# A completely regular space that is very non-normal

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 $\overline{Y}^{\beta X}$)?

• 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). Sep 13, 2013 at 17:16
• Yes, $\overline{Y}^{\beta X}$ refers to the closure of $Y$ in $\beta X$. Sep 13, 2013 at 17:50

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 $Y$.
• Can you also give an example where $Y$ is not locally compact? Sep 20, 2013 at 14:36