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By filters I will mean filters on some set $\mho$.

I define product of an infinite family of filters in two ways. I feel (by analogy with properties of Tychonoff product vs box product of topological spaces) that these two products are different, but don't know how to find a counter-example supporting their non-equality.

The first product of family $a_{i\in n}$ (where $n$ is an (infinite) index set) of filters is defined as the filter induced by the base $$\Pi_1 = \left\{ \prod_{i\in n} A_i \,|\, A\in\prod_{i\in n}a_i \right\}.$$

The second product is the filter induced by the base $$\Pi_2 = \left\{ \prod_{i\in n} B_i \,|\, m\text{ is a finite subset of $n$}, A\in\prod_{i\in n\setminus m}a_i, i\in m\Rightarrow B_i=A_i, i\in n\setminus m\Rightarrow B_i=\mho \right\}.$$

Prove that $\Pi_1 \ne \Pi_2$ for some $a$ (counter-example wanted).

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  • $\begingroup$ I missed $n$ and $n\setminus m$, now they are interchanged. $\endgroup$ – porton Apr 4 '13 at 20:46
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If all the $a_i$ are principal ultrafilters on sets with at least two elements, then $\Pi_1$ will also be principal, since it concentrates on the singleton that picks out the base of each $a_i$. But $\Pi_2$ will not contain any singleton set (if the index set is infinite), since the $\Pi_{i}B_i$ will all contain continuum many elements.

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  • $\begingroup$ It seems for me that the requirement that they contain not less that two elements is superfluous. $\Pi_2$ will be proper even if all $a_i$ are non-proper filters, right? $\endgroup$ – porton Apr 4 '13 at 20:55
  • $\begingroup$ Oh, sorry, I've misunderstood you $\endgroup$ – porton Apr 4 '13 at 20:56
  • $\begingroup$ I had meant that the space should have at least two points, since otherwise, the product space has only one point, and the two product filters are the same, a singleton consisting of that point, the entire product space. $\endgroup$ – Joel David Hamkins Apr 4 '13 at 20:58

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