Products of cofinite spaces are compact exactly under BPI

Assume BPI for the forward direction. If $X=\mathrm{\varnothing }$ for any reason, it is compact; this includes an empty factor and does not assert that nonempty factors have nonempty product. If $I=\mathrm{\varnothing }$, the product is a singleton and compact. In the remaining case $X\mathrm{\ne }\mathrm{\varnothing }$, fix one $\bar{x}\in X$.

By [L1], it is enough to let $\mathcal{H}\subseteq \mathcal{C}(X)$ have the finite intersection property. By [F3], extend the filter generated by $\mathcal{H}$ to an ultrafilter $\mathcal{F}$ of subsets of $X$.

For each $i\in I$, push $\mathcal{F}$ forward through the projection by putting${\mathcal{G}}_i:=\{B\subseteq X_i:{\pi }_i^{-1}[B]\in \mathcal{F}\}\mathrm{.}$ Preimages preserve complements and finite intersections, so ${\mathcal{G}}_i$ is an ultrafilter on $X_i$. Its closed members have the finite intersection property. By compactness of the cofinite factor [L2], their intersection $A_i:=\bigcap \{C\in {\mathcal{G}}_i:C\text{ is closed in }{X}_i\}$ is nonempty.

By [F1], each $A_i$, being an intersection of closed sets, is either all of $X_i$ or finite. If it is finite, ${\mathcal{G}}_i$ contains a finite set: this is $X_i$ itself when $X_i$ is finite, while if $X_i$ is infinite and $A_i\mathrm{\ne }{X}_i$, some closed member occurring in the intersection defining $A_i$ is proper and hence finite. An ultrafilter containing a nonempty finite set contains exactly one of its singleton subsets. If that singleton is $\{a_i\}$, then every member of ${\mathcal{G}}_i$ contains $a_i$, and $\{a_i\}$ is itself a closed member, so $A_i=\{a_i\}$. Define $x_i$ to be the unique point of $A_i$ whenever $A_i$ is a singleton, and otherwise put $x_i:={\bar{x}}_i$; in this latter case $A_i=X_i$. This defines $x=(x_i{)}_{i\in I}\in X$ without making a new choice.

Fix $H\in \mathcal{H}$. By [L1], write $H={\bigcup }_{q\in Q}{\pi }_q^{-1}[C_q]$ with $Q$ finite and each $C_q$ closed. Since $H\in \mathcal{F}$, finite primeness of the ultrafilter gives some $q\in Q$ with ${\pi }_q^{-1}[C_q]\in \mathcal{F}$. Thus $C_q\in {\mathcal{G}}_q$, so $x_q\in A_q\subseteq C_q$ by step 4.1. Hence $x\in H$. Therefore $x\in \bigcap \mathcal{H}$, and [L1] proves that $X$ is compact.

The forward implication is established in step 5.1. For the converse, assume every product of cofinite spaces is compact. Let $S$ be any set carrying a proper filter $\mathcal{D}$, and put $Y=2^{\mathcal{P}(S)}$. The two-point cofinite topology is discrete, so the hypothesis makes $Y$ compact. The constant-zero function shows $Y$ is nonempty in ZF. The finite discrete factor is also the one described in [F4]; no converse for arbitrary compact Hausdorff products is inferred from that fact.

On a point $v:\mathcal{P}(S)\to 2$ impose the constraints $v(S)=1$, $v(\mathrm{\varnothing })=0$, $v(S\setminus A)=1-v(A)$ and $v(A\cap B)=v(A)v(B)$ for every $A,B\subseteq S$, and $v(D)=1$ for every $D\in \mathcal{D}$. Each constraint defines a clopen subset of $Y$: it is a finite union of patterns in its finitely many involved coordinates, and both each pattern and its complement are unions of basic discrete-coordinate cylinders. This family has FIP. For any finite list of constraints, only finitely many filter-members $D_1,\dots ,D_k$ occur. Their intersection is nonempty by properness of the filter, with the empty intersection equal to $S$, also nonempty. Fix $s$ in that intersection and set $v_s(A)=1$ exactly when $s\in A$. This valuation satisfies all Boolean equations, including the listed filter constraints. Thus [F2] and compactness give a single $v$ satisfying every constraint.

Put $\mathcal{U}=\{A\subseteq S:v(A)=1\}$. The constraints include $S$ and exclude $\mathrm{\varnothing }$, close $\mathcal{U}$ under intersections, and make it upward closed: if $A\subseteq B$ and $v(A)=1$, then $1=v(A\cap B)=v(A)v(B)$ forces $v(B)=1$. They also decide exactly one of $A$ and its complement, so [F3] makes $\mathcal{U}$ an ultrafilter extending $\mathcal{D}$. Every proper filter has therefore been extended; if $S$ is empty there is no proper filter and the assertion is vacuous. By the UFL-to-BPI direction of [F3], BPI holds. Combined with step 5.1, this proves the equivalence.