# the Stone-Čech compactification by using ultrafilters

If $X$ is discrete, one can construct $\beta X$ as the set of all ultrafilters on $X$.

But which kind of topology must we use in the above sentence?

How can we define the Stone–Čech compactification by using ultrafilters? What is the form of the topology in terms of ultrafilters?

• Dear fatemeh : I took the liberty of adjusting your question a little. I hope you find the changes helpful. – rschwieb Aug 28 '13 at 20:18

## 1 Answer

For each $A\subseteq X$ let $\widehat A=\{p\in\beta X:A\in p\}$; then $\{\widehat A:A\subseteq X\}$ is a base for the topology of $\beta X$. Note that if $x\in X$, then

$$\widehat{\{x\}}=\left\{p\in\beta X:\{x\}\in p\right\}=\{p_x\}$$

where $p_x$ is the principal ultrafilter over $x$: $p_x$ is an isolated point in $\beta X$. This is why we identify $p_x$ with $x$ and say that $X\subseteq\beta X$, when really it’s $\{p_x:x\in X\}$ that is the subset of $\beta X$ corresponding to $X$.

• $p_ x$ is an isolated point in $β_X$?? for this reason, we say $\beta X$ is as a set of principal ultrafilters? if we consider $\omega$ with discrete topology, can we use mentioned topology for $\beta \omega$? – fatemeh Aug 28 '13 at 20:28
• @fatemeh: No, $\beta X$ is the set of all ultrafilters on $X$, principal and non-principal. And yes, this is exactly the topology on $\beta\omega$. – Brian M. Scott Aug 28 '13 at 20:33
• when you say $p \in \beta X$, is $p$ an ultrafilters on $X$? I mean the topology that you defined means the set of all ultrafilters that contain $A \subset X$? is this the same stone topology? – fatemeh Aug 29 '13 at 11:30
• @fatemeh: Yes, I’m taking $\beta X$ here to be the set of all ultrafilters on $X$, so $p\in\beta X$ is an ultrafilter on $X$. And yes, if $A\subseteq X$, then the set of all ultrafilters on $X$ that contain $A$ is a basic open set in $\beta X$. This is the topology that it gets as the Stone space of the Boolean algebra $\wp(X)$ and as the Čech-Stone compactification of the discrete space $X$. – Brian M. Scott Aug 29 '13 at 17:17