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.


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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