Compactness is one of the most important concepts in point-set topology. However when you first come across it it might not necessarily be obvious why. Paul Crowley asked me yesterday to do a follow-up post to my one about continuous functions to explain compactness, so I spent some time this morning thinking about how one might make it seem more intuitive.

I’m not sure this is that post, but it’s an interesting insight I had while trying to figure that post out. Another more comprehensive one may follow.

There are a variety of induction principles. The most well known one is that if you prove that something is true for 0 and that if it’s true for n it’s true for n + 1 then it must be true for all natural numbers. This is a special case of the induction principle for well ordered sets: In a well-ordered set, if p being true for all y < x implies that p is true for x, then p is true for all x.

The common thread is that induction allows you to conclude that something is true for the larger case by showing it’s true for smaller pieces and that the set of things for which it is true is closed under some operation.

I noticed this morning that there is a topological induction principle that’s fairly straightforwardly equivalent to compactness. What it essentially says is that compactness is the property that lets you conclude that things that are true locally are true globally.

Theorem: A topological space \(X\) is compact if and only if for every property of sets \(p\) such that:

For all \(x \in X\) there is some open set \(U\) with \(x \in U\) and \(p(U)\). If \(p(A_1), \ldots, p(A_n)\) then \(p(\bigcup A_i)\)

Then \(p(X)\).

Proof:

First assume \(X\) is compact. By property 1 the set of open \(U\) such that \(p(U)\) is an open cover. By compactness it has a finite subcover. i.e. we can find \(U_1, \ldots, U_n\) such that \(p(U_i)\) and \(\bigcup U_i = X\). Thus by property 2, \(p(X)\).

Now assume the induction principle holds. Let \(\mathcal{U}\) be an open cover of \(X\). Let \(p(A)\) be the property that \(A\) is covered by a finite union of elements of \(\mathcal{U}\). This satisfies the requisite properties – every point is contained in a single element of \(\mathcal{U}\) which suffices for property 1, and a finite union of finite sets is finite, so it satisfies property 2. Therefore by the induction principle, \(X\) is also covered by a finite union of elements of \(\mathcal{U}\), i.e. a finite subcover. Therefore \(X\) is compact.

QED

Note the restriction to finite unions. If we’re allowed to take arbitrary unions then this is just true of all topological spaces.

I feel like there’s probably a similar induction principle lurking in the intersections of closed sets equivalent form. Perhaps something that looks more like recursive function definition? I’m not totally sure.

For an example of how you could use this, here is a reframing of the proofs of two classic results:

Theorem: A compact metric space is bounded.

Proof: A finite union of bounded sets is bounded, and the open ball \(B(x, 1)\) is a bounded open set containing \(x\). Therefore by topological induction \(X\) must also be bounded. QED

Theorem: A compact subset of a Hausdorff topological space is closed.

Let \(X\) be Hausdorff and \(Y \subseteq X\) be compact. Let \(x \in X \setminus Y\).

Let \(p(A)\) be true if \(x\) is not in the closure of \(A\) considered as a subset of \(X\).

Because \(X\) is Hausdorff for every \(y \in Y\) we can find \(U, V\) open and disjoint with \(y \in U\) and \(x \in V\). So \(\overline{U} \subseteq V^c \subseteq X \setminus \{x\}\) and thus \(p(U)\). The closure of a finite union of sets is the union of the closure of the sets, so if \(p(A_1), \ldots, P(A_n)\) then \(p(\bigcup A_i)\). Therefore \(p(Y)\). i.e. \(x\) is not in the closure of \(Y\). But \(x\) was an arbitrary point not in \(Y\), hence \(Y\) must be closed. QED.

These aren’t really very different from the normal proofs. Virtually identical even. I think they come out slightly more nicely this way, but there’s not much in it. Hopefully it helps clear up some of the intuition around compactness though.