## Topology: More on Compact Spaces

In the previous article, we defined compact spaces as those where every open cover has a finite subcover, i.e. if $X = \cup_i U_i,$ then we can find a finite set of indices $i_1, i_2, \ldots, i_n$ such that $X = \cup_{k=1}^n U_{i_k}.$ On an intuitive level, one should imagine a compact space as being constrained, and is the topological equivalent of a finite set.

We also showed that in the case of metric spaces, compactness is equivalent to sequential compactness. Thus, the two concepts are very similar. In fact, we’ll see a parallel between the properties of compact spaces and the corresponding properties of sequentially compact spaces.

First, a simple result:

Lemma. A subspace $Y\subseteq X$ is compact if and only if :

• for any collection of open subsets $\{U_i\}$ of X such that $\cup_i U_i \supseteq Y,$ there exists a finite set of indices $i_1, i_2, \ldots, i_n$ such that $\cup_{k=1}^n U_{i_k}\supseteq Y.$

Proof

(→) : Y is compact. If $\cup_i U_i\supseteq Y$ we have $\cup_i (U_i\cap Y) = Y$ and each $U_i\cap Y$ is open in Y. Thus, there’s a finite subcover $\cup_{k=1}^n (U_{i_k}\cap Y) = Y$ which gives $\cup_{k=1}^n U_{i_k} \supseteq Y.$

(←) : let $\{V_i\}$ be an open cover of Y. Each $V_i = U_i\cap Y$ for some open subset Ui of X. Now $\cup_i V_i = Y \implies \cup_i U_i \supseteq Y.$ By assumption, there’s a finite set of indices $i_1, \ldots, i_n$ such that $\cup_{k=1}^n U_{i_k}\supseteq Y$ which gives $\cup_{k=1}^n V_{i_k}=Y.$

Thus, $Y\subseteq X$ is compact if and only if every open cover of Y has a finite subcover, where “open cover” means a collection of open subsets of X whose union contains Y.

## Properties of Compact Spaces

Next, we have:

Proposition 1.

• If Y is a closed subset of compact space X, then Y is compact.
• If Y is a compact subset of Hausdorff space X, then Y is closed in X.

Proof

1st statement: suppose $\{U_i\}$ is a collection of open subsets of X such that $\cup_i U_i\supseteq Y.$ Then $\{U_i\}$ together with XY is an open cover of X. Hence, there’s a finite subcover of X:

$U_{i_1}\cup U_{i_2}\cup \ldots \cup U_{i_n} \cup (X-Y) = X \implies \cup_{k=1}^n U_{i_k} \supseteq Y.$

By the lemma, Y is compact.

2nd statement: suppose x is a point of accumulation of Y and $x\not\in Y$.

• For each y in Y, separate x and y by open subsets of X: $x\in U_y, y\in V_y, U_y\cap V_y=\emptyset.$
• Since $Y \supseteq \cup_{y\in Y} V_y$ and Y is compact, we can find a finite subcover: $Y \supseteq \cup_{k=1}^n V_{y_k}.$
• Let $U = \cap_{k=1}^n U_{y_k}$ which is open in and contains x. Since $U_{y_k}\cap V_{y_k} = \emptyset$ for each k, we have $U\cap Y=\emptyset,$ which contradicts the fact that x is a point of accumulation of Y. ♦

Note

Counter-examples exist for the second statement if X is not Hausdorff. E.g. if X = {1, 2} with the coarsest topology and Y = {1}, then Y is compact but not closed in X.

Proposition 2. If f : X → Y is a continuous map and X is compact, then f(X) is compact.

Proof.

Let $\{V_i\}$ be a collection of open subsets of Y such that $\cup_i V_i \supseteq f(X).$ Letting $U_i = f^{-1}(V_i),$ we have

$\cup_i U_i = \cup_i f^{-1}(V_i) = f^{-1}(\cup_i V_i) = X.$

Thus, there’s a finite subcover $\cup_{k=1}^n U_{i_k} = X$ which gives $\cup_{k=1}^n V_{i_k} \supseteq f(X).$ By the above lemma, f(X) is compact. ♦

Proposition 3. If $Y, Z\subseteq X$ are compact, then so is $Y\cup Z.$ Thus a finite union of compact subspaces is compact.

Proof.

Let $\{U_i\}$ be an open cover for $Y\cup Z.$ Then since $\cup_i U_i \supseteq Y$ and Y is compact, there’s a finite set of indices $i_1, \ldots, i_n$ such that $\cup_{k=1}^n U_{i_k} \supseteq Y.$ Likewise, there’re indices $i_{n+1}, \ldots, i_m$ such that $\cup_{k=n+1}^m U_{i_k} \supseteq Z.$ Then $\cup_{k=1}^m U_{i_k} = Y\cup Z$ so $\{U_i\}$ has a finite subcover. ♦

It’s also true that a product of compact spaces is compact. In the case of finite products, the proof isn’t hard, but what we’d like to show is the case of arbitrary products. Known as Tychonoff’s theorem, its proof is tricky and we’ll leave it till later. For now the reader may try to prove as an exercise that if X and Y are compact, then so is X × Y.

