# Finitely many non-convergent ultrafilters

I am trying to prove that if a space $X$ has finitely many non-convergent ultrafilters, then every non-convergent ultrafilter $\mathcal U$ contains a set $A$ that is not contained in any of other non-convergent ultrafilters.

I honestly have no idea why this should be true, why anyone would think of it, and how to go about proving it.

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This has nothing to do with convergence. Given any finite family of distinct ultrafilters, say $U_1,\dots,U_n$, on a set $X$, each $U_i$ contains a set $A$ that is in none of the others. To prove it, fix $i$ and do the following for each $j\neq i$. Since $U_i\neq U_j$, there is a set $B_j$ that is in one of $U_i$ and $U_j$ but not the other. If it's in $U_i$ but not $U_j$, set $C_j=B_j$; if, on the other hand, $B_j$ is in $U_j$ but not $U_i$, then set $C_j=X-B_j$. Either way, $C_j$ is in $U_i$ but not in $U_j$. Now set $A=\bigcap_{j\neq i}C_j$. This $A$ is in $U_i$ because it's the intersection of finitely many sets from $U_i$. If $j\neq i$ then $A\notin U_j$ because $A\subseteq C_j$ and $C_j\notin U_j$.
An alternative proof views the ultrafilters as points in the Stone-Cech compactification $\beta X$ of the space you get by giving $X$ the discrete topology. Then the result says that, given finitely many distinct points, each has a neighborhood containing none of the others. That's true in any $T_1$ space, and $\beta X$, being a compact Hausdorff space, is way better than just $T_1$. (I suspect, though, that this topological proof hides essentially the same argument that I gave in my answer.) – Andreas Blass Jun 30 '13 at 15:13
Thanks. As you can tell, my intuition of ultrafilters sucks. I find your explanation using $\beta X$ intuitive, except that I don't see how the neighborhoods in $\beta X$ relate back to the elements inside the ultrafilters. – echoone Jun 30 '13 at 21:00
@echoone For discrete $X$, the clopen sets in $\beta X$ form a basis for the topology, and the clopen sets are exactly the sets of the form $\overline A=\{U\in\beta X: A\in U\}$ for arbitrary subsets $A$ of $X$. So, once you have a neighborhood of $U_i$ that contains none of the other $U_j$'s, shrink it to a clopen neighborhood $\overline A$, and that gives you the subset $A$ of $X$ that you need. – Andreas Blass Jun 30 '13 at 23:38