Construction of Compactification. Let $U(S)$ be the collection of ultrafilters on a non empty set $S$. And let $O(A) = \{U\in U(S)| A\in U\}$ for $A\subseteq S$.
I am told to show that for $\tau = \{O(A)\subseteq U(S)| A\subseteq U\}$, we have that $(U(S),\tau)$ is a topological space.
The thing is, I am having trouble in seeing why the infinite union of $O(A)^{'s}$ is in $\tau$.
 A: In the definition of $\tau$, the requirement $A\subseteq U$ should presumably have been $A\subseteq S$.  Unfortunately, even with this correction, $\tau$ is not a topology (unless $S$ is finite and therefore all ultrafilters on it are principal).  The problem is that the operation $A\mapsto O(A)$ fails to respect infinite unions, and $\tau$ is in fact not closed under infinite unions.  Brian Rushton suspected that (in a comment on his answer), but his guess that the $O(A)$ are supposed to be the closed sets won't work either, because $A\mapsto O(A)$ doesn't respect infinite intersections either.  The best that can be said for $\tau$ is that it is a base for the usual topology on $U(S)$, which makes this set the Stone-Čech compactification of the discrete space $S$.
A: The topology refers to subsets of $U(S)$, not ultrafilters (but rather sets of ultrafilters). Of course you can take the union of a family of subsets of $U(S)$ (or any set).
A: But you're not taking the union of two ultra filters; you're taking the union of two sets of ultra filters, which is well-defined. $O(A)\cup O(B)=O(A\cup B)=$ the set of all ultra filters containing $A$ or $B$ (the equalities use the definition of ultrafilter).
