How do we show the following?

Let $X$ be a topological space and let $x \in X$. Show that if $x$ has a countable neighborhood basis in $X$ then $x$ has a countable neighborhood basis in $\beta X$. Here $\beta X$ denotes the Stone–Čech compactification of $X$.

  • 1
    $\begingroup$ One thinks about what the topology on $\beta X$ is. $\endgroup$ – Arturo Magidin Dec 1 '10 at 17:47
  • $\begingroup$ @Arturo: The Tychonoff product topology. Then what to do? $\endgroup$ – user10 Dec 1 '10 at 18:47

If $X$ is locally compact, then it is easy, since $X$ is open in $\beta X$, hence the basis of $x$ in $X$ is also a basis in $\beta X$.

Here's an argument for normal spaces (using my favorite of the many characterizations of $\beta X$ (for normal spaces)):

We're going to define $\beta X$ as the space of ultrafilters in the algebra of closed subsets of $X$. That is, an element of $\beta X$ is a maximal set $\mathfrak{a}$ of closed sets that is closed under finite intersections and such that if $f\in\mathfrak{a}$ and $g\supseteq f$ is closed, then $g\in\mathfrak{a}$. A base for the closed sets consists of sets of the form $F_f=\{\mathfrak{a}\in\beta X : f\in\mathfrak{a}\}$ for closed sets $f\subseteq X$ (meaning the closed sets are intersections of such $F_f$s.) Finally, define the embedding $X\to\beta X$ by letting $\hat{x}=\{f\subseteq X:x\in f\}$.

Now, suppose that $\{U_n\}_{n\in\omega}$ is a local base at $x$ in $X$. Then $\{Z_n\}_{n\in\omega}$ where $Z_n=X\setminus U_n$ is a local closed base at $x$, meaning that $x\notin Z_n$ for all $n$ and if $f\subseteq X$ is any closed set with $x\notin f$ then there is an $n$ with $f\subseteq Z_n$.

Now I claim that $\{F_{Z_n}\}_{n\in\omega}$ is a closed base at $\hat{x}$ in $\beta X$. For, let $A\subseteq\beta X$ be closed with $\hat{x}\notin A$. Thus there is some closed $f\subseteq X$ with $\hat{x}\notin A_f$ and $A\subseteq A_f$. This implies that $x\notin f$ hence $f\subseteq Z_n$ for some $n$. But $x\notin Z_n$ so $\hat x\notin F_{Z_n}$ and $F_{Z_n}\supseteq F_f\supseteq A$.

  • 1
    $\begingroup$ for completely regular spaces we use ultrafilters of zerosets. The proof should go through, I think, mutatis mutandis. $\endgroup$ – Henno Brandsma Feb 10 '11 at 18:22
  • $\begingroup$ Yeah, and for general spaces, i guess the same argument should work (if one is okay with non-Hausdorff compactification.) $\endgroup$ – Apollo Feb 10 '11 at 19:11

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.