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}

1 2 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 neigh- borhoods: 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

−1 f (OY ) ⊂ OX , 2.3. COMPACT SETS 3 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 S0 ⊂ 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 Cc along with S. Then S ∪ {Cc} is an open cover of the compact set K and so has a finite subcover S0. The S0 is also automatically a finite cover of C, and remains so if we discard Cc from S0 in case S0 does contain Cc. This yields a finite subset of S which still covers C. QED

Lemma 2 In a Hausdorff space X: 4 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 Wy 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 ⊂ Vy = Ux1 ∪ · · · ∪ UxN

On the other hand we have the open set

Wy = Fy1 ∩ · · · ∩ FyN 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 Kc 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 neigh- borhoods, 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 6 CHAPTER 2. TOPOLOGICAL SPACES 2.5 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 ≤ 1U

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 8 CHAPTER 2. TOPOLOGICAL SPACES 2.6 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 −1 −1 πa (Ua) = Ua × Xb, and πb (Ub) = Xb × Ub are open subsets of Xa × Xb in the product topology, and hence so is their intersection −1 −1 πa (Ua) ∩ πb (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 of the element x ∈ Qα∈I Xα. The map

πα : Y Xα → Xα : x 7→ xα α∈I is the α–th coordinate of x. The product topology on Qα∈I Xα is the smallest topology on Qα∈I Xα for which the projection maps πα are all continuous. A subset of Qα∈I Xα is open in the product topology if and only if it is the union of sets of the form −1 −1 πα1 (Uα1 ) ∩ · · · ∩ παn (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 9

Theorem 2 Suppose Xα is a compact Hausdorff space for each α ∈ I, where I is any non-empty indecing set. Then the product space Qα∈I Xα is a com- pact Hausdorff space.