(H) Lecture 6 Lecturer: Zuoqin Wang Time: April 20, 2020

COMPACTNESS: DEFINITIONS AND BASIC PROPERTIES

Last week we learned: • Basic conceptions characterizing topology: – Open sets, closed sets, neighborhoods – the closure, the interior, the boundary, the derived set – Convergence of a sequence, continuity of a map – Equivalent ways to characterize continuity

Methods: via Objects: topological spaces, Morphisms: continuous maps = sets satisfying = maps satisfying a open sets (O1) - (O3) f −1(open) = open. b b a closed sets (C1) - (C3) f −1(closed) = closed. b b a −1 b neighborhoods (N1) - (N4) f (neighborhoods of f(x)) b = neighborhoods of x. a closure (K1) - (K4) f(A) ⊂ f(A). b b a −1 −1 b interior (I 1) - (I 4) f (Int(A)) ⊂ Int(f (A)) b • Examples of – Tmetric, Tdiscrete, Ttrivial, Tcofinite, TSorgenfrey, – Different topologies on the product space: Tp.c. = Tproduct, Tbox – topologies defined by bases, by subbases and by maps You should be aware of new phenomena that appear in topological spaces, e.g. • a convergent sequence could have more than one limit. • limit point need not be the limit of any sequence (called a sequential limit) This week we will learn compactness: • Various kinds of compactness • Tychonoff theorem and its applications • Compactness in metric space

1 2 COMPACTNESS: DEFINITIONS AND BASIC PROPERTIES

1. Compactness: various definitions and examples As we have mentioned in Lecture 1, compactness is a generalization of finiteness. The simplest compact sets are finite sets. The next simplest compact set are the bounded closed intervals. Let’s compare finite sets, [0, 1] and (0, 1):

a X = a finite set b X = [0, 1] X = (0, 1) b a b Any real valued function (Extremal value property) continuous functions b f : X → R is bound- Any real valued continu- f : X → R could be ed, and attains its maxi- ous function f : X → R is unbounded, or bounded

mal/minimal values. bounded, and attains its withouta extremal value. maximal/minimal values. what is b b a b Any sequence x1, x2, ··· (Bolzano-Weierstrass) ∃ “divergent sequence” b in X has a constant subse- Any sequence x1, x2, ··· whose limit is not in X, 1 1 quence xn1 = xn2 = ··· = in X has a conver- e.g. 1, 2 , 3 , · · · → 0 ∈/ X. c ∈ X. gent subsequence

xn1 , xn2 , · · · → c ∈ X a b Any infinite subset b A ⊂ X has a limit point c ∈ X. S a S b ∞ 1 If X = α Vα, then b S(Heine-Borel) If X = X = n=1( n , 1) but no fi- ∃ V , ··· ,V s.t. X = V , where each V is nite covering. S α1 αk α α α k open, then ∃ V , ··· ,V i=1 Vαi . S α1 αk k s.t. X = i=1 Vαi .

Remark. (1) Why the finiteness/compactness is important? Because you can get global information from local information. (local-to-global principal) For example, the extremal value property: local (pointwise) bounded (continuity) + compact ⇒ global bounded! (2) It turns out that all the four properties for X = [0, 1] are equivalent. But they are not equivalent in general topological spaces.

Before we give the definition of various compactness, we need the following con- ception of covering:

Definition 1.1. Let (X, T ) be a , and A ⊂ X be a subset. S • A family of subsets U = {Uα} is called a covering of A if A ⊂ α Uα. • A covering U is called a finite covering if it is a finite collection. • We say a covering V is a sub-covering of U if V ⊂ U . • A covering U is called an open covering if each Uα is open. COMPACTNESS: DEFINITIONS AND BASIC PROPERTIES 3

Definition 1.2. Let (X, T ) be a topological space.

