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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.

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    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).2014-04-01

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