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It is trivial that every set theoretic filter with added empty set is a topology (a collection of open sets).

What if we consider products of filters considered as topological spaces? (both Tychonoff product and box product)?

Let $A=A_{i\in n}$ and $B=B_{i\in n}$ are indexed families of filters where $n$ is an (infinite) index set.

Does it hold that there is a non-discrete topology $C$ finer both than $\prod A$ and $\prod B$ if an only if there is an $n$-indexed family $C'$ of proper filters such that $C'_i$ is finer both than $A_i$ and $B_i$ for every $i\in n$? (The question should be answered for both Tychonoff product and box product.)

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    It seems to me that the backward directio$n$ should be fairly straightforward, whichever product you're using. I will have to think on the forward direction.2012-06-07

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