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I came across the following fact in Alex. Kechris' Classical Descriptive Set Theory (Ex. 3.12, pg. 17, 1994):

Let $0^n$ be a string of $n$ $0$s. Then the map $f(x)=0^{x_0}10^{x_1}10^{x_2}\cdots$, where $x=\langle x_0, x_1, \ldots\rangle$, is a homeomorphism of the Baire space $\mathcal{N}=\omega^\omega$ with a countable $G_\delta$ set in the Cantor space $\mathcal{C}=2^\omega$.

I don't think I understand the statement. The way I understand it is that there is a homeomorphism between $\mathcal{N}$ and a countable $G_\delta$ subset $A$ under the relative topology. But how can the uncountable set $\omega^\omega$ be in a one-to-one correspondence with a countable set $A$ in the first place?

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You've misread the exercise: it says cocountable. A cocountable set is one whose complement is countable. (You are of course quite right that an uncountable space can never be homeomorphic to a countable one.)

It's a bit annoying - the typesetting means that "co-countable" is split across two lines, with the "co" at the top and then the "countable." This makes it really easy to miss the "co" and read it as simply "countable."

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  • $\begingroup$ Oh no! What a blunder! $\endgroup$ – Benjamin T Apr 18 '18 at 3:41
  • $\begingroup$ It is unfortunate typesetting - I missed it too at first glance. $\endgroup$ – Noah Schweber Apr 18 '18 at 3:42
  • $\begingroup$ I missed it all the while until I saw your comment! $\endgroup$ – Benjamin T Apr 18 '18 at 16:31
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As Noah said here, it's a co-countable subset of $2^\omega$ (so easily a $G_\delta$); in $2^\omega$ you only miss those points of $2^\omega$ with finitely many $1$'s which is a countable dense set of the Cantor space (homeomorphic to $\mathbb{Q}$ in fact). So we split the Cantor set into a copy of "rationals" and "irrationals" as it were, just like the reals.

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  • $\begingroup$ Yes, that became obvious once I realized I have misread the statement. Interesting exercise! $\endgroup$ – Benjamin T Apr 18 '18 at 16:29

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