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Let $X$ be a topological space and $N$ a subset of $X$. Is it true that the closure of

$N$ in $X$ is homotopy equivalent to $N$. I think it is not. take for example $N=\mathbb Q\subset \mathbb R=X$. Then $\bar Q=\mathbb R$ is contractible while $\mathbb Q$ is not even connected. This question came to my mind when i read that a submanifold with boundary keeps its homotopy type after removing its boundary.

Are there conditions where the closure of a subset has the homotopy type of the subset?

Thank you in advance.

palio
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  • You will certainly need closures of path connected subspaces to remain path connected. This problem was briefly discussed in this question and it doesn't seem to have a nice resolution. – Miha Habič Oct 15 '12 at 19:42
  • keeping the path connectivity property when passing to the closure, only ensures having the same $\pi_0$ and it is far from proving having the same homotopy type – palio Oct 15 '12 at 19:58
  • I still beleive that if $N$ deformation retracts onto a subset $V$ then $\bar N$ also deformation retracts onto $V$. – palio Oct 15 '12 at 20:03
  • I mentioned the path connectedness thing as a necessary condition for what you want to achieve. Also, your claim in the second comment cannot possibly be correct as you main post shows. $\mathbb{Q}$ deformation retracts onto $\mathbb{Q}$ but $\mathbb{R}$ doesn't. – Miha Habič Oct 15 '12 at 21:21
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    This is true for any open subset of R^n... and probably for any open subset of a smooth manifold... – Dylan Wilson Oct 17 '12 at 23:13
  • what is true the homotopy equivalence or the deformation retraction? please do you know where i can read something on this? – palio Oct 18 '12 at 09:04
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    The homotopy equivalence. And I don't know a good reference for this general phenomenon- I was just thinking about the existence of collars on manifolds with boundary. For that you can look at any book on smooth manifolds, probably. – Dylan Wilson Oct 18 '12 at 15:27
  • There are examples where everything is path-connected, too. Consider the $\mathbb{R}^n$ minus the origin (homotopy type of a sphere), and its closure $\mathbb{R}^n$ (homotopy type of a point). – jdc Apr 01 '14 at 05:07

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