In the previous article, we defined compact spaces as those where every open cover has a finite subcover, i.e. if then we can find a finite set of indices such that 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 is compact if and only if :

- for any collection of open subsets of X such that there exists a finite set of indices such that

**Proof**

(→) : *Y* is compact. If we have and each is open in *Y*. Thus, there’s a finite subcover which gives

(←) : let be an open cover of *Y*. Each for some open subset *U _{i}* of

*X*. Now By assumption, there’s a finite set of indices such that which gives ♦

Thus, 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 is a collection of open subsets of *X* such that Then together with *X*–*Y* is an open cover of *X*. Hence, there’s a finite subcover of *X*:

By the lemma, *Y* is compact.

2nd statement: suppose *x* is a point of accumulation of *Y* and .

- For each
*y*in*Y*, separate*x*and*y*by open subsets of*X*: - Since and
*Y*is compact, we can find a finite subcover: - Let which is open in
*X*and contains*x*. Since for each*k*, we have 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 be a collection of open subsets of *Y* such that Letting we have

Thus, there’s a finite subcover which gives By the above lemma, *f*(*X*) is compact. ♦

Proposition 3.If are compact, then so is Thus a finite union of compact subspaces is compact.

**Proof**.

Let be an open cover for Then since and *Y* is compact, there’s a finite set of indices such that Likewise, there’re indices such that Then so 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 in *X* must have a subsequence which converges to *a* in *X*. Since is Cauchy, this implies ♦

**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 ofRis compact if and only if it’s closed inRand 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 , continuity at *a* means there’s a δ>0 such that Now the collection of the open balls covers *X*, so it has a finite subcover

where

Let Now whenever

- for some 1 ≤
*i*≤*n*; - then ;
- thus and ;
- so

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 which covers X, has a finite subcover.

**Proof**.

Suppose the given condition holds. To prove compactness, let {*U _{i}*} be an open cover of

*X*. Write each as a union of basic open sets Then and there must be a finite subcover

Since we have ♦

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 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 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 the collection 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 satisfy then we have for some
*i*. [ Indeed, if each then maximality implies has a finite subcover, say where each Then is a finite subcover of Σ. Contradiction. ]

Now we claim that Σ ∩ *S* covers *X*. For if we can find a such that Since *V* is a basic open set, write for The above condition then implies for some *i*. Since we see that

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 is a collection of compact spaces, then is compact.

**Proof**.

For index *i* and open subset the collection of open slices forms a subbasis for *X*. Suppose we have an open cover of *X* of the form {*V _{i}*}. Then for some index

*i*, the union of

*U*‘s which appear must be the whole of

_{i}*X*; otherwise, for each

_{i}*i*, pick some and the tuple is not in the union of {

*V*}. By compactness of

_{i}*X*, there’s a finite subcover via the {

_{i}*U*}, so

_{i}*X*has a finite subcover via the {

*V*}. ♦

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

Corollary (Heine-Borel Theorem). A closed and bounded subset X ofRis compact.

**Proof**.

By scaling, we assume 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 Since:

and

we have Then there exist *i*, *j* such that so there’s a finite subcover 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 then either α ≤ β or β ≤ α.

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

- whenever is a totally ordered subset, there’s an upper bound of T, i.e. s ≥ t for all
Then S has a maximal element u, i.e. u ≥ s for all

It can be proven that Zorn’s lemma is equivalent to the **Axiom of Choice**: if is a collection of non-empty sets, then the product 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:

(*) is an open cover of *X* without any finite subcover.

- First, order all Σ satisfying (*) by inclusion, i.e.
- Suppose we have a collection of which is totally ordered and each satisfies (*). Totally ordered means for any
*i*,*j*, either or - Let We’ll prove Σ satisfies (*) as well.
- Indeed, if Σ had a finite subcover {
*V*}, then each_{i}*V*is from some Since there’re only finitely many such_{i}*i*‘s and is totally ordered, one is bigger than all others and contains all*V*‘s, and thus has a finite subcover (contradiction)._{i}

- Indeed, if Σ had a finite subcover {
- By Zorn’s lemma, there’s a maximal Σ satisfying (*). For any the cover 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).