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Let $X$ be an infinite dimensional topological space such that :

$ \forall n \in \mathbb{N}$, $ \exists X_{n} \subset X$, $n$-dimensional subspaces verifying :

  • $\forall r<n$, the homotopy groups $\pi_{r}(X_{n})$ are trivial.
  • $X_{n} \subset X_{n+1}$
  • $\bigcup_{n \in \mathbb{N}} X_{n}$ is dense in $X$.

Question : Is $X$ weakly contractible ?

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  • $\begingroup$ What does dimension mean here? Something like Lebesgue covering dimension, or do you really only care about CW complexes or something? $\endgroup$
    – Qiaochu Yuan
    Commented Aug 19, 2013 at 9:38
  • $\begingroup$ @QiaochuYuan : thank you for your comment. My needs are close to CW complexes, but I ask the question in general. The Lebesgue covering dimension seems coincide with the dimension of a CW-complex. I could add the condition that every open subspaces have the same Lebesgue covering dimension (to exclude the disjoint union of $\mathbb{S}^{1}$ and $\mathbb{S}^{2}$, for example). Do you know a name for such spaces? $\endgroup$
    – Sebastien Palcoux
    Commented Aug 19, 2013 at 11:32
  • $\begingroup$ I have posted a specification here. $\endgroup$
    – Sebastien Palcoux
    Commented Aug 19, 2013 at 15:09

2 Answers 2

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Of course not. Let $X = \mathbf R^\infty \times S^1$, and let $X_n = \mathbf R^{n-1} \times I_n$ where $\{I_n\}$ is an increasing sequence of intervals in $S^1$ whose union is $S^1 \setminus \{\mathrm{point}\}$.

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  • $\begingroup$ oups... your $X_{n}$ is not $n$-dimensional. $\endgroup$
    – Sebastien Palcoux
    Commented Aug 19, 2013 at 9:11
  • $\begingroup$ But you can take $H=R^{\infty}$ and $H_n =R^{n-1}$ instead. $\endgroup$
    – Johannes Ebert
    Commented Aug 19, 2013 at 9:15
  • $\begingroup$ Oh, I missed the condition on the dimension of $X_n$. I modified my example by your suggestion, Johannes. $\endgroup$
    – Dan Petersen
    Commented Aug 19, 2013 at 9:22
  • $\begingroup$ I have posted a specification here. $\endgroup$
    – Sebastien Palcoux
    Commented Aug 19, 2013 at 15:10
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As Dan pointed out, this is wrong in general, but there are situations where an approximation result does hold. One example for such results is due to Palais ('On the homotopy type of certain groups of operators', Topology, Vol 3). It applies when $X$ is an open subset of a Banach space, with some extra properties.

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  • $\begingroup$ Thank you. You enlighten me that I need to find some extra properties. I have some concrete examples in mind. I will read your reference, and perhaps I will open a more specific post. $\endgroup$
    – Sebastien Palcoux
    Commented Aug 19, 2013 at 11:45
  • $\begingroup$ I have posted a specification here. $\endgroup$
    – Sebastien Palcoux
    Commented Aug 19, 2013 at 15:10

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