Why is $\beta \omega$ compact? I was trying to show to construct and prove the basic properties of the Stone-Čech compactification of $\omega$ using ultrafilters. I defined $\beta \omega$ as the set of all ultrafilters on $\omega$ and showed that $\{\hat A: A \subset \omega\}$ is an basis for some topology on $\omega$, where $\hat A=\{p \in \beta \omega: A \in p\}$. We identify $n \in \omega$ with the filter generated by $n$. Then I showed that $\omega$ is discrete, that $\omega$ is a dense subset of $\beta \omega$ and that $\beta \omega$ is Hausdorff. It remains to show that $\beta \omega$ is compact and that if $K$ is a compact Hausdorff space and $f: \omega\rightarrow K$ is given then there exists a countinuous $F: \beta \omega \rightarrow K$ such that $F(n)=f(n)$ for each $n \in \omega$.
I'm still trying to show that $\beta\omega$ is compact. How should I construct a finite subcover of an arbitrary open cover of $\beta \omega$? Also, can you give me some advice of how to build the function $F$ as above?
 A: If suffices to prove that a cover of $\beta\omega$ that consists of basic open sets $\hat A_i, i \in I$ has a finite subcover. 
The fact that this is a cover means that every ultrafilter on $\omega$ (i.e. every point of $\beta\omega$) contains some $A_i$ (i.e. $\in \hat A_i$).
This implies that $\omega \setminus A_i, i \in I$ does not have the finite intersection property, or else there would be an ultrafilter that would be a superset of this family, and this ultrafilter would contain no member $A_i$. So a finite intersection of them is empty, and the corresponding $A_i$ then cover $\omega$, so their corresponding $\hat A_i$ cover $\beta\omega$.
As to the extension of functions: there is no choice (also clear as $\omega$ is dense...), but to define $F$ for some ultrafilter $\mathcal{U}$, show that $\{f[U]: U \in \mathcal{U}\}$ is a base for an ultrafilter and so must have a (unique!) limit in $K$ by compactness of $K$. This is the only choice that will make $F$ continuous if it is to extend $f$.
