Showing that a topology of ultrafilters over $\mathbb{N}$ is compact

Question

I have a set $\beta$ of all ultrafilters on $\mathbb{N}$, and I have shown that the sets $\mathcal{O}_A:= \{\mathcal U \in \beta: A \in \mathcal U\}$ form a basis of a hausdorff topology $\mathcal{T}$, in which all sets of the basis are closed and open.

I would now like to show that the space $(\beta, \mathcal{T})$ is compact, by showing that every cover of $\beta$ by basis sets have a finite subcover.

I am not really sure how to do this.. I have tried proving by contradiction, and got:

Suppose there exist sets $A_i$, $i \in I$ with $\cup_{i \in I} \mathcal{O}_{A_i} \supseteq \beta$, such that for all finite subsets $E$ of $I$ the finite subcover $\cup_{i \in E} \mathcal{O}_{A_i}$ does not overlap $\beta$. Then, for arbitrary $E$ we have that there exists an ultrafilter $U \in \beta$ with $A_e \not \in U$ for all $e \in E$, which is equivalent to $(A_e)^C \in U$, since ultrafilters contain a set, or the complement of that set.

I tried intersecting these sets, but I cannot seem to find any contradiction.. Any help on what I could do would be greatly appreciated!

Answer 1

Suppose that $\mathscr{A} = \{ A_i : i \in I\}$ does not have a finite subset covering $\mathbb{N}$. Then the family

$$\mathscr{C} = \{ \mathbb{N}\setminus A_i : i \in I\}$$

generates a filter $\mathscr{F}$ on $\mathbb{N}$. Let $\mathscr{U}$ be an ultrafilter containing $\mathscr{F}$. Then

$$\mathscr{U} \notin \bigcup_{i \in I} \mathcal{O}_{A_i},$$

which contradicts the assumption that $\{ \mathcal{O}_{A_i} : i \in I\}$ is a cover of $\beta$.

Hence there is a finite $E \subset I$ such that

$$\mathbb{N} = \bigcup_{i \in E} A_i.$$

Conclude that

$$\beta = \bigcup_{i \in E} \mathcal{O}_{A_i}.$$