2
$\begingroup$

Take a completely regular Hausdorff topological space $X$ considered as a subset of its Stone-Čech compactification $\beta X$. If $X$ is not normal, we can find a closed subset $Y$ of $X$ and a continuous function $f:Y\rightarrow[0,1]$ with no continuous extension to $\overline{Y}^{\beta X}$. Can $f$ fail to have a continuous extension to any $Z\subseteq\overline{Y}^{\beta X}$ properly containing $Y$? Can this even happen when $Y$ is not locally compact (so $Y$ is not open in $\overline{Y}^{\beta X}$)?

$\endgroup$
2
  • 2
    $\begingroup$ I take it you meant "closure of $Y$ as a subspace of $\beta X$". But I found the notation a little confusing at first (looks like an exponential). $\endgroup$
    – Todd Trimble
    Sep 13, 2013 at 17:16
  • 1
    $\begingroup$ Yes, $\overline{Y}^{\beta X}$ refers to the closure of $Y$ in $\beta X$. $\endgroup$ Sep 13, 2013 at 17:50

1 Answer 1

2
$\begingroup$

If I did not miss something, the following is a simple example of such space. The idea is to have a space such that $\beta X$ is equal to the one point compactification of $X$.

Consider the Tychonov plank $T=(\omega_1+1)\times(\omega+1)$, and $X$ be its subspace $T - \{\langle \omega_1,\omega\rangle\}$. Then $T=\beta X$, as well known, see for instance http://dantopology.wordpress.com/2009/10/21/the-tychonoff-plank/. Take the usual closed subset $Y=\{\langle \omega_1,y\rangle :y\in\omega\}\cup\{\langle x,\omega\rangle :x\in\omega_1\}$, then $\overline{Y}^{\beta X}-Y$ is the singleton $\{\langle \omega_1,\omega\rangle\}$. Thus the function $f$ which has value $0$ on one part of $Y$ and $1$ on the other cannot be continuously extended to any subset of $\bar{Y}^{\beta X}$ properly containing $Y$.

$\endgroup$
4
  • $\begingroup$ Great! That's exactly the kind of example I was after. $\endgroup$ Sep 20, 2013 at 14:28
  • $\begingroup$ Can you also give an example where $Y$ is not locally compact? $\endgroup$ Sep 20, 2013 at 14:36
  • $\begingroup$ I don't have time to check it now, but maybe Example 92 in "Counter-example in Topology" does the job (it's a regular space such that any real valued function is constant, so its Stone-Cech compactification is equal to the one-point compactification). $\endgroup$ Sep 23, 2013 at 8:23
  • $\begingroup$ Ah, but of course, since this space is not even Tychonoff, it does not embed in its Stone-Cech compactification (and does not have a one-point compactification by the way), so what I wrote in the previous comment does not make sense. I do not have an example offhand, I am afraid. $\endgroup$ Sep 23, 2013 at 9:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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