(1) We say X is compact in X if any open covering U = {Uα} of X admitsS a finite k 1 sub-covering, i.e. there exists {Uα1 ,Uα2 , ··· ,Uαk } ⊂ U s.t. X = i=1 Uαi . (2) We say X is sequentially compact if any sequence x1, x2 · · · ∈ X admits a

convergent subsequence xn1 , xn2 , · · · → x0 ∈ X. (3) We say X is limit point compact if for any infinite subset S ⊂ X, S0 6= ∅. Remark. Suppose A ⊂ X be a subset, then we say A is compact/sequentially com- pact/limit point compact if, when endowed with the subspace topology, (A, Tsubspace) is compact/sequentially compact/limit point compact. Note that by definition,

A ⊂ X is compact if andS only if for any family of open sets U = {Uα} in X satisfying A ⊂ U , one can find U , ··· ,U ∈ U such that S α α α1 αk k A ⊂ j=1 Uαj . Example. (1) Unbounded subsets in Rn, and non-closed subsets in Rn are not compact, not sequentially compact, not limit point compact. (2) bounded closed subsets in Rn are compact, sequentially compact, limit point compact. (3)( X, Tcofinite) is compact:

Suppose X ⊂ ∪αUα. Choose arbitrary α1. By definition, Uα1 is open,

so its complement X \Aα1 is a finite set. Now one only need to choose finitely many elements in A to cover it. It is also sequentially compact: This is a consequence of Example 3(a) in Lecture 3, page 6: Consider any sequence x1, x2, ··· . If no element repents infinitely many times, then the whole sequence converges to any point. If at least one el- ement repent infinitely many times, then that gives us a “constant” subsequence which converges to that element itself. And it is limit point compact: Reason: For any infinite set S, S0 = X since U∩S 6=∅ for any open U. (4) X = (N, Tdiscrete) × (N, Ttrivial) is NOT compact: Reason: Let Un = {n} × N. Then {Un}n∈N is an open covering which has no finite sub-covering. It is also NOT sequentially compact. Reason: The sequence {xn = (n, 1)} has no convergent subsequence. BUT: it is limit point compact: 0 Reason: In fact, for any S 6= ∅, we have S 6= ∅, since if (m0, n0) ∈ S 0 0 and n1 6= n0, then (m0, n1) ∈ {(m0, n0)} ⊂ S .

1The definition of compactness via covering property of open sets was first introduced by Alexan- droff and Urysohn in 1924. 4 COMPACTNESS: DEFINITIONS AND BASIC PROPERTIES

It is not too hard to see that “limit point compact” is the weakest among the three: Proposition 1.3. Let X be any topological space. (1) If X is compact, then it is limit point compact. (2) If X is sequentially compact, then it is limit point compact. Proof. (1) Suppose X is compact, and S ⊂ X is any subset. Suppose S has no limit point. Then S is closed since S0 = ∅ ⊂ S. For each point a ∈ S, since a 6∈ S0, c one can find an open set Ua ⊂ X s.t. S ∩ Ua = {a}. Now {S ,Ua | a ∈ S} is an open covering of X. By compactness, there exists a1, ··· , ak ∈ S such that c k X = S ∪ (∪i=1Uai ). It follows k S = S ∩ X = S ⊂ (∪i=1Uai ) ∩ S = {a1, ··· , ak},

which implies S = {a1, ··· , ak} is finite. (2) Suppose X is sequentially compact, and S ⊂ X is any infinite set. Take any infinite sequence {x1, x2, · · · } ⊂ A s.t. xi 6= xj for i 6= j. Then there exists a

subsequence xn1 , xn2 , · · · → x0 ∈ X. It follows from definition that 0 0 0 x0 ∈ {xn1 , xn2 , ···} ⊂ {x1, x2, ···} ⊂ S . So S0 6= ∅.  Remark. (1) We will see later: for topological spaces, • “compact : sequentially compact”, • “compact ; sequentially compact”. (2) We will prove: for metric spaces, “compact ⇔ sequentially compact ⇔ limit point compact”.

