Topological restrictions on cardinality

Question

From what I know, a Polish (completely metrizable separable) space has a cardinality at most of $\mathbb R$. Completeness assumption can be omitted here, because a completion of a metrizable separable space is Polish. On the other hand, without separability the cardinality of the space can be greater than that of $\mathbb R$ - we just can endow any set with a discrete topology.

My question is the following: can a separable topological space has the cardinality greater than that of $\mathbb R$?

You want to include some separation conditions, otherwise the answer is no: any set with the trivial topology is separable (the closure of a point is the entire set). If you add some separation condition like Hausdorff then there are non-trivial upper bounds. — t.b., May 25 '12 at 08:34
For Hausdorff spaces you have $|X|\le 2^{2^{d(X)}}$, this result was shown in this question. Another estimate, which works for Hausdorff spaces is $|X|\le d(X)^{\chi(X)}$. — Martin Sleziak, May 25 '12 at 08:45
An upper bound is $2^{2^{\omega}}$ points by a theorem of Pospíšil, and the Stone-Čech compactification of $\mathbb{N}$ shows that this estimate is sharp. — t.b., May 25 '12 at 08:45
@tb: and thank you! I've deleted my request to provide non-trivial bounds you've mentioned in the first comment since when I've posted it, Brian has already updated his answer to include this case. — SBF, May 25 '12 at 08:48

score 8 · Accepted Answer · edited Apr 13 '17 at 12:20

8

Yes: $\beta\omega$ is a separable space of cardinality $2^{2^\omega}=2^\mathfrak{c}$. There’s a complete proof in this answer to an earlier question.

Assuming that the separable space $X$ is $T_2$, this is an upper bound on its cardinality. Let $D$ be a countable dense subset. For each $x\in X$ let $\mathscr{D}(x)=\{D\cap V:V\text{ is an open nbhd of }x\}$. Then $\mathscr{D}(x)$ is a family of subsets of $D$, so $|\mathscr{D}(x)|\le 2^\omega=\mathfrak{c}$. If $x,y\in X$ and $x\ne y$, disjoint nbhds of $x$ and $y$ have disjoint traces on $D$, so $\mathscr{D}(x)\ne\mathscr{D}(y)$. Thus, $$|X|=|\{\mathscr{D}(x):x\in X\}|\le|\wp(\wp(D))|=2^{2^\omega}=2^{\mathfrak c}\;.$$

Added: Originally I had written that $T_1$ separation was sufficient, but as t.b. points out, this is false: the cofinite topology on a set of any cardinality is a compact, separable, $T_1$ topology.

edited Apr 13 '17 at 12:20

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answered May 25 '12 at 08:33

Brian M. Scott

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Thanks! I'm not well-familiar with Greeks outside of finance: does $\beta$ mean Stone-Cech compactification? And if yes, what is $\omega$ here? – SBF May 25 '12 at 08:35
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$\omega=\mathbb N$. – Asaf Karagila May 25 '12 at 08:36
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@Ilya: Yes, that’s the Čech-Stone compactification of $\omega$, the set of non-negative integers; you can also call it $\beta\Bbb N$. – Brian M. Scott May 25 '12 at 08:36
@Brian: thank you, and thanks for updating the answer for $\mathrm T_1$ space. – SBF May 25 '12 at 08:44
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I think you need $T_2$ instead of $T_1$. Isn't the cofinite topology separable for any set? – t.b. May 25 '12 at 08:51
@t.b.: Yep; fixed. Thanks. – Brian M. Scott May 25 '12 at 08:55

score 4 · Answer 2 · answered May 25 '12 at 08:39

It depends if you require separation axioms to hold (regularity, Hausdorff, etc.)

If you simply require separability, let $X$ be a set as large as you would like it to be, along with the trivial topology.

Fix any $x\in X$. Now for every $y\in X$, and an open environment $U$ of $y$ we have that $x\in U$. Therefore $\operatorname{cl}(\{x\})=X$.

Similarly you can consider the topology of all sets $U$ such that $U=\varnothing$ or $x\in U$. In this topology $\{x\}$ is also dense.

In both topologies we have a dense singleton.

thanks, I was too focused on the discrete topology – SBF May 25 '12 at 08:46 — SBF, May 25 '12 at 08:46

Topological restrictions on cardinality

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