This post here is a specification of this [post](https://mathoverflow.net/questions/139790/homotopy-problem-for-infinite-dimensional-topological-space).


Let $(X\_{n},d\_{n})\_{n \in \mathbb{N}}$ be a sequence of [intrinsic metric](http://en.wikipedia.org/wiki/Intrinsic_metric) spaces verifying :


* $X\_{n}$ have [topological dimension](http://en.wikipedia.org/wiki/Lebesgue_covering_dimension) $n$.
* $X\_{n+1}$ is [n-connected](http://en.wikipedia.org/wiki/N-connected).
* $X\_{n} \subset X\_{n+1}$
* The distance $d\_{n}$ and $d\_{n+1}$ generate the same topology on $X\_{n}$.
* $\forall x,y \in X\_{n}$ : $d\_{n+1}(x,y) \le d\_{n}(x,y)$.
* $(X\_{n},d\_{n})$ is [quasi-isometric](http://en.wikipedia.org/wiki/Quasi-isometry) to $(X\_{n+1},d\_{n+1})$


**Definition** : Let $d$ be a distance on $ \bigcup{X\_{n}}$, defined as follows : $d(x,y) = lim\_{n \to \infty} d\_{n}(x,y)$. 


**Remark** : There is a small abuse in the previous definition because $d\_{n}(x,y)$ is defined only for $x, y \in X\_{n}$. But because we take $n \to \infty$, there is no problem.


**Definition** : Let $X:=\overline{\bigcup{X\_{n}}}$, the **complete metric space** obtained as a [completion](http://en.wikipedia.org/wiki/Complete_space#Completion) of $\bigcup{X\_{n}}$ for $d$.



> 
> **Question** : Is $X$ [weakly contractible](http://en.wikipedia.org/wiki/Weakly_contractible) ?
> 
> 
> 


**Remark :** If yes, perhaps some of these conditions are useless in the proof, and perhaps the useful conditions can be highly generalized.



Homotopy problem for infinite dimensional topological space II
The intristic metric assumption seems to be added to exclude $S^1$ from the previous post. But what about $S^d$? It can be approximated with bigger and bigger disks
$\{x\in S^d\ |\ x\_1\leq 1-\frac{1}{n} \}$ (with the natural intristic metric) and in the limit you get the sphere. (Of course dimension of such a disk is $d$ so one needs some obvious bypasses for $n<d$.)


As it was already noticed quasi-isometry requirement is trivial for bounded spaces.


Similarly with the dimension assumption: we can always take cartesian product with some nice space like $\prod\_{i=1}^n [0,\frac{1}{2^i}]$ with the $\ell\_2$-metric ($\frac{1}{2^i}$ is unnecessary but this way all the bounds are uniform).


