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?


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 Oct 9 '13 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 $X$. $\endgroup$ – Brian M. Scott Oct 9 '13 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 Oct 9 '13 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 $\beta X$. $\endgroup$ – Brian M. Scott Oct 9 '13 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 Oct 9 '13 at 10:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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