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I've been having conflicting thoughts about the following problem, and I was wondering if anyone could help me out.

Is is true that the cardinality of every regular separable space does not exceed $2^{\omega}$?

Thank you for your time.

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That's false the Stone-Cech compactification of the naturals has size $2^{2^\omega}$ and satisfies all your requirements (it is even compact). –  azarel Sep 2 '12 at 23:57
    
See also here and here for more details on azarel's comment. –  t.b. Sep 3 '12 at 0:03

2 Answers 2

up vote 8 down vote accepted

This is false: $\beta\omega$ is a separable compact Hausdorff space of cardinality $2^{2^\omega}>2^\omega$. (Here is a proof that $|\beta\omega|=2^{2^\omega}$.)

In general the best we can say is that if $\langle X,\tau\rangle$ is a Hausdorff space with a dense subset $D$, then $|X|\le 2^{2^{|D|}}$. In particular, is $X$ is separable, $|X|\le2^{2^\omega}$. To see this, for each $x\in X$ let $\mathscr{D}(x)=\{V\cap D:x\in V\in\tau\}$. Since $X$ is Hausdorff, the map $X\to\wp\big(\wp(D)\big):x\mapsto\mathscr{D}(x)$ is injective, so, $|X|\le 2^{2^{|D|}}$.

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This is false. I don't know how you're using the assumption of regularity; let me talk about Hausdorff separable spaces $X$ with countable dense subsets $S$. Then a naive argument might run as follows: since $S$ is dense, every element of $X$ is the limit of a sequence with values in $S$, and since $X$ is Hausdorff, limits are unique if they exist. Consequently the cardinality of $X$ is at most the cardinality of the set of sequences with values in $S$, which is $2^{\omega}$.

This argument fails at the very first step: we cannot conclude that every element of $X$ is the limit of a sequence with values in $S$. This is true if $X$ is also first-countable, and then the argument works, but false in general.

Regular and separable does not imply first-countable. Indeed, any compact Hausdorff space is regular (in fact completely regular by Urysohn's lemma), and an example which is not first-countable and also separable, namely $\beta \mathbb{N}$, was given in the comments.

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Without regularity, can't I make a space that is $\mathbb R$ with the standard topology, but with each irrational point replicated $\kappa$ times for any cardinal $\kappa$? It's not Hausdorff, but I think it is separable. –  Ross Millikan Sep 3 '12 at 4:09

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