# Extending uniformly continuous functions on subspaces to non-metrizable compactifications

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 is a uniformly continuous extension of $f$ from $X$ to the closure of $X$ in $Y$.

Can we extend f to a uniformly continuous function on the closure of $X$ in $Z$?

For this we equip $Z$ with the unique uniform structure which all compact Hausdorff spaces posses.

It seems that this question boils down to the question whether the from $Z$ induced uniform structure on $X$ coincides with the one coming from the metric on $X$. But sadly, I can't answer this on my own since I'm not familiar with uniform spaces.

If the answer does depend on $Z$, I'm interested in $Z = \beta Y$ (the Stone-Cech compactification).

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 function on the closure which can be extended to a bounded continuous function on $Y$ by the Tietze extension theorem. This, in turn, extends to a continuous function on the Stone-čech compactification. You can then restrict the latter to the closure of $X$ in $Z$.
• Ok, this settles the case $Z = \beta Y$. Thanks. But what about other compactifications of $Y$? – AlexE Apr 23 '13 at 11:20
• If you have $Y$ and $Z$ as in your question with the extension property for each subspace $X$ and each bounded, uniformly continuous function thereon, then by using $X=Y$ one sees that $Z$ has the uiversal property for each bounded, uniformly continuous function on $Y$. This means that $Z$ is the so-called Samuel compactification of $Y$. So the only leeway is in the case of $Y$ for which the two notions of compactification diverge. – jbc Apr 23 '13 at 12:34