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Let $(X_{n},d_{n})_{n \in \mathbb{N}}$ be a sequence of complete geodesic metric spaces such that:

$X_{n}$ is a regular$^1$ CW-complex of constant local dimension$^3$ $n$, it is of finite type$^4$, boundaryless$^2$, unbounded, uniform$^5$, and it is the $n$-skeleton of $X_{n+1}$, which is n-connected. Moreover, the distances $d_{n}$ , $d_{n+1}$ generate the same topology on $X_{n}$ and $\forall x,y \in X_{n} \ d_{n+1}(x,y) \le d_{n}(x,y)$.
Finally $(X_{n},d_{n})$ is quasi-isometric to $(X_{n+1},d_{n+1})$, through the inclusion map $X_{n} \subset X_{n+1}$, and a distance $d$ on $ \bigcup{X_{n}}$ is defined (for $x, y \in X_{n_0}$) by $d(x,y) := lim_{n (\ge n_0) \to \infty} d_{n}(x,y)$.

Definition : Let $X:=\overline{\bigcup{X_{n}}}$ be the completion of the metric space $\bigcup{X_{n}}$ with $d$.
Question : Is $X$ weakly contractible ?

Remark: Some of these conditions could be useless for a proof, and others, highly generalized.
Motivation: See here for applications to geometric group theory and noncommutative geometry.

$^1$Regular (for a CW complex) : the attaching maps are homeomorphism (see this post).
$^2$Boundaryless (for a regular CW complex) : the boundary of each closed cell is contained is the union of the boundaries of other closed cells.
$^3$Constant local dimension : the topological dimension of all neighborhood of all point, is constant.
$^4$Finite type : finitely many $r$-cells ending in a fixed $(r-1)$-cell.
$^5$Uniform : For all $r$-cell $c_{1}$ and $c_{2}$, there is a neighborhood $n_{1}$ of $c_{1}$ and $n_{2}$ of $c_{1}$, such that $n_{1}$ is homeomorphic to $n_{2}$.

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    $\begingroup$ As a math.SE moderator I have closed the math.SE version of this question. It this is deemed inappropriate for MO, leave a comment to me on the math.SE question of this question and I/we can re-open it there. $\endgroup$
    – user642796
    Commented Sep 19, 2013 at 8:08
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    $\begingroup$ The limit distance $d(x,y)$ may be zero for some $x\ne y$, so it is not a metric in the usual sense. Do you disallow this, or use a generalized notion of a metric? $\endgroup$
    – Sergei Ivanov
    Commented Oct 2, 2013 at 21:19
  • $\begingroup$ We have posted this new question counteracting the answer below by adding a rigidity assumption. $\endgroup$
    – Sebastien Palcoux
    Commented Jul 15, 2017 at 14:41

1 Answer 1

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This is not true even in finite dimensions. There exists a decreasing sequence of complete Riemannian metrics on the plane, pairwise Lipschitz equivalent, such that the pointwise limit is isometric to the standard sphere without one point. Then the completion is the sphere.

To construct such a sequence, consider the metric of the punctured sphere in geodesic polar coordinates: $ds^2= dr^2+\sin^2 r\,d\varphi^2$ and add a term like $2^{-n}f(r)dr^2$, where $f(r)=1/r$ for $r$ near 0. This makes the distance to the origin infinite, so the metric is complete. But the additional term goes to zero as $n\to\infty$, so the limit is the standard metric of the punctured sphere.

To make an infinite-dimensional example, take a metric product with your favorite contractible infinite-dimensional cell complex.

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  • $\begingroup$ Thank you Sergei for your answer. Unfortunately your counter-example doesn't check all the axioms of the problem, in particular, your $X_{n}$ is not $n$-dimensional (but $2$-dimensional). Now, if you take a product with a contractible $(n-2)$-dimensional contractible space, then we will have many other problems with the other axioms. $\endgroup$
    – Sebastien Palcoux
    Commented Oct 3, 2013 at 11:42
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    $\begingroup$ I meant you take a product with a fixed infinite-dimensional space and let $X_n$ be the $n$-skeleton of the product. $\endgroup$
    – Sergei Ivanov
    Commented Oct 3, 2013 at 15:56
  • $\begingroup$ Let $(Y_{n})$ be your sequence of metric spaces (with your additional term on the metric structure, depending on $n$), and let $M$ be a contractible infinite dimensional cell complex. So you suggest to take $X_{n}$ as the $n$-skeleton of $Y_{n} \times M$. But why $X_{n} \subset X_{n+1}$ ? Perhaps this question reduces to : why $Y_{n} \subset Y_{n+1}$? Because as regular CW complex geodesic metric spaces, the second is a deformation of the first. Is it right, or do I forget an assumption ? Each one ? Perhaps : the restrictions of $d_{n}$ and $d_{n+1}$ on a $r$-cell ($r \le n$) are equal, right? $\endgroup$
    – Sebastien Palcoux
    Commented Oct 3, 2013 at 17:15
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    $\begingroup$ @SébastienPalcoux $Y_n$ is a constant sequence of CW-complexes (the CW structure is the same, only the metric is slightly different), so there is no problem with the inclusion $X_n\subseteq X_{n+1}$. Your remark about "deformation as geodesic metric spaces" probably refers to the fact that the Riemannian metrics generating metrics $d_n$ differ. But it's completely fine with the assumptions. I think it is finally a correct solution. $\endgroup$
    – savick01
    Commented Oct 4, 2013 at 7:26
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    $\begingroup$ @SébastienPalcoux No, that way you require the metric to be constant. In my opinion it is not natural if you want your metric to be geodesic. If you add an $n+1$-cell to an $n$-skeleton then geodesic distances usually decrease (consider a circle and add a disk inside). But formally it should be fine (even if you will have to struggle to define metrics and show that they are geodesic). $\endgroup$
    – savick01
    Commented Oct 4, 2013 at 7:56

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