A compactification of $X$ is a pair $( h, Y)$ where $Y$ is a compact space and $h\colon X ‎\to Y$ is an embedding such that $h(X)$ is dense in $Y$.

  1. According to definition of mentioned compactification, is the Stone-Čech compactification $\beta X$ Hausdorff?

  2. How can we introduce Stone-Čech compactification by means of ultrafilter?


1 Answer 1


The Čech-Stone compactification $\beta X$ is defined only for Tikhonov spaces $X$ and is always Hausdorff.

The ultrafilter construction of $\beta X$ technically applies only to discrete $X$; for an arbitrary Tikhonov space $X$ one uses maximal filters of zero sets, sometimes called $z$-ultrafilters. If $X$ is discrete, every subset of $X$ is a zero set, so $z$-ultrafilters are just ordinary ultrafilters. In this construction we let $\beta X$ be the set of $z$-ultrafilters on $X$. Let $\mathscr{Z}$ be the family of all zero sets in $X$. For each $Z\in\mathscr{Z}$ let $\hat Z=\{\mathscr{F}\in\beta X:Z\in\mathscr{F}\}$; then $\{\hat Z:Z\in\mathscr{Z}\}$ is a base for a topology on $\beta X$, and that topology turns out to be compact and Hausdorff.

Since $X$ is Tikhonov, for each $x\in X$ there is a unique $z$-ultrafilter $\mathscr{F}_x$ that converges to $X$, so we can define a map $$h:X\to\beta X:x\mapsto\mathscr{F}_x\;.$$ This map turns out to be an embedding of $X$ as a dense subset of $\beta X$.

That’s just a brief outline; for the details you’d be better off consulting a text.

For $T_1$ spaces $X$ there is something called the Wallman compactification $\omega X$, which is not necessarily Hausdorff, though if $X$ is normal it coincides with $\beta X$; it uses a similar construction, but with maximal filters of closed sets instead of maximal filters of zero sets.

  • $\begingroup$ (1)Is there relationship between principal ultrafilter and z-ultrafilters? (2) Do you mean that "the Stone-Čech compatification on discrete space $X$ is the set of all ultrafilters on $X$ or $\beta X$ is a ultrafilter? $\endgroup$
    – Ebi
    Commented Oct 9, 2013 at 7:28
  • $\begingroup$ @Ebi: (1) Not really. The principal ultrafilters on a discrete space correspond to the $z$-ultrafilters $\mathscr{F}_x$ in my answer: if $X$ is discrete, the $z$-ultrafilter converging to $x\in X$ is the principal ultrafilter over $x$, i.e., $\{A\subseteq X:x\in A\}$. (2) The former: if $X$ is discrete, $\beta X$ is the set of all ultrafilters on $X$. $\endgroup$ Commented Oct 9, 2013 at 7:48
  • $\begingroup$ Is it true "The appropriate topology on $\beta X$ for every $D \subset X$ is that the set of ultrafilters containing $D$."?I mean "$ ‎‎\widehat{D} = \{ \mathcal{P} \in \beta X : D \in \mathcal{P} \}$ is a topology"‎‎.‎‎ $\endgroup$
    – Ebi
    Commented Oct 9, 2013 at 8:30
  • $\begingroup$ @Ebi: If $X$ is discrete, $\{\widehat{D}:D\subseteq X\}$ isn’t a topology on $\beta X$, but it is a base for a topology on $\beta X$. $\endgroup$ Commented Oct 9, 2013 at 9:28
  • $\begingroup$ Excuse me. you are right. I wrote wrong.So, by means of this set ($\{\widehat{D}$) a topology for $\beta X$ is defined by the set of ultrafilters containing $D$? and The set $X$ corresponds to the set of principal ultrafilters? $\endgroup$
    – Ebi
    Commented Oct 9, 2013 at 10:01

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