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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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One point I'd like to clarify is that if $\mathscr{F}$ is a set theoretic filter on a set $X$, then $\mathscr{F}$ fails to be a topology on $X$, but $\mathscr{F}\cup\{\emptyset\}$ is a topology on $X$. –  Cameron Buie Jun 7 '12 at 19:50
@CameronBuie: Edited. –  porton Jun 7 '12 at 19:57
It seems to me that the backward direction should be fairly straightforward, whichever product you're using. I will have to think on the forward direction. –  Cameron Buie Jun 7 '12 at 20:36
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