# Is it possible to take “limits up to homotopy”?

Before I begin, an apology: it's been a while since I've done much analysis, so I might be misusing (or just missing) terminology.

I have a chain complex $(V_\bullet,\partial)$ of topological vector spaces (in particular, the differential is continuous). My vector spaces are Cauchy-complete in the following sense: If $x_n \in V_i$ is a sequence of vectors of fixed degree such that $\lim_{\min(m,n) \to \infty} (x_n - x_m) = 0$, then $\lim_{n\to \infty} x_n$ converges.

Actually, I have a particular sequence $x_n \in V_0$ that I care about. Unfortunately, $\lim_{n\to \infty} x_n$ does not converge. But it is "a Cauchy sequence up to homotopy." Specifically, for any two $m,n$, I have a specific vector $y_{m,n} \in V_1$ such that: $$(*) \quad\quad\quad \lim_{\min(m,n) \to \infty} (x_n - x_m - \partial(y_{m,n})) = 0$$ Also, I should mention that each individual $x_n$ is not closed, so I can't talk about their classes in homology. But I do know that $\lim_{n\to \infty}(\partial x_n) = 0$. (Going in the other direction, none of the natural limits of the $y_{m,n}$ converge, but I can prove variations of equation $(*)$ for them too, and so on ad infinitum.)

My question is whether $\lim_{n\to \infty} x_n$ exists in some homotopical sense, and how to define it. Of course, I don't expect that there is a specific (closed) element $x_\infty = \lim x_n$. But I would expect that there is some natural set of such elements $x_\alpha$, for $\alpha$ ranging over some indexing set, along with specific homotopies $y_{\alpha,\beta}$ satisfying $\partial(y_{\alpha,\beta}) = x_\beta - x_\alpha$.

If not, are there additional reasonable conditions that would assure such a limit? I have pretty good control over my particular example. For instance, much stronger than the Cauchy-completeness, I know that if $\lim_{n\to \infty} (x_{n+1} - x_n) = 0$, then $\lim_{n\to \infty} x_n$ converges. The reason I know things like that is because the specific vector spaces $V_\bullet$ that I care about are essentially power-series algebras over $\mathbb Q$. So if there's some natural condition that's necessary for convergence, I can easily check it, and it's probably satisfied.

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Sorry if this is silly, but: let $B_* \subset V_*$ be the subspace given by the image of $\partial_{*+1}$. Then, consider the quotient $B_*^\perp = V_*/B_*$ and let $\rho_*:V_* \to B_*^\perp$ be the obvious quotient map. What exactly is the difference between $x_n \subset V_\ast$ "converging up to chain-homotopy" and standard convergence of $\pi_*(x_n)$ to zero in $B_*^\perp$? –  Vidit Nanda Jul 25 '13 at 4:26
@ViditNanda: Um, good question. For some reason, I'm very trained to think that non-closed things aren't "real" in some sense. Anyway, the only difference between my data and a sequence that converges in $V/B$ is that I have a little extra data, e.g. the $y$s. –  Theo Johnson-Freyd Jul 25 '13 at 19:18
One thing I didn't explicitly ask for, but that one should always hope for, is good behavior under homotopy equivalences of complexes. –  Theo Johnson-Freyd Jul 25 '13 at 19:20
This is a fascinating question, but could you clean it up a bit? Particularly, (1) the use of $x_n$ as both a sequence and the $n$-th element of that sequence makes it hard to parse, and (2) it'd help to know what variations on equation $*$ hold for the $y$'s. The importance of (2) is that we have no control whatsoever on the $y$s as stated, so for instance things can get really hideous and Cantor-set like. Finally, what is your index-set on the chain complex? Wouldn't *everything* be closed in $V_0$ if you use $\mathbb{N}$? –  Vidit Nanda Jul 27 '13 at 15:24