U an open subset of X with K ⊂ U . Then there is
a function f ∈ Cc (X), continuous of compact support, such that
1K ≤ f ≤ 1 U
Proof. Since X is locally compact Hausdorff and the compact set K is
contained in the open set U , there is an open set U1 with compact closure
U 1 and
K ⊂ U1 ⊂ U 1 ⊂ U
Applying this argument again to the compact set U 1 lying inside the open
set U , we have an open set U0 , with compact closure, such that
U 1 ⊂ U0 ⊂ U 0 ⊂ U
Thus we have the open sets U0 and U1 , each with compact closure, satisfying
K ⊂ U1 ⊂ U 1 ⊂ U0 ⊂ U 0 ⊂ U
We shall produce for each rational number r ∈ (0, 1) an open set Ur , with
compact closure, in such a way that
U r ⊂ Us whenever s < r.
(2.1)
To this end, let r1 , r2 , r3 , ... be an enumeration of the rationals in [0, 1], taking
r1 = 0 and r2 = 1. Suppose n ≥ 2 and that we have defined open sets
Ur1 , ..., Urn , each having compact closure and satisfying the condition (2.1).
We want to define Urn+1 to be an open set with compact closure which will
continue to respect the requirement (2.1). So let p be the largest element
of {r1 , ..., rn } less than rn+1 and q the smallest element larger than rn+1 ; in
particular,
p < rn+1 < q
Since p < q we have U q ⊂ Up and so there is an open set Urn+1 , with compact
closure satisfying
U q ⊂ Urn+1 ⊂ U rn+1 ⊂ Up
2.5. URYSOHN’S LEMMA
7
Thus, inductively [or recursively], this allows us to construct the entire family
of sets Ur , with r running over all the rationals in [0, 1], satisfying (2.1).
Now define the function f : X → R as follows. On U 1 we set f equal
to 1, and outside U0 we set f equal to 0. [In particular, this ensures that
f equals 1 on K and equals 0 outside U .] More generally, let f (x) be the
supremum of all the rationals r ∈ [0, 1] for which x ∈ Ur :
f (x) = sup{r ∈ Q ∩ [0, 1] : x ∈ Ur }
and we take the sup of the empty set (i.e. if x ∈
/ U0 ) to be 0.
Let r be any rational in [0, 1]. If x ∈ U r then x also belongs to all “lower”
Uq (i.e. for q < r, because then U r ⊂ Uq ), and so f (x) ≥ r:
if x ∈ U r then f (x) ≥ r
On the other hand, if x ∈
/ Ur then x is certainly not in any “higher” Uq (i.e.
for q > r), and so f (x) ≤ r:
if x ∈
/ Ur then f (x) ≤ r
In particular, if r < s are rationals in [0, 1] then on the set Ur − Us , the
function f takes values in [r, s].
Now consider any point x0 ∈ X, and any neighborhood (a, b) of f (x0 ).
Assume for the moment that 0 < f (x0 ) < 1. Let r, s ∈ (0, 1) be rationals
with a < r < f (x0 ) < s < b. Since f (x0 ) > r it follows that x0 ∈ Ur . On the
other hand, since f (x0 ) < s it also follows that x0 ∈
/ U s . Thus x0 belongs
to the open set Ur − U s , and on this open neighborhood of x0 the function
f takes values in [r, s] ⊂ (a, b). This proves that f is continuous at x0 if
f (x0 ) ∈ (0, 1).
Now suppose f (x0 ) = 0. Consider any neighborhood (a, b) of 0. Let s be
any rational number with 0 < s < b. Since f (x0 ) = 0 < s, the point x0 is
c
outside U s , i.e. x0 is in the open set U s . Moreover, on this open set f has
c
values ≤ s < b. Thus on the open neighborhood U s of x0 , the function f
takes values in (a, b). So f is continuous at x0 . The argument for the case
f (x0 ) = 1 is similar. QED
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CHAPTER 2.
2.6
TOPOLOGICAL SPACES
Product Spaces and Tikhonov’s Theorem
Consider two topological spaces Xa and Xb . Then for the Cartesian product
Xa × Xb we have two projection maps
πa : Xa × Xb → Xa : (xa , xb ) 7→ xa ,
πb : Xa × Xb → Xb : (xa , xb ) 7→ xb
The smallest topology on Xa × Xb which makes both the maps πa and πb
continuous is called the product topology on Xa × Xb .
Thus if Ua is any open subset of Xa and Ub is any open subset of Xb then
πa−1 (Ua ) = Ua × Xb ,
and
πb−1 (Ub ) = Xb × Ub
are open subsets of Xa × Xb in the product topology, and hence so is their
intersection
πa−1 (Ua ) ∩ πb−1 (Ub ) = Ua × Ub
Indeed, the product topology on Xa × Xb is precisely the collection of all
subsets of Xa × Xb which are unions of sets of the form Ua × Ub , with Ua
open in Xa and Ub open in Xb .
More generally, consider an indexing set I 6= ∅ and a topological space
Xα for each α ∈ I. The Cartesian product set
Y
Xα
α∈I
is the set of all maps
x : I → ∪α∈I Xα : α 7→ xα
for which xα ∈ Xα for all α ∈ I. The element xα ∈ Xα is the α–th coordinate
Q
of the element x ∈ α∈I Xα . The map
πα :
Y
Xα → Xα : x 7→ xα
α∈I
Q
is the α–th coordinate of x. The product topology on α∈I Xα is the smallest
Q
topology on α∈I Xα for which the projection maps πα are all continuous.
Q
A subset of α∈I Xα is open in the product topology if and only if it is
the union of sets of the form
πα−11 (Uα1 ) ∩ · · · ∩ πα−1n (Uαn )
with n running over positive integres, α1 , ..., αn running over I, and Uαi runs
over open sets Xαi .
2.6. PRODUCT SPACES AND TIKHONOV’S THEOREM
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Theorem 2 Suppose Xα is a compact Hausdorff space for each α ∈ I, where
Q
I is any non-empty indecing set. Then the product space α∈I Xα is a compact Hausdorff space.