TITLE: Locally complete space is topologically equivalent to a complete space
QUESTION [8 upvotes]: Can someone please tell me where I can find a citeable reference for the following result:
Call the metric space $(X, d)$ "locally complete" if for every $x \in X$ there a neighbourhood of $x$ which is complete under $d$.
If $(X,d)$ is locally complete and separable then there exists a metric $d'$ on $X$ such that $(X,d) \to (X,d')$ is a homeomorphism and $(X, d')$ is complete.
This result follows immediately from Alexandrov's theorem that a $G_\delta$ subset of a Polish space is Polish, but I'd rather find a statement in the literature of the above more straightforward result if there is one.

REPLY [3 votes]: For a reference: this paper has a reference [30] that has a proof. The author cites your result and refers to it. I don't have access to these papers, so I cannot verify exactly.