While it is well known for metric spaces, being separable is equivalent to be second-countable. In [this](http://www.mathcounterexamples.net/a-separable-space-that-is-not-second-countable/) [post](http://bit.ly/1HgwCdl) I give a counterexample for a non metric space.


What are other topological properties that added to separability ensure second-countability?



Spaces for which separable is equivalent to second-countable
According to [this paper](http://arxiv.org/abs/1301.5297) under $MA+\neg CH$ a scattered compact space is metrizable if and only if it is separable and hereditarily supercompact. It therefore seems as if to obtain metrizability from separability and some other property, one needs quite strong conditions and one needs to use strong set theoretic assumptions like $MA+\neg CH$.


Let me now define what all of these terms mean. A space $X$ is said to be supercompact if it has a subbasis such that every cover from this subbasis has a two element subcover. For example, the unit interval $[0,1]$ is supercompact. By Alexander's subbase theorem, every supercompact space is compact. A space $X$ is said to be hereditarily supercompact if every closed subspace of $X$ is supercompact. If $X$ is a space, then let $X'$ be the set of all non-isolated points of $X$. We define the Cantor-Bendixson derivatives $X^{(\alpha)}$ for all ordinals $\alpha$ as follows. Let $X^{(0)}=X$, let $X^{(\alpha+1)}=(X^{(\alpha)})'$, and let $X^{(\lambda)}=\bigcap\_{\alpha<\lambda}X^{(\alpha)}$ for limit ordinals $\lambda$. A topological space $X$ is said to be scattered if $X^{(\alpha)}=\emptyset$ for some ordinal $\alpha$.


[1](http://arxiv.org/abs/1301.5297): <http://arxiv.org/pdf/1301.5297v1.pdf> Hereditarily supercompact spaces


Taras Banakh, Zdzislaw Kosztolowicz, Slawomir Turek


