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Let $U$ be an open bounded subset of $\mathbb{R}^n$, and $K$ a compact subset of $U$. Does there always exist a compact subset $L$ of $U$ that contains $K$, and such that $L$ is a retract of $U$.

Looks to me true as I can get as close as I wish with $L$ to the to the limit boundary of $U$ and contain any compact $K$. As $L$ is close enough it will have all the connectivity properties of $U$ and will be a retract... but how do I prove it?

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  • $\begingroup$ Are you making any smoothness assumptions about the boundary of U? $\endgroup$
    – Yemon Choi
    Commented Mar 23, 2017 at 3:13
  • $\begingroup$ Have you looked at the section of Dold's "Lectures on Algebraic Topology" on ENRs? It might give you some ideas of how to prove (or disprove) what you want. $\endgroup$
    – Mark Grant
    Commented Mar 23, 2017 at 11:19
  • $\begingroup$ If you want $L$ to be a subpolyhedron then this is already false for domains in $R^3$: Use $U$ homeomorphic to the Whitehead manifold $W$. (A regular neighborhood of such a retract $L\subset W$ would have to be a PL 3-ball but $W$ cannot be exhausted by 3-balls.) I am not sure about general compact subsets $L$. $\endgroup$
    – Misha
    Commented Mar 23, 2017 at 15:14
  • $\begingroup$ For what it's worth, it is true for $n=2$. $\endgroup$
    – Mizar
    Commented Mar 24, 2017 at 11:26

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