How big can a separable Hausdorff space be? 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.
 A: 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 $S$.
A: 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 $2^{2^{\aleph_0}}$.
