Metric space completeness is a restricted form of topological compactness

This is one of those facts that is obvious once you’ve seen it but I sometimes forget isn’t well known.

Topological compactness admits the following slightly less common but directly equivalent definition:

A family of sets has the finite intersection property if every intersection of a finite subfamily of it has non-empty intersection. A topological space is said to be compact if every family of closed sets with the finite intersection property has non-empty intersection.

(The reason this is directly equivalent is that a family of closed sets with empty intersection is precisely the set of complements of some open cover)

Starting from this definition, we can get an alternative definition for what it means for a metric space to be complete that makes the relationship between completeness and compactness much more obvious:

A metric spaces is complete if and only if for every family of closed sets with the finite intersection property that contains sets of arbitrarily small diameter has non-empty intersection.

i.e. metric completeness is “compactness for arbitrarily small sets”.

Let’s show that this is equivalent to the normal Cauchy sequence definition.

First, assume a metric space has this property, let’s show it’s complete in the Cauchy sequence sense.

So, let \(x_n\) be a cauchy sequence. Now, define \(F_n = \overline{\{x_m: m \geq n\}}\).

The family \(\{F_n\}\) is nested and non-empty, so certainly has the finite intersection property. Its elements are closed by construction.

To see that it contains sets of arbitrarily small diameter, we just need to show that the tail sets \(\{x_m: m \geq n\}\) can have arbitrarily small diameter, as the diameter of the closure of a set is the same as the diameter of the set. But \(x_n\) is a cauchy sequence, so for \(\epsilon > 0\) we can find \(N\) such that for \(m, n \geq N\), \(d(x_n, x_m) < \epsilon\). Necessarily then the tail set \(\{x_m: m \geq N\}\) has diameter \(\leq \epsilon\).

Now suppose \(x \in \bigcap F_n\). Then \(x_n \to x\), as if we pick \(N\) such that \(diam(F_N) < \frac{1}{2}\epsilon\), then because \(x\) is a limit point we can find \(x_m\) with \(m \geq n\) and \(d(x_m, x) < \frac{1}{2}\epsilon\), so necessarily for all \(n \geq N\), \(d(x_n, n) \leq d(x_m, x_n) + d(x_m, x) < \epsilon\), and the sequence converges to \(x\) as desired.

The other direction now. Suppose we have a Cauchy sequence complete metric space and a family of closed sets with the finite intersection property and arbitrarily small diameter sets.

Let \(E_n\) be a sequence of sets in that family with \(diam(E_n) \to 0\).

Pick \(x_n \in \bigcap\limits_{m \leq n} E_n\) (we can do this because of the finite intersection property). Then \(x_n\) is a Cauchy sequence: Pick \(N\) such that \(diam(E_N) < \epsilon\), then for \(m, n \geq N\), \(x_m, x_n \in E_N\), so necessarily \(d(x_m, x_n) \leq diam(E_N) < \epsilon\).

Because it’s a Cauchy sequence, it converges, say to \(x\). Because the \(E_n\) are closed, \(x\) is necessarily a member of \(\bigcap E_n\).

Suppose now that \(F\) is some other member of our family. We can repeat the above construction replacing \(E_n\) with \(E_n \cap F\), and get an element \(y\) with \(y \in F \cap \bigcap E_n\). But \(x, y \in E_n\), so \(d(x, y) \leq diam(E_n) \to 0\). So \(d(x, y) = 0\) and thus \(x = y\). Hence \(x\) must be in the intersection of the whole family.

QED

Why does this characterisation matter?

Well, clearly it doesn’t matter that much. The vast majority of people who do analysis have probably never encountered it and their lives are not substantially worse off for that lack.

There are basically two major reasons I like this fact:

  1. I think it makes the relationship between compactness and completeness much clearer.
  2. What Zorn’s lemma does for transfinite induction it does for sequences in analysis – it essentially lets you factor out a repetitive mechanical part of the proof and just reuse it without having to repeat the machinery over and over again. I’m slightly allergic to sequences, so this tends to play well with how I like to do analysis.
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