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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?

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    $\begingroup$ 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. $\endgroup$ – Joseph Van Name Mar 8 '15 at 14:39
  • $\begingroup$ @JosephVanName: ...and hence we didn´t need the extra condition after all :) $\endgroup$ – Ramiro de la Vega Mar 9 '15 at 11:18
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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 $\alpha$.

1: http://arxiv.org/pdf/1301.5297v1.pdf Hereditarily supercompact spaces

Taras Banakh, Zdzislaw Kosztolowicz, Slawomir Turek

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Some examples of properties which imply second-countability (or, equivalently, metrizability) for separable regular spaces can be extracted from the work on the Normal Moore Space Conjecture.

Recall that a regular space $X$ is said to be a Moore Space if there is a sequence $\{\mathcal{C}_n: n \in \mathbb{N}\}$ of open covers of $X$ such that for every $x \in X$, the family $\{st(x, \mathcal{C}_n): n \in \mathbb{N}\}$ is a local base at $x$, where $st(x, \mathcal{C}_n)=\bigcup \{U: U \in \mathcal{C}_n \wedge x \in U \}$.

Jones proved that if $2^{\aleph_0} < 2^{\aleph_1}$ then every separable normal Moore space is metrizable.

The assumption $2^{\aleph_0} < 2^{\aleph_1}$ is consistent with and independent of the usual axioms of set theory.

Is the set-theoretic assumption needed? Zenor proved that if you take the Moore plane and replace the x-axis with a Q-set (that is, an uncountable subset of the reals whose every subset is a relative $G_\delta$), then you get a separable normal Moore space which is not metrizable. Since every second-countable space has at most $2^{\aleph_0}$ many $G_\delta$ subsets and every uncountable set has at least $2^{\aleph_1}$ many subsets, it is clear that if there is a $Q$-set then $2^{\aleph_0}=2^{\aleph_1}$.

Furthermore, Zenor proved that if there is no Q-set then every separable Normal Moore space is metrizable, so the non-existence of a Q-set is equivalent to metrizability for separable Normal Moore space.

Traylor proved that every metacompact normal separable Moore space is metrizable.

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It was proved by S. Todorcevic that it is consistent with ZFC that (S) every regular hereditarily separable space is lindelof which can imply metrizability in some cases. It follows from hypothesis S that every hereditarily separable pseudocompact topological group is compact and first countable and hence metrizable. See [1. p. 230]

  1. Topological Groups and Related Structures, An Introduction to Topological Algebra. Authors: Arhangel’skii, Alexander, Tkachenko, Mikhail.
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