For $a$ an $I$-indexed family of filters and $S$ an $I$-indexed family of subsets of $U$ such that $U\smallsetminus S_i\notin a_i$ for every $i\in I$, define the restricted product $\prod^Sa$ by
$$\left(\prod\nolimits^Sa\right)R\Leftrightarrow\left(\prod a\right)R\land\{i\in I:R_i\ne S_i\}\text{ is finite.}$$
This is again a nonempty multifuncoid.

Then:

Every nonempty multifuncoid $f$ contains a restricted product of ultrafilters. Fix $S$ such that $f(S)$. For every $J\subseteq I$ finite, let $A_J$ be the set of sequences $a$ of ultrafilters such that $S_i\in a_i$ for every $i$, and $f(R)$ holds for every $R$ where $R_i\in a_i$ for $i\in J$, and $R_i=S_i$ for $i\notin J$. Then $A_J$ is closed in $(\beta U)^I$, $A_J\cap A_{J'}\supseteq A_{J\cup J'}$, and $A_J\ne\varnothing$ by the finite case, hence there exists $a\in\bigcap_JA_J$ by compactness of $(\beta U)^I$. Then $f\supseteq\prod^Sa$.

The restricted product of an infinite family of ultrafilters does not contain any product of a family of ultrafilters (assuming $U$ has more than one element), thus refuting the original wording of your conjecture. Indeed, if $f=\prod^Sa$ and $f(R)$, then $R_i=S_i$ for all but finitely many $i$, whereas if $g=\prod b$ is a product of a family of ultrafilters, we can for every $i\in I$ fix $R_i\in b_i$ such that $R_i\ne S_i$; then $g(R)$, but not $f(R)$, so $g\nsubseteq f$.

Point 1 says that the intuition behind the conjecture is basically sound, but the notion of the product has to be modified to make it really work to take into account that the axioms of multifuncoids only concern local behaviour when a single (or finitely many, by iteration) coordinate is changed, they do not imply anything about what happens when infinitely many coordinates change.

Since the proof above refers to the case of finitely many coordinates in a stronger form than what is claimed to hold in the question, I may as well give a self-contained proof of 1.

As before, fix $S$ such that $f(S)$. By definition, $S_i\ne\varnothing$ for every $i$. If $a$ is a family of filters such that $S_i\in a_i$ for all $i\in I$, consider a modified product
\begin{align}
\left(\prod\nolimits_ma\right)R&\Leftrightarrow(\forall i\in I)\,R_i\in a_i,\\
\left(\prod\nolimits_m^Sa\right)R&\Leftrightarrow\left(\prod\nolimits_ma\right)R\land\{i\in I:R_i\ne S_i\}\text{ is finite.}
\end{align}`
Note that if all $a_i$ are ultrafilters, then $\prod_ma=\prod a$, and $\prod_m^Sa=\prod^Sa$. It thus suffices to find $a$ such that $\prod_m^Sa\subseteq f$, and all $a_i$ are ultrafilters.

Let $P$ be the set of all families $a$ of proper filters such that $S_i\in a_i$ for all $i$, and $\prod_m^Sa\subseteq f$. We define a partial order on $P$ by $a\le b$ iff $a_i\subseteq b_i$ for all $i\in I$. It is easy to see from the definition of a multifuncoid that:

(*) Whenever $f(R)$, $R_i\subseteq R'_i$ for every $i$, and $R_i=R'_i$ for all but finitely many $i$, then $f(R')$.

It follows that $P$ is nonempty, since $a\in P$, where $a_i$ is the filter generated by $S_i$. Since the pointwise union of any chain in $P$ is an element of $P$, Zorn’s lemma implies that there exists a maximal element $a\in P$.

I claim that every $a_j$ is an ultrafilter. Assume for contradiction that it is not, and let $X\subseteq U$ be such that $X,U\smallsetminus X\notin a_j$. Define $b$ by $b_i=a_i$ for $i\ne j$, and $b_j$ is the filter generated by $a_j\cup\{X\}$. Since $a< b$, we have $b\notin P$, thus there exists $R$ such that $\neg f(R)$, $R_i=S_i$ for all but finitely many $i$, $R_i\in a_i$ for all $i\ne j$, and $X\cap Y\subseteq R_j$ for some $Y\in a_j$. Symmetrically, there exists $R'$ and $Y'\in a_j$ such that $\neg f(R')$, $R'_i=S_i$ for all but finitely many $i$, $R'_i\in a_i$ for $i\ne j$, and $(U\smallsetminus X)\cap Y'\subseteq R'_j$. Using (*) and the closure of $a_i$ under intersections, we can replace $R_i$ with $R_i\cap R'_i$ for all $i\ne j$, and the same for $R'_i$. Thus, without loss of generality, $R_i=R'_i$ for all $i\ne j$. But then by the definition of a multifuncoid, $\neg f(R'')$, where $R''_i=R_i=R'_i$ for $i\ne j$, and $R''_j=R_j\cup R'_j$. However, $R''_j\supseteq Y\cap Y'\in a_j$, hence $R''\in\prod_m^Sa\subseteq f$, a contradiction.

$%%%%$

`$I=\{0,1\}$`

when`$a_0$`

and`$a_1$`

are ultrafilters. Take`$A_0=B_1=U$`

and`$A_1=B_0=\emptyset$`

. $\endgroup$ – Andreas Blass Apr 7 '11 at 14:31