Ultrafilters over product spaces

Question

Suppose that for $i\in I$, $X_i$ are topological spaces and $U_i$ is an ultrafilter over $X_i$. Consider the space $\Pi_i X_i$ with the product topology.

I want to know when, if ever, it is possible to construct an ultrafilter $U$ over $\Pi_i X_i$ such that for every open $\Pi_i A_i \subseteq \Pi_i X_i$:

• $\Pi_i A_i \in U$ iff $A_i\in U_i$ for every $i$.

Answer 1

In fact, it is possible to construct an ultrafilter $U$ such that the condition $\prod_{i \in I} A_i \in U$ iff $A_i \in U_i$ for every $i$ holds for all $(A_i)_{i \in I} \in \prod_{i \in I} P(X_i)$ - regardless of whether or not $\prod_i A_i$ is open. Here is an outline of how you would construct it - let me know if you need help in filling in any of the details.

• $\{ \prod_{i\in I} A_i \mid (A_i) \in \prod_{i \in I} P(X_i) \}$ forms a basis for some filter.
• Therefore, there is some ultrafilter $U$ containing this set.
• By construction, the reverse implication in the condition on $U$ is easy. For the forward implication, suppose that $\prod_{i\in I} A_i \in U$. Then for each $i$, either $A_i \in U_i$ or $X_i \setminus A_i \in U_i$; let $A_i' := A_i$ or $X_i \setminus A_i$ accordingly so that $A_i' \in U_i$. Then $\prod_{i\in I} A_i' \in U$. On the other hand, if $A_i \notin U_i$ for some $i$, then $\left( \prod_{i\in I} A_i \right) \cap \left( \prod_{i\in I} A_i' \right) = \emptyset$, leading to a contradiction.