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

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

Let U be an open bounded subset of 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?

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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E.G.
E.G.
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Approximating an open bounded set by compact set

Let U be an open bounded subset of 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?