# Are locally metrizable topological spaces sequential?

Suppose $$(X,\tau)$$ is a topological space which is locally metrizable, that is every point $$x \in X$$ has an open neighbourhood $$U$$ in which we can define a metric that gives rise to the same topology of $$U$$ (as a subset of $$X$$).

Suppose now that we have a function $$f: (X,\tau) \to \mathbb{R}$$ (or any metric space in place of $$\mathbb{R}$$). Can I reason with sequences in order to show that $$f$$ is continuous? That is, is sequential continuity equivalent to continuity in locally metrizable topological spaces?

• What happens when you try to prove it? Questions showing no effort may be closed here. Dec 25, 2021 at 15:04
• sorry, maybe I'm not getting the point here. Does this imply that I can reason with sequences when try to prove weak continuity of a functional f from a Banach space X whose dual is separable (hence the weak topology can be induced by a metric on norm-bounded subsets)? Dec 25, 2021 at 16:47

Let $$X$$ be locally metrisable.
If $$A$$ is sequentially closed, then if $$x \in \overline{A}$$, we'd have a metrisable neighbourhood $$U_x$$ of $$x$$ by assumption. It would easily follow that there is a sequence $$a_n \in A\cap U_x$$ so that $$a_n \to x$$ and sequential closedness of $$A$$ would imply $$x \in A$$ and hence $$A$$ is closed.
So indeed sequential continuity with domain $$X$$ would imply ordinary continuity e.g. (a standard consequence of being a sequential space).