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Let $a$ and $b$ are filters. The product $a\times b$ is defined as the filter (on the set of pairs) induced by the base $\{ A\times B | A\in a, B\in b \}$.

It is simple to show that product of a non-trivial ultrafilter with itself is not an ultrafilter (as it is not finer than the principal filter corresponding to the identity relation).

My question: Is product of every two (different) non-trivial ultrafilters always not an ultrafilter?

Note: non-trivial ultrafilter is the same as non-principal ultrafilter.

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    $\begingroup$ For the benefit of anyone who (like me) happens accross this page as an ultrafilter novice: Although the filter mentioned in this question is not an ultrafilter, nevertheless there IS an ultrafilter on $A \times B$ canonically induced by ultrafilters $\mu$ on $A$, $\nu$ on $B$. Call it $\mu \otimes \nu$. It's the "outer measure" generated by $\mu,\nu$. To wit, if $W \subset A \times B$, then $W \in \mu \otimes \nu$ iff for every finite cover $W \subset \bigcup_{i=1}^n U_i \times V_i$ of $W$ by "rectangles", there exists an $i$ with $U_i \in \mu$ and $V_i \in \nu$. $\endgroup$
    – Tim Campion
    Commented Mar 20, 2013 at 17:09
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    $\begingroup$ Note that $\mu \otimes \nu$ refines the filter the OP describes. Also, although this definition of $\mu\otimes \nu$ is obviously symmetric in $A$ and $B$, an alternative description can be given which is not so obviously symmetric. Namely, $W \in \mu \otimes \nu$ iff $\{a \in A \mid W|_{a, -} \in \nu\} \in \mu$. Here $W|_{a,-} \subset B$ is meant to denote the set $\{b \in B \mid (a,b) \in W\}$. $\endgroup$
    – Tim Campion
    Commented Mar 20, 2013 at 17:13
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    $\begingroup$ @TimCampion: I don't understand why you think the ultrafilter monad is commutative. Do you have a citation? Here I gave an argument for why the ultrafilter monad is not commutative: nforum.mathforge.org/discussion/5039/compactum/… $\endgroup$
    – Todd Trimble
    Commented Jul 27, 2013 at 21:17
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    $\begingroup$ @TimCampion Maybe a more direct way of pointing out the problem is to take a nonprincipal ultrafilter $\mu$ on $\omega$ and then asking: what is the outer measure of $\{(i, j): i < j\}$ under $\mu \otimes \mu$? Alternatively: if the measure of $\{(i, j): i < j\}$ is 1 under one and therefore both of the non-symmetric-looking definitions, then it would follow that $\{(j, i): i < j\}$ has measure 1 in the other and therefore both, and I guess it's clear why this leads to contradiction. $\endgroup$
    – Todd Trimble
    Commented Jul 29, 2013 at 20:21
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    $\begingroup$ @ToddTrimble I had convinced myself that this worked out, and was surprised. But it did seem to me that morally the ultrafilter monad should be commutative because it is a submonad of the monad of all measures, which I thought was commutative. On a second look at Kock's "Commutative Monads as a Theory of Distributions", I see he adds the disclaimer that this is false. It's only when you linearize things that measures become (in some sense...) commutative with respect to vector space $\otimes$. Perhaps my comments are outlandishly false enough that I should delete them... $\endgroup$
    – Tim Campion
    Commented Sep 30, 2013 at 19:51

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The product $a\times b$ of two ultrafilters is an ultrafilter if and only if, for every function $f$ from the underlying set of $a$ into $b$ (that's not a typo for "into the underlying set of $b$"), there is a set $A\in a$ such that $\bigcap_{x\in A}f(x)\in b$. One way for this to happen is for the underlying set of $a$ to be small enough and $b$ to be complete enough, as in Joel's answer. Notice, though, that the condition is, despite its appearance, symmetrical between $a$ and $b$. In particular, if $b$ lives on $\omega$ while $a$ is countably complete, then the condition is satisfied because $f$ will be constant on some set in $a$ (because countably complete ultrafilters are closed under intersection of continuum many sets). [Archaeologists may be interested to know that this characterization of ultrafilters whose product is ultra occurs on page 22 of my 1970 Ph.D. thesis, which is, thanks to patient scanning, available from my web page.]

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No. If $a$ and $b$ are principal ultrafilters, then so is the product filter as you have defined it. If $a$ and $b$ contain $\{x\}$ and $\{y\}$, respectively, then the base of your product includes the singleton $\{(x,y)\}$, and hence it is the principal ultrafilter.

But perhaps by "nontrivial" you meant nonprincipal. In this case, here is another example. Let $\mu$ be any ultrafilter on $\omega$ and let $\nu$ be a $\kappa$-complete ultrafilter on a measurable cardinal $\kappa$. If we consider the product filter $\mu\times\nu$ on $\omega\times\kappa$, as you have defined it, then it is an ultrafilter, since for any $X\subset \omega\times\kappa$, there are fewer than $\kappa$ many possible horizontal slices $X_\alpha=\{n\mid (n,\alpha)\in X\}$, and so there is some $A\subset \omega$ such that $\{\alpha\lt\kappa\mid X_\alpha=A\}\in \nu$. If $A\in\mu$, then $X$ is in the product filter, and if $A\notin\mu$, then the complement of $X$ is in the product filter. So $\mu\times\nu$ as you have defined it is an ultrafilter.

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