A space is metrisable if and only if it admits a countable basis

I have a topology question about metrisable and countable basis. Please indicate whether the argument below is true or not:

"A space is metrizable if and only if it admits a countable basis. " Determine and proof whether it's true for each of the following spaces:

(i) Let (X,T) be a Hausdorff space.

(ii) Let (X,T) be a regular space.

(iii) Let (X,T) be a normal space.

This is what I have right now:

i) False. Proof: For metrizable --> countable basis, let X be an uncountable space with discrete topology T, then X is metrizable but doesn't have a countable basis (the basis will be singletons). Since X is uncountable, basis is uncountable.

For metrizable <-- countable basis, consider real number R with lower limit topology, so the base is {[a,b) | a$<$b, a,b $\in$ Q}. The basis is countable, but it's not metrizable.

ii) False. Proof: For metrizable --> countable basis, let X be an uncountable space with discrete topology T, then X is normal and metrizable. Every normal space is regular space, so X is regular and metrizable, but its basis {{x} | x $\in$ X} is uncountable.

For metrizable <-- countable basis, it's true because of Urysohn's Metrization theorem.

iii) False. Proof: For metrizable --> countable basis, let X be an uncountable space with discrete topology T, then X is normal and metrizable, but its basis {{x} | x $\in$ X} is uncountable.

For metrizable <-- countable basis, it's true because of Urysohn's Metrization theorem.

Are my proof correct? If not, what is the correct approach? Any help is greatly appreciated!

• Your example for (i) is false, $\mathbb{Q}$ in the lower limit topology is homeomorphic to standard $\mathbb{Q}$, so metrisable. – Henno Brandsma Apr 4 '18 at 4:00

You could stop at your first example: If $X$ is uncountable and discrete, then $X$ is Hausdorff, regular, and normal. It is metrisable (using the discrete metric $d(x,y) = 1$ for $x \neq y$) but it has a minimal base of size $|X|$, as all singletons sets $\{\{x\}: x \in X\}$ must be part of any base for $X$. So $X$ is not second countable.

So the equivalence is already proven false in all three classes of spaces. You could have stopped there, because an equivalence is already false when one implication is. But to go a bit deeper:

The reverse implication is even false for Hausdorff spaces: there are second countable Hausdorff spaces that are not metrisable. (Your example of $\mathbb{Q}$ in the lower limit subspace topology does not work, but look at $\pi$-base for examples, e.g.)

Urysohn's metrisation theorem says only that $X$ $T_3$ (regular and $T_1$) and second countable implies that $X$ is metrisable. It says nothing about the (as we saw, false) converse.

In the last case, if by normal is meant $T_4$ (normal and $T_1$) then Urysohn gives us one implication (as $T_4 \Rightarrow T_3$) but again not the reverse.

If however normal and regular allow for non-$T_1$ spaces, neither implication needs to hold, as indiscrete spaces are normal and regular and second countable but not metrisable unless $X$ is a singleton.

Within the class of $T_3$ spaces, the Bing-Nagata-Smirnov theorem gives us an equivalence:

1. $X$ is metrisable.
2. $X$ has a $\sigma$-locally finite base.
3. $X$ has a $\sigma$-discrete base.

which properly generalises Urysohn (as a countable base is both).