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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?

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1 Answer 1

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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.

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  • $\begingroup$ Hi professor, could I ask your assistance here, please? $\endgroup$ Aug 2, 2020 at 18:16

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