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It is just an idea (might be wrong) but, i think that if a Hausdorff space, say $X$, contains too many elements, then a countable subset cannot be dense in it.

Does there exist a cardinality that any space with that (or bigger) cardinality cannot have a countable dense subset?

Thanks.

Martin Sleziak
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ThePortakal
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2 Answers2

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The maximum possible cardinality of a separable Hausdorff space is $2^{2^{\aleph_0}}$.

Let $X$ be a separable Hausdorff space and let $A$ be a countable dense subset of $X$. Define a function $f:X\to\mathcal P(\mathcal P(A))$ by setting $f(x)=\{B\subseteq A:x\in\overline{B}\}$. Using the fact that $X$ is Hausdorff, it's easy to see that $f$ is injective, whence $|X|\le|\mathcal P(\mathcal P(A))|\le2^{2^{\aleph_0}}$. (The Hausdorff separation axiom cannot be replaced by T$_1$ here; an arbitrary set with the minimal T$_1$ topology is a separable T$_1$-space.)

The product of continuum many separable spaces is separable. In particular, then, the product of $2^{\aleph_0}$ copies of the discrete space $\{0,1\}$ is a separable compact Hausdorff space of cardinality $2^{2^{\aleph_0}}$.

The Stone-Čech compactification of $\mathbb N$ is another famous example of a separable compact Hausdorff space of cardinality $2^{2^{\aleph_0}}$.

bof
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    Proof of 1-1: if $x \neq y$ find disjoint open sets $U_x$ of $x$ and $U_y$ of $y$. Then $x \notin \overline{U_y \cap A}$, as witnessed by $U_x$, so $U_y \cap A \notin f(x)$, while (as $A$ is dense), $y \in \overline{U_y \cap A}$ or $U_y \cap A \in f(y)$, so $f(x) \neq f(y)$. – Henno Brandsma Dec 31 '19 at 08:58
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Consider a arbitrary set $S$ with the indiscrete topology: the only open sets are $\emptyset$ and $S$. Then even a one-point set is dense in $S$.

Robert Israel
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    OP edited the question to rule out this example; the space must now be Hausdorff. – MJD Mar 23 '14 at 23:30