Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb R$?

share|improve this question
4  
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 8:34
1  
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 8:45
2  
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 8:45
    
@Martin: thank you! –  Ilya May 25 '12 at 8:47
    
@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. –  Ilya May 25 '12 at 8:48
add comment

2 Answers

up vote 7 down vote accepted

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.

share|improve this answer
    
Thanks! I'm not well-familiar with Greeks outside of finance: does $\beta$ mean Stone-Cech compactification? And if yes, what is $\omega$ here? –  Ilya May 25 '12 at 8:35
    
$\omega=\mathbb N$. –  Asaf Karagila May 25 '12 at 8:36
    
@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 8:36
    
@Brian: thank you, and thanks for updating the answer for $\mathrm T_1$ space. –  Ilya May 25 '12 at 8:44
1  
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 8:51
show 1 more comment

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.

share|improve this answer
    
thanks, I was too focused on the discrete topology –  Ilya May 25 '12 at 8:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.