Proposition 4. A compact metric space X is complete.

Proof.

X is sequentially compact. Thus, a Cauchy sequence $(x_n)$ in X must have a subsequence which converges to a in X. Since $(x_n)$ is Cauchy, this implies $(x_n)\to a.$ ♦

Note.

Clearly compactness is a stronger condition than completeness. E.g. R is complete but not compact. On an intuitive level, completeness says “if a sequence of points get successively closer, they must converge” while compactness says “in every sequence, some of the points must converge”.

Heine-Borel Theorem. A subspace X of R is compact if and only if it’s closed in R and bounded.

Proof.

Since X is a metric space, it is compact iff it’s sequentially compact. And by corollary 4 of previous article, this is true iff X is closed in R and bounded. ♦

Finally for metric spaces, we have:

Theorem. If f : (X, d) → (Y, d’) is a continuous map of metric spaces and X is compact, then f is uniformly continuous.

Proof.

Let ε>0. For each $a\in X$, continuity at a means there’s a δ>0 such that $f(N_X(a, 2\delta)) \subseteq N_Y(f(a), \epsilon).$ Now the collection of the open balls $N_X(a, \delta)$ covers X, so it has a finite subcover

$X = N_X(a_1, \delta_1) \cup\ldots \cup N_X(a_n, \delta_n),$ where $f(N_X(a_i, 2\delta_i))\subseteq N_Y(f(a),\epsilon).$

Let $\delta = \min(\delta_i).$ Now whenever $d_X(u, v) <\delta,$

• $u\in N_X(a_i, \delta_i)$ for some 1 ≤ i ≤ n;
• then $d_X(a_i ,v) \le d_X(a_i, u) + d_X(u, v) < \delta_i + \delta \le 2\delta_i$;
• thus $d_Y(f(a_i), f(u)) < \epsilon$ and $d_Y(f(a_i), f(v)) < \epsilon$;
• so $d_Y(f(u), f(v)) < 2\epsilon.$

And we’re done. ♦

## Alexander’s Subbase Theorem

If we fix a basis B of a given topological space X, in determining whether every open cover of X has a finite subcover, we may assume this cover is from B.

Proposition 5. X is compact if and only if:

• any collection of basic open sets $\{V_i\}\subseteq B$ which covers X, has a finite subcover.

Proof.

Suppose the given condition holds. To prove compactness, let {Ui} be an open cover of X. Write each $U_i = \cup_j V_{ij}$ as a union of basic open sets $V_{ij}\in B.$ Then $X = \cup_{i,j} V_{ij}$ and there must be a finite subcover

$X = V_{i_1 j_1} \cup V_{i_2 j_2} \cup \ldots \cup V_{i_n j_n}.$

Since $V_{i_k j_k}\subseteq U_{i_k}$ we have $X = \cup_{k=1}^n U_{i_k}.$ ♦

Alexander’s subbase theorem says we can push this to the level of a subbasis!

Alexander’s Subbase Theorem. If S is a subbasis of X, then X is compact if and only if:

•  any collection of subbasic open sets $\{W_j\} \subseteq S$ which covers X, has a finite subcover.

Proof.

The proof relies on a non-trivial set-theoretic result known as Zorn’s lemma. Let B be the basis obtained from taking finite intersections of elements from S. If X is not compact, by proposition 5, there’s a collection $\Sigma := \{V_i\} \subseteq B$ of basic open sets which covers X but with no finite subcover. We’ll assume Σ is “maximal”, i.e.

• Σ has no finite subcover, but for any $V\in B-\Sigma,$ the collection $\Sigma\cup \{V\}$ has a finite subcover, which must include V. [ Note: we don’t assume there’s a unique maximal Σ. ]

We’ll defer its discussion till later and urge the reader to just accept it for now. Maximality of Σ then implies:

• If $V_1, \ldots, V_n\in B$ satisfy $V_1\cap \ldots\cap V_n\in \Sigma,$ then we have $V_i\in \Sigma$ for some i. [ Indeed, if each $V_i\in B-\Sigma,$ then maximality implies $\Sigma\cup \{V_i\}$ has a finite subcover, say $X = (\cup_j V_{ij}) \cup V_i$ where each $V_{ij} \in \Sigma.$ Then $X = (\cup_{i, j} V_{ij})\cup V$ is a finite subcover of Σ. Contradiction. ]

Now we claim that Σ ∩ S covers X. For if $x\in X,$ we can find a $V\in\Sigma$ such that $x\in V.$ Since V is a basic open set, write $V = W_1\cap \ldots \cap W_n$ for $W_i\in S.$ The above condition then implies $W_i\in \Sigma$ for some i. Since $x\in W_i,$ we see that

$X = \cup_{W\in \Sigma\cap S} W,$

By the given condition, Σ ∩ S has a finite subcover, so Σ has a finite subcover, which is a contradiction. ♦

Corollary (Tychonoff’s Theorem). If $\{X_i\}$ is a collection of compact spaces, then $X:=\prod_i X_i$ is compact.

