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I would like to know if the extension theorem of uniformly continuous functions can be generalized to pseudometric spaces.

That is, let $X,Y$ be a pseudometric spaces and $D\subset X$ a dense subset. If $f:D\rightarrow Y$ is uniformly continuous then can it be extended uniquely to a continuous function?

Uniform continuity can be defined analogously to the metric case. I'm not sure the same proof works because we don't have continuity iff preservation of convergence.

I guess we may need completeness of $Y.$

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A completeness of $Y$ is stated even in the metric case.

It seems that an analog on the extension theorem for metric spaces holds for pseudometric spaces too, and the adapted proof should work. For each $x\in X$ as $\hat f(x)$ you can put an arbitrary point $y$ of the limit of the filter $\{f(Ox\cap D):D$ is a neighborhood of $x\}$.

we don't have continuity iff preservation of convergence.

Are you sure? It is an other subject that a sequence in a non Hausdorff space can have more than one limit, so, in a pseudometric case a continuous extension of $f$ may be not unique.

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  • $\begingroup$ Thanks for your answer. I'm not sure about that statement anymore, the example I had in mind doesn't work. $\endgroup$ – Robert Apr 18 '13 at 16:44
  • $\begingroup$ Yes, I was confused about something. Thanks. $\endgroup$ – Robert Apr 18 '13 at 17:29

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