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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!

Calculix
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    @ε-δ a space with a finer topology than a normal space need not be normal, you also have more closed sets to separate (!). – Henno Brandsma Apr 08 '20 at 21:27

3 Answers3

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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.).

Henno Brandsma
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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.

Brian M. Scott
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    For a more elementary example (incidentally avoiding AC), instead of an ultrafilter you could use some filter on $\mathbb N$ which is not countably generated. E.g. the complements of sets $S\subseteq\mathbb N$ such that $\sum_{n\in S}\frac1n\lt\infty$. – bof Apr 08 '20 at 22:22
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    @bof: Argh! I forgot that I wasn’t working with $\omega+1$, as I usually do; thanks! For me your example is actually less elementary, though I realize that I’m in a minority. (Besides, AC is true. :-)) – Brian M. Scott Apr 08 '20 at 22:29
  • How about the converse? – Focus Jan 12 '24 at 21:07
  • @Focus: What are you thinking of as the converse? – Brian M. Scott Jan 12 '24 at 23:03
  • A coarser topology of a finer metrizable topology is metrizable; but I realized a simple fact that every discrete topology is metrizable, hence the converse cannot be true. – Focus Jan 14 '24 at 06:13
  • @Focus: You can answer it from the other end, too: the indiscrete topology on a set $X$ is the coarser than any other topology on $X$, and for $|X|>1$ it’s not metrizable. – Brian M. Scott Jan 14 '24 at 07:15
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Additional counterexamples can be found in functional analysis. Let $X$ be a (real or complex) Banach space, and let $\mathcal T_{\text{norm}}$ denote the norm topology on $X$. If $X$ is infinite-dimensional, then there exist linear topologies $\mathcal T$ on $X$ that are complete and strictly finer than $\mathcal T_{\text{norm}}$, for example:

  • One may take $\mathcal T$ to be the finest locally convex topology on $X$ — see [Sch99, p. 56, example before 6.3].
  • For additional examples, see this answer on MathOverflow.

Such a topology $\mathcal T$ can never be metrizable, for otherwise it would follow from the open mapping theorem (cf. [Rud91, Corollary 2.12(d)]) that $\mathcal T = \mathcal T_{\text{norm}}$.

Added later: you don't even need the open mapping theorem for this — it can be shown by elementary means that the finest locally convex topology on an infinite-dimensional vector space is never metrizable (see the answers to this question).

Every infinite-dimensional vector space admits a norm (see here), and can therefore be equipped with two locally convex topologies $\mathcal T_{\text{norm}} \subset \mathcal T_{\text{finest lc}}$, where $\mathcal T_{\text{norm}}$ is metrizable but $\mathcal T_{\text{finest lc}}$ is not.

References.

[Rud91]: Walter Rudin, Functional Analysis, Second Edition (1991), McGraw–Hill.

[Sch99]: H.H. Schaefer, M.P. Wolff (translator), Topological Vector Spaces, Second Edition (1999), Graduate Texts in Mathematics 3, Springer.

Josse van Dobben de Bruyn
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