Let $(X,\tau)$ be a second-countable $T_{2.5}$ space, where with $T_{2.5}$ I mean that any distinct points are separated by closed neighborhoods. Does there have to be some metrizable second-countable $\tau' \subseteq \tau$?

The typical examples of $T_{2.5}$ spaces that are not metrizable seem to be constructed by adding additional open sets to some metrizable topology, so I would be interested in a potential example of a space which is constructed differently -- or maybe a proof that we can always find a coarser metrizable second-countable topology.

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    $\begingroup$ For what it's worth, the existence of such a $\tau'$ is equivalent to the existence of a countable set of continuous functions $X\to [0,1]$ that separate points of $X$. $\endgroup$ – Eric Wofsey Feb 26 '17 at 6:31
  • $\begingroup$ @EricWofsey why just points, not points and closed sets as well? $\endgroup$ – Henno Brandsma Mar 8 '17 at 18:21
  • $\begingroup$ @HennoBrandsma: Because you only need $\tau'\subseteq\tau$, not $\tau'=\tau$. $\endgroup$ – Eric Wofsey Mar 8 '17 at 18:32
  • $\begingroup$ @EricWofsey So continuous in the $\tau$-topology on $X$. $\endgroup$ – Henno Brandsma Mar 8 '17 at 18:35
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    $\begingroup$ @EricWofsey Arens square .(See counterexamples) This has points that cannot be separated by continuous functions. $\endgroup$ – Henno Brandsma Mar 8 '17 at 21:53

As pointed out by Henno Brandsma in the comments, an example is the "Arens square" as modified by Brian Scott in his answer to this question. This space $(X,\tau)$ is $T_{2.5}$ (thanks to Brian Scott's modification), and is second-countable since it has only countably many points and it is clearly first-countable.

However, there is no continuous function $f:X\to [0,1]$ such that $f(0,0)=0$ and $f(1,0)=1$. Indeed, given such a function $f$, there would be $\epsilon>0$ such that $f(x,y)<1/3$ whenever $0<x<1/4$ and $0<y<\epsilon$ and $f(x,y)>2/3$ whenever $3/4<x<1$ and $0<y<\epsilon$. It then follows by continuity that $f(1/4,y)\leq 1/3$ and $f(3/4,y)\geq 2/3$ for $0<y\leq \epsilon$. Now fix $q\in Q_{1/2}$ such that $q<\epsilon$. Note that every closed neighborhood of $(1/2,q)$ contains a point of the form $(1/4,y)$ and also a point of the form $(3/4,y')$ for $y,y'<\epsilon$. It follows that $f(1/2,q)$ must be both $\leq 1/3$ and $\geq 2/3$, which is a contradiction.

If a coarser metrizable topology $\tau'\subseteq\tau$ existed, then there would exist such a function $f$ that is continuous with respect to $\tau'$, and hence also with respect to $\tau$. Thus no such $\tau'$ exists.


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