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Answer by jbc for Extending uniformly continuous functions on subspaces to...
The closure of $X$ in $Z$ is compact , so there is no hope if $f$ is not bounded. If it is bounded then so is its extension to the closure of $X$ in $Y$ and this gives a bounded uniformly continuous...
View ArticleExtending uniformly continuous functions on subspaces to non-metrizable...
I have a complete metric space $Y$, some non-metrizable(!) Hausdorff compactification $Z$ of it and a subspace $X \subset Y$.Furthermore, I do have a uniformly continuous function $f$ on $X$. So there...
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