completeness of the Tensor product of vector spaces which are complete for filtration topology The following question actually came up (in a minor role) in research recently and I am having quite a hard time deciding it. Still the straightforward nature of the question made me decide to post it to stackexchange instead of overflow. Any help would be appreciated. 
Suppose $A$ is a filtered vector space with filtration 
$$\ldots \supset F^{-l}A\supset F^{1-l}A\supset\ldots F^{-1}A\supset F^0A\supset F^1A\supset\ldots\supset F^kA\supset F^{k+1}A\supset\ldots$$
such that $\bigcup_kF^kA=A$ and $\bigcap_kA=\{0\}$. Then we define the corresponding metric $d\colon A\times A\rightarrow \mathbb{R}_{\geq0}$ by setting $d(v,w)=0$ if $v=w$ and $d(v,w)=2^{-|v-w|}$ is $v\neq w$ where for all $x\neq 0$ we have $|x|=\max\{k\mid x\in F^kA\}$. We now suppose additionally that $A$ is complete. Finally consider another vector space $B$ with a similar filtration and again such that $B$ is complete for the corresponding metric. 
Now the (algebraic) tensor product is naturally equipped with the filtration 
$$F^k(A\otimes B)=\bigoplus_{\substack{i,j\in \mathbb{Z}\\ i+j=k}}F^iA\otimes F^jB$$ and it is striaghtforward to show that this is again a filtration such as the one above. The question is now simple. Is the metric corresponding to this filtration on $A\otimes B$ complete? Of course I also want to know why/why not. 
I think that it is in fact complete, but I can't quite come up with a proof. The first thing to try would be to show that Cauchy sequences in $A\otimes B$ are somehow "built" out of Cauchy sequence in $A$ and $B$ respectively, but this is not straightforward, since for instance taking any sequences $a_k\in F^{2k}A$ and $b_k\in F^{-k}B$ yields a Cauchy sequence $a_k\otimes b_k$ in $A\otimes B$. The sequence $B_k$ however does not need to be Cauchy at all! of couse in this specific case the sequence converges to $0$. Even just considering Cauchy sequences of pure tensors $a_k\otimes b_k$ I seem to be having some trouble. 
 A: Well, don't I feel silly answering my own question before anyone even commented on it and of course the answer comes from the most obvious example. 
The tensor product is not in general complete. Namely if we denote the base-field by $L$ and we let $A=L[\![t]\!]$ and $B=L[\![s]\!]$ both the ring of formal power series. We consider the filtrations $F^kA=t^kA$ and $F^kB=s^kB$ for $k\geq 0$, while  $F^kA=A$ and $F^kB=B$ for $k\leq 0$. Then these satisfy the requirements. Now the tensor product $A\otimes B$ is given by sums 
$$\sum_{k,l\geq 0}\sum_{i=1}^f a_{k,i}b_{l,i} t^k\otimes s^l.$$ It is filtered by 
$F^k(A\otimes B)=\bigoplus_{l=0}^kt^lA\otimes s^{k-l}B$. Thus we find the Cauchy sequence $c_n=\sum_{m=0}^n t^m\otimes s^m$. A limit of this sequence would be given in the form above by the system of equations 
$$\sum_{i=1}^f a_{k,i}b_{l,i}=\delta_{kl}$$ for all $k,l\geq 0$ and where $\delta_{kl}$ denotes the Kronecker delta. Now given $f\in \mathbb{N}$ we find that no such limit can exist since it would mean the matrices $A=(a_{k,i})_{\substack {1\leq k\leq f+1\\ 1\leq i\leq f}}$ and $B=(b_{l,j})_{\substack {1\leq l\leq f+1\\ 1\leq j\leq f}}$ determining linear maps $A, B \colon L^f\rightarrow L^{f+1}$ would satisfy $AB^T=I_{f+1}$, i.e. the identity of $L^{f+1}$ would factor through $L^f$. 
