Maybe this is too technical and elementary, but I cannot make up my mind, nor find a reference.

The situation is the following: let $X$ be a double cochain (right half-plane) complex of abelian groups and let

$$ (\mathbf{Tot}^{\prod} X)^n = \prod_{p+q = n}X^{pq} $$

denote its total-product complex. The first filtration on $X$,

$$ {}_I F^s(X) = \begin{cases} X^{pq} & \ \text{if} \ p \geq s , \\\ 0 & \ \text{otherwise} \end{cases} $$

gives you the filtration on $\mathbf{Tot}^\prod X$:

$$ (F^s \mathbf{Tot}^\prod X)^n = \prod_{p+q=n , p\geq s} X^{pq} $$

and you have, as with any filtered differential complex $(A, F, d)$, an induced filtration in cohomology:

$$ F^pHA = \mathbf{im} (HF^pA \longrightarrow HA) \ . $$

My question is the following: is the filtration induced by ${}_I F$ on $H(\mathbf{Tot}^\prod X)$ Hausdorff? That is,

$$ \bigcap_p F^pH(\mathbf{Tot}^\prod X ) = 0 \ ? $$

I couldn't find an answer in the literature. Weibel's book says that in this situation the spectral sequence arising from the first filtration is "convergent". Unfortunately, for Weibel this only means that you have an isomorphism $E_0 HA = E_\infty A$. Cartan-Eilenberg "Homological Algebra" doesn't work with the total-product complex, but with the total-sum one:

$$ (\mathbf{Tot}^{\bigoplus} X)^n = \bigoplus_{p+q = n}X^{pq} $$

For this one, I think, the answer is "yes": if I had some cohomology class

$$ [x] \in \bigcap_p F^pH^n(\mathbf{Tot}^\bigoplus X ) \quad \Longleftrightarrow \quad [x] \in F^pH^n(\mathbf{Tot}^\bigoplus X ) \ \text{for all} \ p $$

then I could find representatives for $[x]$ like

$$ (0, \stackrel{p-1}{\dots}, 0, x^{p,n-q}, x^{p+1, n-q-1}, \dots ) $$

for all $p \geq 0$. Since there is only a finite number of $x^{pq} \neq 0$ for each element of $(\mathbf{Tot}^\bigoplus X)^n$, in a finite number of steps, I can be sure that I can find a representative for $[x]$ which is zero, so $[x]=0$.

But this reasoning doesn't work with $\mathbf{Tot}^\prod X$: you can only state with certainty that, for every $p$ there is some $x_p \in F^p$ and $b_p$ such that

$$ x - x_p = db_p \quad \Longleftrightarrow \quad x- db_p \in F^p \ . $$

These equations have a nice interpretation: if you topologize $\mathbf{Tot}^\prod X$ taking as basic open sets $x + F^p$ for all $x \in \mathbf{Tot}^\prod X$ and $p$, they read:

$$ (db_p) \longrightarrow x \ . $$

That is, $x$ is a limit of coboundaries. But, unless the filtration is finite, this doesn't imply that $x$ itself *is* a coboundary, does it?

So I could ask my question this way: in this situation, is the set of coboundaries closed?