Take a completely regular Hausdorff topological space $X$ considered as a subset of its StoneČech 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 $\overline{Y}^{\beta X}$)?

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 $\beta X$. $\endgroup$– Tristan BiceSep 13, 2013 at 17:50
1 Answer
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/thetychonoffplank/. 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 $Y$.

$\begingroup$ 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 "Counterexample in Topology" does the job (it's a regular space such that any real valued function is constant, so its StoneCech compactification is equal to the onepoint 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 StoneCech compactification (and does not have a onepoint 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