Here’s a result I’m going to need later for

When I’ve previously proved it I’ve used Fodor’s lemma, but I’ve just realised that there’s actually a simpler and slightly nicer proof by directly using some of the machinery you’d normally use to prove Fodor’s lemma. It’s not really substantially different, I just think it’s a little tidier.

The key result of this post is that every continuous function from an uncountable cardinal to a metric space is eventually constant. We’ll need to start with some set theoretic machinery about cardinals which we’ll use to prove this and then we’ll move on to some more purely topological arguments.

Let \(\kappa\) be an uncountable cardinal. The cofinality of \(\kappa\) is the smallest cardinality of an unbounded subset of \(\kappa\).

A club set in \(\kappa\) is an unbounded subset of \(\kappa\) that is closed under taking suprema of subsets.

Theorem: Let \(\kappa\) be an uncountable cardinal with \(cf(\kappa) > \aleph_0\). Let \(A, B \subseteq \kappa\) be club. Then \(A \cap B\) is club.

Proof:

It’s clear that the intersection is closed under suprema, so it remains only to show that that it is unbounded.

Let \(x \in \kappa\). Pick \(a_0 \in A\) with \(a \geq x\). Now recursively pick \(b_n \geq a_n\) and \(a_{n+1} \geq b_n\) with \(b_n \in B\) and \(a_n \in A\). We can do this because both sets are unbounded.

Because \(cf(\kappa) > \aleph_0\) any sequence must be bounded above (by definition: Otherwise it would be a countable unbounded set). Now let \(y = \sup a_n = \sup b_n\) (by construction the two sequences must have the same supremum). Then because \(A\) and \(B\) are closed under suprema we must have \(y \in A \cap B\). Therefore for any \(x \in \kappa\) there exists \(y \in A \cap B\) with \(y \geq x\). Therefore \(A \cap B\) is unbounded. QED

We can now extend this straightforwardly:

Theorem: Let \(\kappa\) be an uncountable cardinal with \(cf(\kappa) > \aleph_0\). Let \(A_n \subseteq \kappa\) be club. Then \(\bigcap A_n\) is club.

Proof:

By induction on the previous theorem finite intersections of club sets are club. Therefore \(A_1 \cap \ldots \cap A_n\) is club. Therefore for \(x \in \kappa\) we may pick \(a_n \in A_1 \cap \ldots \cap A_n\) with \(a_0 \geq x \) and \(a_{n+1} \geq a_n\).

Then \(a = \sup a_n \in \bigcap A_n \) and \(a \geq x\). Therefore \(\bigcap A_n\) is club. QED

This will give us a surprising (at least the first time you see it, but it’s fairly widely known) result about continuous functions.

Theorem: Let \(\kappa\) be a cardinal with uncountable cofinality. Let \(X\) be a metric space and let \(f : \kappa \to X\) be continuous when \(\kappa\) is given the order topology. Then there exists \(\alpha \in \kappa\) such that \(f|_{[\alpha, \kappa)}\) is constant.

Proof:

First note that any closed subset of \(\kappa\) is closed under suprema, so any unbounded subset closed under the order topology is club.

Secondly note that \(\kappa\) is sequentially compact (as any countable subset is contained in a compact subset). Therefore its image under \(f\) is sequentially compact and, since \(X\) is a metric space, thus compact. In particular it is separable. We may assume \(f\) is surjective and hence that \(X\) is separable.

Let \(x_1, \ldots, x_n, \ldots\) be a dense subset of \(x\). For \(\epsilon > 0\) the sets \(A(n, \epsilon) = \{ \alpha : f(\alpha) \in \overline{B}(x_n, \epsilon) \}\) form a countable cover of \(\kappa\) so by the pigeonhole principle at least one of them must be uncountable. Let \(C_\epsilon\) pick some \(A(n, \epsilon)\) such that it is uncountable. Then \(C_\epsilon\) is a club-set with \(\mathrm{diam}(f(C_\epsilon)) \leq 2 \epsilon\).

Let $$C = \bigcap C_{\frac{1}{n}}$$

By the preceding theorem, \(C\) is a club set. Necessarily \(\mathrm{diam}(f(C)) = 0\), so \(f(C) = \{x\}\) for some \(x\).

Now, consider the set \(M_\epsilon = \{ \alpha : d(f(\alpha), x) \geq \epsilon \}\). This set must necessarily be bounded, as otherwise it would intersect \(C\), which would contradict the definition. But \( \{ \alpha : f(\alpha) \neq x \} = \bigcup M_{\frac{1}{n}}\), which is a countable union of countable sets and thus bounded. Hence the set of points where f is not equal to x is bounded, and thus f is eventually constant as desired.

QED

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