# Silly Proofs 3

Woo hoo. Blog is back. :-)

Here’s a new silly proof I spotted recently.

Theorem: Let $$f : \omega_1 \to \mathbb{R}$$ be continuous. Then $$f$$ is eventually constant.

Proof:

This proof assumes the that $$2^{\aleph_0} > \aleph_1$$. The result doesn’t actually need this though, which is one of the main reasons this proof is silly.

So, $$f$$ is continuous. $$\omega_1$$ is countable compact, thus so is $$f(\omega_1)$$. But $$\mathbb{R}$$ is a metric space, so countably compact subsets are compact. But every compact subspace of $$\mathbb{R}$$ has cardinality $$\aleph_0$$ or $$2^{\aleph_0}$$. We know that $$2^{\aleph_0} > \aleph_1$$, so it can’t be $$2^{\aleph_0}$$. Hence it has cardinality $$\aleph_0$$. By the pigeon hole principle we must have $$f$$ being constant on some uncountable set. But for any $$t$$ the set $$\{ x : f(x) = t \}$$ is closed, with these being disjoint for distinct values of $$t$$. You can’t have disjoint uncountable closed sets in $$\omega_1$$, so all but one of these sets must be countable. Thus take an upper bound for all the countable subsets, say $$y$$. $$f$$ is constant on $$[y, \omega_1)$$.

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