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).



Two different products of filters
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. 


