The Nagata-Smirnov Metrization Theorem states that
$X$ is metrizable iff it is $T_3$ and has a $\sigma$-locally finite base
So, I was wondering if this holds for pseudometric spaces too, if we remove the $T_0$ condition.
That is, is -
$X$ is pseudometrizable iff it is regular and has a $\sigma$-locally finite base
true?
Yes, this is true and the result follows from the metric case. If $(X,d)$ is pseudometric, then there is the associated metric space (called the metric identification here) has a $\sigma$-locally finite base and the corresponding base in $(X,d)$ is then as required, and conversely if we have a regular space with a $\sigma$-locally finite base, then its Kolmogorov quotient is $T_3$ (from $T_0$ plus regular) and has a $\sigma$-locally finite base too and so is metrisable and we pull back that metric to get a pseudometric on the original space.
