Product of ultrafilters, is it an ultrafilter? - MathOverflow

Product of ultrafilters, is it an ultrafilter?

2011-08-19T19:05:37Z

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.

Answer by Joel David Hamkins for Product of ultrafilters, is it an ultrafilter?

2011-08-19T19:11:24Z

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.

Answer by Andreas Blass for Product of ultrafilters, is it an ultrafilter?

2011-08-19T20:22:47Z

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.]