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Why is the following true? I saw it on Wikipedia but found no proof in the referenced book. Let $U\subseteq \Bbb R^n$ be an open subset. A subset $A\subseteq U$ is relatively compact in $U$, if and only if $A$ is bounded and the closure of $A$ in $\Bbb R^n$ does not intersect the boundary of $U$.

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    Ugh, this is why I secretly hate mathematics.2012-12-10

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The closure of $A$ relative to $U$ denoted by $\text{cl}_U(A)$ equals $\text{cl}(A)\cap U$. Since $\text{cl}(A)\subseteq\text{cl}(U)=U\bigsqcup\partial U$ and does not intersect the boundary, it is contained in $U$. Hence $\text{cl}_U(A)=\text{cl}(A)$. Since $A$ is bounded, so is the closure, and therefore it is compact.

Conversely, if the boundary of $A$ intersects $\partial U$, then it is not contained in $U$, and we have $\text{cl}_U(A)\subset\text{cl}(A)$. In this case $\text{cl}_U(A)$ is not closed and consequently not compact.

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As people pointed out, you can apply the Heine-Borel theorem to the closure of $A$, you just need to make sure the closure stays inside $U$ and that is why you need the condition that the closure does not intersect boundary of $U$.