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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.

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  • $\begingroup$ If $X$ is a non-empty set equipped with the trivial topology, any singleton will be dense in it, so you should probably require $X$ to be Hausdorff, to make the question more interesting. $\endgroup$
    – Dejan Govc
    Mar 23, 2014 at 22:56
  • $\begingroup$ Of course, thanks. I edited the question. $\endgroup$ Mar 23, 2014 at 22:59
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    $\begingroup$ See this answer. $\endgroup$
    – Hanul Jeon
    Mar 23, 2014 at 23:06

2 Answers 2

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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 $2^{2^{\aleph_0}}$.

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    $\begingroup$ 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 $f(x) \neq f(y)$. $\endgroup$ Dec 31, 2019 at 8:58
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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 $S$.

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    $\begingroup$ OP edited the question to rule out this example; the space must now be Hausdorff. $\endgroup$
    – MJD
    Mar 23, 2014 at 23:30

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