Let $X$ be a Tychonoff space and let $\beta X$ is the Stone-Čech compactification of $X$. Assume $f:X\longrightarrow \mathbb{R}$ is a bounded function. Then there exists a function $f^{\beta }:\beta X\longrightarrow \mathbb{R}$ (the map $f\longrightarrow f^{\beta }$ is an isomorphism of $C^{\ast }\left( X\right) $ onto $C\left( \beta X\right) $).

Moreover, every point $p$ of $\beta X$ is the limit of a unique $z$-ultrafilter $\mathcal{A}^{p}$ on $X$. If $p\in X$, then $\mathcal{A}^{p}=\left\{ Z\in Z\left( X\right) :p\in Z\right\} $.

My question is that for $p\in \beta X$, how to define $f^{\beta }\left( p\right) $ in terms of $z$-ultrafilter $\mathcal{A}^{p}$ on $X$?

Actually, my purpose is to check whether $f^{\beta }$ also provides a property that $f$ provides (for example, $f$ is constant on some subsets) using ultrafilters.


1 Answer 1


Of course $f^\beta(p)$ is the limit of $f(x)$ along the z-ultrafilter $\mathcal A^p$. For $t \in \mathbb R$, we write $$ \lim_{\mathcal A^p} f(x) = t $$ iff for every neighborhood $U$ of $t$ there is $V \in \mathcal A^p$ so that $\forall x \in V, f(x) \in U$.
In your setting, there exists a unique $t$ with that property.
The map $f^\beta$ is the unique continuous extension of $f$ from the dense subset $X$ to all of $\beta X$.


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