Woo hoo. Blog is back. :-)
Here’s a new silly proof I spotted recently.
Theorem: Let [tex]f : \omega_1 \to \mathbb{R}[/tex] be continuous. Then [tex]f[/tex] is eventually constant.
Proof:
This proof assumes the that [tex]2^{\aleph_0} > \aleph_1[/tex]. The result doesn’t actually need this though, which is one of the main reasons this proof is silly.
So, [tex]f[/tex] is continuous. [tex]\omega_1[/tex] is countable compact, thus so is [tex]f(\omega_1)[/tex]. But [tex]\mathbb{R}[/tex] is a metric space, so countably compact subsets are compact. But every compact subspace of [tex]\mathbb{R}[/tex] has cardinality [tex]\aleph_0[/tex] or [tex]2^{\aleph_0}[/tex]. We know that [tex]2^{\aleph_0} > \aleph_1[/tex], so it can’t be [tex]2^{\aleph_0}[/tex]. Hence it has cardinality [tex]\aleph_0[/tex]. By the pigeon hole principle we must have [tex]f[/tex] being constant on some uncountable set. But for any [tex]t[/tex] the set [tex]\{ x : f(x) = t \}[/tex] is closed, with these being disjoint for distinct values of [tex]t[/tex]. You can’t have disjoint uncountable closed sets in [tex]\omega_1[/tex], so all but one of these sets must be countable. Thus take an upper bound for all the countable subsets, say [tex]y[/tex]. [tex]f[/tex] is constant on [tex] \[y, \omega_1) [/tex].