Proof.

For index i and open subset $U_i\subseteq X_i,$ the collection of open slices $V_i := (\prod_{j\ne i} X_j) \times U_i$ forms a subbasis for X. Suppose we have an open cover of X of the form {Vi}. Then for some index i, the union of Ui‘s which appear must be the whole of Xi; otherwise, for each i, pick some $x_i\in X_i-\cup U_i$ and the tuple $(x_i)\in X$ is not in the union of {Vi}. By compactness of Xi, there’s a finite subcover via the {Ui}, so X has a finite subcover via the {Vi}. ♦

Curiously, we also get a much easier proof of the Heine-Borel theorem.

Corollary (Heine-Borel Theorem). A closed and bounded subset X of R is compact.

Proof.

By scaling, we assume $X\subseteq [0, 1].$ By proposition 1, it suffices to prove X = [0, 1] is compact. Since we can pick a subbasis of R comprising of (-∞, b) and (a, ∞), we can pick a subbasis of X comprising of [0, b) and (a, 1]. Now suppose there’s an open cover of X of the form $\{[0, b_i)\}_i \cup \{(a_j, 1]\}_j.$ Since:

$\cup_i [0, b_i) = [0, \sup_i b_i)$ and $\cup_j (a_j, 1] = (\inf a_j, 1]$

we have $\sup_i b_i > \inf_j a_j.$ Then there exist ij such that $b_i > a_j$ so there’s a finite subcover $\{[0, b_i), (a_j, 1]\}.$ By Alexander’s Subbase Theorem, X is compact. ♦

## Zorn’s Lemma

It’s time to patch our proof of the subbase theorem. First, let’s state our main tool. Recall that a partially ordered set (S, ≤) is totally ordered if any two elements are comparable, i.e. if $\alpha, \beta \in T,$ then either α ≤ β or β ≤ α.

Zorn’s Lemma. Suppose (S, ≤) is a partially ordered set satisfying:

• whenever $T\subseteq S$ is a totally ordered subset, there’s an upper bound $s\in S$ of T, i.e. s ≥ t for all $t\in T.$

Then S has a maximal element u, i.e. u ≥ s for all $s\in S.$

It can be proven that Zorn’s lemma is equivalent to the Axiom of Choice: if $\{A_i\}$ is a collection of non-empty sets, then the product $\prod_i A_i$ is non-empty. This looks deceptively obvious, so we’ll just take it at face value while sweeping the philosophical ramifications under the rug (or continue the discussion another time). Read the wikipedia entry for an idea of how involved the discussion can get.

In any case, we had already surreptitiously used the Axiom of Choice without mentioning it, e.g. while proving that the closure of the product of subsets is the product of the closure (proposition 6 here). We won’t mention exactly where, and leave it to the reader to find out. 🙂 ]

Anyway, assuming Zorn’s Lemma, we’ll show that there’s a maximal Σ which satisfies:

(*) $\Sigma\subseteq B$ is an open cover of X without any finite subcover.

• First, order all Σ satisfying (*) by inclusion, i.e. $\Sigma\le \Sigma' \iff \Sigma \subseteq \Sigma'.$
• Suppose we have a collection of $\{\Sigma_i\}$ which is totally ordered and each $\Sigma_i$ satisfies (*). Totally ordered means for any ij, either $\Sigma_i\subseteq \Sigma_j$ or $\Sigma_j \subseteq \Sigma_i.$
• Let $\Sigma = \cup_i \Sigma_i.$ We’ll prove Σ satisfies (*) as well.
• Indeed, if Σ had a finite subcover {Vi}, then each Vi is from some $\Sigma_i.$ Since there’re only finitely many such i‘s and $\{\Sigma_i\}$ is totally ordered, one $\Sigma_i$ is bigger than all others and contains all Vi‘s, and thus has a finite subcover (contradiction).
• By Zorn’s lemma, there’s a maximal Σ satisfying (*). For any $V\in B-\Sigma$ the cover $\Sigma\cup \{V\}$ fails (*) and must have a finite subcover (which must include V since Σ has no subcover).

That’s all for now. We’ll have plenty of opportunities to use Zorn’s Lemma in other topics, e.g. in proving every vector space has a basis (if you think this is obvious, try finding a basis for the space V of all functions N → R; and no, the collection of delta functions don’t work since we can’t use infinite sums).

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