A problem on infinite dimensional metric space

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 $n_{2}$.

• 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. – user642796 Sep 19 '13 at 8:08
• 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? – Sergei Ivanov Oct 2 '13 at 21:19
• We have posted this new question counteracting the answer below by adding a rigidity assumption. – Sebastien Palcoux Jul 15 '17 at 14:41

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.
• 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. – Sebastien Palcoux Oct 3 '13 at 11:42
• I meant you take a product with a fixed infinite-dimensional space and let $X_n$ be the $n$-skeleton of the product. – Sergei Ivanov Oct 3 '13 at 15:56
• 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? – Sebastien Palcoux Oct 3 '13 at 17:15
• @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. – savick01 Oct 4 '13 at 7:26
• @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). – savick01 Oct 4 '13 at 7:56