It is from the Handbook of Set-theoretic topology:
We consider $\beta \omega$ as the set of all ultrafilters on $\omega$ with the topology which is generated by taking as a base all sets of the form $\overline{V}=\{u\in \beta \omega: V\in u\}$ where $V\in [\omega]^\omega$. The integers are identfied with the fixed(=principal) ultrafiters, and $\omega$ is dense in $\beta \omega$.
Could somebody help me to understand the structure of the space $\beta \omega$?
How to understand that the integers are identfied with the fixed(=principal) ultrafiters?
Thanks for your help.
To construct the Stone-Čech compactification $\beta X$ of a Tychonoff space using ultrafilters, you let $\beta X$ be the set of all ultrafilters on the zero sets of $X$. The base mentioned above (where the sets $V$ are zero sets) would then be the $closed$ base for its topology. Of course when dealing with discrete spaces like $\omega$, any subset is a zero set, so you consider ultrafilters on $\mathcal P(X)$. Prove that the natural assignment of points of $\omega$ to principal ultrafilters $x\to\{V\subseteq\omega:x\in V\}$ is one-to-one. Then you need to prove this defines an embedding, that $\beta\omega$ is compact Hausdorff, and that $\omega$ is dense in $\beta\omega $. It turns out that the base set $\overline V$ is the closure of $V$ in $\beta\omega$. From here you can show that disjoint zero sets (arbitrary subsets in this case) have disjoint closures in $\beta\omega$. This is one of several equivalent properties which uniquely determine $\beta\omega$ up to homeomorphism.
Remember that ultrafilters are maximal collections w.r.t. the finite intersection property.
If $u$ is a filter on $\mathcal P(X)$, then $u$ is an ultrafilter iff every subset of $X$ that intersects every element of $u$ is an element of $u$ iff for every $A\subseteq X$, either $A\in u$ or $X\setminus A \in u$.