A topology that is finer than a metrizable topology is also metrizable?

If $\tau_{1}$ and $\tau_{2}$ are two topologies on a set $\Omega$ such that $\tau_{1}$ is weaker than $\tau_{2}$ (i.e. $\tau_{1}\subset\tau_{2}$) and $\tau_{1}$ is metrizable, is it then true that $\tau_{2}$ is also metrizable?

My guess is that it is not true, since we do not know what the sets in $\tau_{2}$ look like, let alone whether they can contain 'open metric balls' or not. But maybe it is possible to adapt the metric so that the sets in $\tau_{2}$ can contain 'open metric balls'.

Also, it may be the case that the converse is easier to work with: If $\tau_{2}$ is not metrizable, then $\tau_{1}$ is not metrizable.

I find it very hard to think of any (counter)examples. Any help or hints would be greatly appreciated!

Solution 1:

No, take $X=\Bbb R$ and $\tau_1$ the usual topology, clearly metrisable, and $\tau_2$ the Sorgenfrey (aka "lower limit") topology (generated by the sets of the form $[a,b)$). Then $\tau_1 \subsetneq \tau_2$ but $\tau_2$ is not metrisable, for several reasons, the most accessible of which are (and the reason it's covered in some many topology text books and papers): its square is not normal, or it's separable but it doesn't have a countable base. See Wikipedia for more info.

Another example is $\mathbb{R}^\omega$ in the (metrisable) product topology and the finer box topology (which is not even first countable).

Also Munkres' $\Bbb R_K$ space which is $\Bbb R$ in the usual topology but with one extra closed set $K=\{\frac{1}{n}: n \in \Bbb N^+\}$, is not even regular (and all metrisable spaces are normal and perfectly normal) is another elementary example (see 2nd edition, p.197/198.).

Solution 2:

A countable example is furnished by the subspace $X=\{0\}\cup\{2^{-n}:n\in\Bbb N\}$ of $\Bbb R$ with topology $\tau_0$ that it inherits from the usual topology on $\Bbb R$. Let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$, and let $\tau=\{U\subseteq X:0\notin U\text{ or }\{n\in\Bbb N:2^{-n}\in U\}\in\mathscr{U}\}$; it is easily verified that $\tau$ is a topology on $X$ that is finer than $\tau_0$, but $\langle X,\tau\rangle$ is not even first countable.