If a set is bigger than a dense set in cardinality, does it have a limit point?

Question

Suppose $X$ is a separable T$_4$ space and $A\subset X$, then $|A|\geq2^{\aleph_0}$ implies $A$ has a limit point. I know a proof of this using the power set concept. Does this theorem still hold if $2^{\aleph_0}$ is replaced by the smallest uncountable cardinal $\aleph_1$, without assuming $\aleph_1=2^{\aleph_0}$? Thanks.

Answer 1

Jones' lemma (I prove it here) states for in the separable case :

For normal , separable $X$, if $C$ is closed and discrete (as a subspace), then $2^{|C|} \le 2^{\aleph_0}$.

This implies that a set $A$ of $2^{\aleph_0}$ in $X$ always has a limit point, or else $A$ would be closed and discrete and $2^{|A|} =2^{2^\aleph_0} > 2^{\aleph_0}$ by Cantor's theorem. We cannot say (in ZFC) that $2^{\aleph_1} > 2^{\aleph_0}$, they could very well be equal (e.g. under MA and $\lnot$CH), so the argument does not work for sets of size $\aleph_1$.

IIRC, under MA($\omega_1$) $\Psi$-spaces based on a AD family of size $\aleph_1$ are normal and form a counterexample for sets of that size, and other examples also exist. See Fremlin's book on the consequences of MA. I also found this old paper, from results in which it also follows, I think.

So for sets of size $\aleph_1$ the question seems independent of ZFC. This often happens in general topology: one way under CH, the other way under MA and $\lnot$CH, because the latter is a "positive way" to fail CH, namely in a way where we can still prove things about the topological behaviour of $\aleph_1$. And the $\diamondsuit$-axiom often gives strong examples, that CH does not suffice for to construct.