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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-Cech compactification of $X$.

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

1 Answer 1

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 $F_{Z_n}\supseteq F_f\supseteq A$.

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

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