Corollary

If the union of a finite sequence $(A_i)_{1 \leq i \leq n}$ of subsets of $X$ belongs to an ultrafilter $\mathfrak{F}$, then at least one of the $A_i$ belongs to $\mathfrak{F}$.

Proof is by induction on $n$. In particular, if $(A_i)_{1 \leq i \leq n}$ is a covering of $X$, then at least one of the $A_i$ belongs to $\mathfrak{F}$. Proposition 5 characterizes the ultrafilters; more generally, we have: