# Spaces for which separable is equivalent to second-countable

While it is well known for metric spaces, being separable is equivalent to be second-countable. In this post I give a counterexample for a non metric space.

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

• You probably know this already, but any second countable regular space is metrizable, and in this context most interesting topological spaces are regular. Therefore, any condition that we can add to separability to guarantee second countability will also guarantee metrizability. – Joseph Van Name Mar 8 '15 at 14:39
• @JosephVanName: ...and hence we didn´t need the extra condition after all :) – Ramiro de la Vega Mar 9 '15 at 11:18

According to this paper 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 $\alpha$.