Let $X$ be a completely regular space (that is, any closed subset and point not in the closed set can be separated by continuous function).

Let $C = C(X,[0,1])$ be the collection of continuous functions $f$ defined from $X$ to $[0,1].$

Recall the construction of Stone-Cech Compactification of $X$ via embedding in a cube.

For any $x\in X,$ defined a map $\phi:X\to [0,1]^C$ by $\phi(x) = (f(x))_{f\in C}.$ By Tychonoff's theorem, the space $[0,1]^C$ is compact in Tychonoff product topology. Note that $\overline{\phi(X)},$ which is a closed subset of a compact set, is compact in $[0,1]^C$. Then we let $\overline{\phi(X)}$ to be the Stone-Cech compactification of $X,$ denoted as $\beta X.$

Question: Let $V$ be an open subset in $\beta X.$ Is it true that $U = V\cap X$ open in $\beta X?$

  • $\begingroup$ math.stackexchange.com/questions/353413/… $\endgroup$ – Eric Wofsey Mar 4 '18 at 8:44
  • $\begingroup$ @EricWofsey: How does the link answer my question? $\endgroup$ – Idonknow Mar 4 '18 at 8:49
  • $\begingroup$ The answer to your question is yes iff $X$ is open in $\beta X$. $\endgroup$ – Eric Wofsey Mar 4 '18 at 8:50
  • $\begingroup$ @EricWofsey: I see. By your link, $X$ is open in $\beta X$ iff $X$ is locally compact. $\endgroup$ – Idonknow Mar 4 '18 at 8:53

Firstly, the intersection is not well-defined as it stands. $X$ is not a subset of $\beta X$ in this setup. What does hold is that $\phi[X]$ is a subset of $\beta X \subseteq [0,1]^C$ which is homeomorphic to $X$ (this is what $\phi$ being an embedding means). Often $X$ is identified with this image, also in notation. So $V$ open in $\beta X$ (so relatively open in the product) means that $V \cap \phi[X]$ is open in $\phi[X]$ (or $X$), not in $\beta X$, unless $X$ or $\phi[X]$ is open in $\beta X$ (which happens iff $X$ is locally compact Hausdorff).

So making the identification $X \sim \phi[X]$ : for all open $V \subseteq \beta X$: $V \cap X$ open in $\beta X$ iff $X$ is locally compact Hausdorff.

  • $\begingroup$ To the proposer: Consider the case $V=\beta X, $ so $V\cap \phi[X]=\phi[X].$... A completely regular space $X$ is locally compact iff $c[X] $ is open in $cX$ for every compactification $c:X\to cX$ iff $c[X]$ is open in $cX$ for some compactification $c:X\to cX.$............+1 $\endgroup$ – DanielWainfleet Mar 5 '18 at 10:07

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