2. Compactness v.s. continuous map Among the three different compactness, compactness and sequentially compactness are more important because they are preserved under continuous maps: Proposition 2.1. Let f : X → Y be continuous. (1) If A ⊂ X is compact, then f(A) is compact in Y . (2) If A ⊂ X sequentially compact, then f(A) is sequentially compact.

Proof. (1) Suppose A is compact. Given any open covering V = {Vα} of f(A), the −1 pre-image U = {f (VαS)} is an open covering of A.S By compactness, there exists k −1 k α1, ··· , αk such that A⊂ i=1 f (Vαi ). It follows f(A)⊂ i=1 Vαi , i.e. f(A) is compact. (2) For any sequence y1, y2, ··· in f(A), there exists x1, x2, ··· in A such that f(xi) = yi. Since A is sequentially compact, there exists a convergent subsequence COMPACTNESS: DEFINITIONS AND BASIC PROPERTIES 5

xn1 , xn2 , · · · → x0 ∈ A. It follows from the continuity of f that yn1 , yn2 , · · · → f(x0) ∈ f(A). So f(A) is sequentially compact.  Remark. The image of a limit point need not be limit point compact. For example, as we have just seen, the product space X = (N, Tdiscrete) × (N, Ttrivial) is limit point compact. We also know that the projection

π1 :(N, Tdiscrete) × (N, Ttrivial) → (N, Tdiscrete) is continuous. However, the image (N, Tdiscrete) is not limit point compact because it 2 is an unbounded subset of R.

Since a subset in R is compact (with respect to the usual topology) if and only if it is sequentially compact if and only if it is bounded and closed, we immediately get

Corollary 2.2 (The extremal value property). Let f : X → R be any continuous map. If A ⊂ X is compact or sequentially compact in X, then f(A) is bounded in R. Moreover, there exists a1, a2 ∈ A s.t. f(a1) ≤ f(x) ≤ f(a2) for ∀x ∈ A.

Proof. By the previous proposition, f(A) is bounded and closed in R. The conclusion follows.  In general, the pre-image of a compact set under a continuous map is no longer compact. Such examples are easy to construct. Definition 2.3. Let X,Y be topological spaces. A map f : X → Y is called a proper map if for any compact set B ⊂ Y , its pre-image f −1(B) is compact in X. Remark. Why do we study proper maps? Here is one reason: Recall that the morphism between topological spaces are continuous maps, since they pull-back open sets to open sets. On the other hand, usually topological properties of compact sets are easier (since we have the “local-to-global” principle). As a result, some topological invariants are defined only for compact objects, or for “compactly- supported” objects. For the latter case, the correct “morphism” should be continuous proper maps, since proper maps can pull-back compactly-supported objects to compactly supported objects. This is the case, for example, in studying compactly supported cohomology groups.

3. New compact spaces from old As usual, we would like to construct new compact spaces from old compact spaces, or even non-compact spaces. The first candidates one can look at is: subspaces of a compact space. Unfortunately, it is easy to see that a compact space could have non-compact subspace, e.g. (0, 1) is a subspace of [0, 1].

2 Note: The discrete topology on N is the same as the subspace topology on N when viewed as a subset in R. 6 COMPACTNESS: DEFINITIONS AND BASIC PROPERTIES

We can take a closer look at the problem: which subsets of [0, 1] remain to be compact? We know that a set in R is compact if and only if it is bounded and closed. If A is a subset of [0, 1], it is automatically bounded. So for a subset A ⊂ [0, 1] to be compact, it is enough to require A to be closed. It turns out that for more general topological spaces, it is also enough to require closedness for a subset to be compact: Proposition 3.1. Let X be compact, and A ⊂ X be closed, then A is compact.

