# Proving Non-metrizable spaces [duplicate]

Maybe this question seems easy, but is there a strategy or a property to look for when demostrating a space is non-metrizable? Because a part from the fact that I can't find a metric generating the space, I cannot seem to find a way to prove that such does not exist.

## marked as duplicate by Tomek Kania, Lee Mosher, Community♦Mar 2 '17 at 21:18

• It depends on the space, doesn't it? That question is very vague. – user251257 Mar 2 '17 at 19:49

You could prove that the space fails some sort of separation axiom that metric spaces have (basically any of the $T$'s here, or show it is not first countable.

• There is something I do not remember properly tickling my mind. Something about having a closed, uncountable discrete subspace. Unfortunately I am not sure the conclusion was "it's not metrizable" but I wonder if anyone recognizes this weird ghost memory. – rschwieb Mar 2 '17 at 19:50
• Any uncountable discrete space is metrisable by the discrete metric. – Tomek Kania Mar 2 '17 at 19:59
• @TomekKania It could be totally unrelated. I just remember something like that being used to rule out a topological property. – rschwieb Mar 2 '17 at 20:00
• @rschweib, this rules out separability of metrics spaces. – Tomek Kania Mar 2 '17 at 20:01
• @rschwieb E.g. the square of the Sorgenfrey line has a uncountable discrete subset, so cannot be second countable, while it is separable which for metric spaces would imply second countable. Etc. – Henno Brandsma Mar 2 '17 at 20:12

Disprove for a space $X$ a property P such that all metric spaces have property P.

Such P include $X$ is Hausdorff, $X$ is normal, $X$ is perfectly normal, $X$ is first countable.

Also for metric spaces we have that being ccc, separable, Lindelöf, second countable are all equivalent (if $X$ is metrisable and has one of these properties, it has all the other ones). If for some $X$ we have that e.g. $X$ is separable but not second countable (as for the Sorgenfrey line (aka the lower limit topology)) then $X$ cannot be metrisable.

For metric spaces we also have that countably compact, pseudocompact, compact and sequentially compact are all equivalent, so here the same strategy can be applied. e.g. $\omega_1$ is countably compact but not compact, so cannot be metrisable.

I think most proofs of non-metrisability are done along these lines. A totally general method is not really possible I think.