Compactification of a discrete space using ultrafilters.

Question

I want to show for the collection of ultra filters on a (non-empty) set $A$, $U(A)$. That $U(A)$ is compact where the topology is derived from the base $U_B = \{F\in U(A)|B\in F\}$.

Seeing as $A$ can be an infinite set, I am having a hard time showing this.

Answer 1

Consider an arbitrary open cover $\bigcup_{i \in I} U_{A_i}$ of the space and assume the contrary - i.e. every finite subset does not cover the whole space. Observe, that the complements of the $A_i$ then generate a filter, which can be extended to an ultrafilter. But this one is not covered by any $U_{A_i}$.