Proof. Let A ⊂ X be closed. Then for any open covering U of A, U ∪ {Ac} is c an open covering of X, which admits a finite sub-covering U1, ··· ,Um,A . It follows A ⊂ U1 ∪ · · · ∪ Um, i.e. {U1, ··· ,Um} is a finite sub-covering of U.  Remark. (1) According to Proposition 2.1 and Lemma 3.2 in Lecture 5, it is easy to see: • If X is sequentially compact, and A ⊂ X is closed, then A is sequentially compact. • If X is limit point compact, and A ⊂ X is closed, then A is limit point compact. (2) However, it is important to point out that a compact set could have non-closed compact subsets. For example, for (X, Ttrivial), any subset is compact. Next let’s consider products of compact spaces. We will show that “compactness” is preserved by the (finite) product operation.3 We need Lemma 3.2 (The tube lemma). If A ⊂ X,B ⊂ Y are compact, N ⊂ X × Y is open, and A × B ⊂ N, then there exists open sets U in X and V in Y s.t. A × B ⊂ U × V ⊂ N.

Proof. We first prove a “basic tube lemma”: the tube lemma holds for A = {x0}. For y0 any (x0, y0) ∈ X × Y, one can find an open set Ux0 in X and an open set Vy0 in Y such that (x , y ) ∈ U y0 × V ⊂ N. S S 0 0 x0 y0 Since B = y∈B{y} ⊂ y∈Y Vy, we can find y1, ··· , yk ∈ Y s.t. [k

B ⊂ Vyi =: V. i=1 T k yi yi Let U = i=1 Ux0 , then U is an open neighborhood of x0 and U ⊂ Ux0 , ∀1 ≤ i ≤ k. Moreover, [ [k [k [k y yi N ⊃ (Ux0 × Vy) ⊃ (Ux0 × Vyi ) ⊃ (U × Vyi ) = U × Vyi = U × V. y i=1 i=1 i=1

3Amazingly, the same conclusion holds for the product of infinitely many compact spaces, as long as we endow the product space with the ! We will prove this next time. COMPACTNESS: DEFINITIONS AND BASIC PROPERTIES 7

Now we prove the general tube lemma. For each x0 ∈ A, by the basic tube lemma, there exists open sets Ux0 and Vx0 s.t.

{x0} × B ⊂ Ux0 × Vx0 ⊂ N.

Since A is compact, there exist x1, ··· , xm ∈ A s.t.

A ⊂ Ux1 ∪ · · · ∪ Uxm =: U. m Let V = ∩i=1Vxi , then B ⊂ V and V is open. So [m

A × B ⊂ U × V ⊂ (Uxi × Vxi ) ⊂ N. i=1  Remark. You should see how the “local-to-global principal” is used in the proof. As a consequence, we prove Proposition 3.3. If A ⊂ X,B ⊂ Y are compact, so is A × B.

Proof. Let W be any open covering of A × B. For any x0 ∈ A, since {x0} × B is 4 (0) (0) compact , one can find W1 , ··· ,Wk ∈ W s.t. (0) (0) {x0} × B ⊂ W1 ∪ · · · ∪ Wk .

By the basic tube lemma, there exists an open set Ux0 in X containing x0 s.t. (0) (0) Ux0 × B ⊂ W1 ∪ · · · ∪ Wk .

Now {Ux0 | x0 ∈ A} is an open covering of A. By compactness, there exist x1, ··· , xm s.t. A ⊂ Ux1 ∪ · · · ∪ Uxm . According to the argument above, for 1 ≤ i ≤ m, (i) (i) we already find W1 , ··· ,Wk(i) ∈ W s.t. (i) (i) Uxi × B ⊂ W1 ∪ · · · ∪ Wk(i). It follows that [ (i) A × B ⊂ (Ux1 ∪ · · · ∪ Uxm ) × B ⊂ Wj , 1≤i≤m,1≤j≤k(i) (i) i.e. {Wj | 1 ≤ i ≤ m, 1 ≤ j ≤ k(i)} is a finite sub-covering of W . 

Corollary 3.4. If X1, ··· ,Xk are compact, so is X1 × · · · × Xk.

4 You can prove this directly by definition, or by realizing {x0} × B as the image of the compact set B under a continuous map, namely, the embedding map from Y to X × Y mapping y to (x0, y).