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 $U$.
Relatively compact subset of open set in $\mathbb{R}^n$
3
$\begingroup$
general-topology
-
0Isn't $A$ relatively compact because it is bounded? Just apply Heine-Borel, or am I missing something? If the coordinates of elements of $A$ are bounded by $M$, then $\overline{A}$ is bounded by $2M$. – 2012-12-10
-
0Agreed. The closure of a set is closed, and we assume it is bounded. In a metric space, this is equivalent to compact. It is necessary to assume it doesn't intersect the boundary otherwise you can construct a sequence converging to the boundary for which there is no convergent subsequence, which would make the set $A$ not compact. – 2012-12-10
-
1Relative compact in $U$, not in $\mathbb{R}^n$. – 2012-12-10
-
0@Cantor: if something is compact in $\mathbb{R}^n$, it is definitely compact in $U$. edit: ah yes, $\overline{A}$ math is simply not a subset of $U$ if it intersects the boundary. – 2012-12-10
-
1What if closure of $A$ is not contained in $U$? – 2012-12-10
-
1Ugh, this is why I secretly hate mathematics. – 2012-12-10