Let $X_i$ be compact topological spaces and let $X = \prod_{i \in I}X_i$ and let $\mathscr F$ be ultrafilter on $X$. Define $\mathscr F_i = \{Y \subseteq X_i : \pi_i^{-1}Y \in \mathscr F\}$. Here $\pi_i : X \to X_i$ is projection mapping.
I am trying to show that if $X_i$ are Hausdorff then $|\bigcap \mathscr F_i| = 1$. So far all my tries fail and I am starting to doubt if it is true. But I also fail to make counterexample.
Thank you for any help.
(Some things I have proved so far:
If $X_i$ are Hausdorff then $X$ is also Hausdorff.
If $\mathscr F$ is ultrafilter then $|\bigcap \mathscr F| \le 1$. I
f $X_i$ is compact then $|\bigcap \mathscr F_i| > 0$.)