2
$\begingroup$

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).

$\endgroup$
3
$\begingroup$

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 $Z$.

$\endgroup$
  • $\begingroup$ Ok, this settles the case $Z = \beta Y$. Thanks. But what about other compactifications of $Y$? $\endgroup$ – AlexE Apr 23 '13 at 11:20
  • $\begingroup$ 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. $\endgroup$ – jbc Apr 23 '13 at 12:34

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.