Closure of an open set in manifold I have a question. During a proof of a proposition, the following is stated:
Let $K$ be a compact set in a manifold $M$ of dimension $n$. Then there exists an open set $U$ such that $K\subset U$ and such that $\bar U$ is compact.
Why can we find this open set? Is it true that every closed is compact in a manifold? (I'd say that this is somehow implied by being locally homeomorphic to $\mathbb{R}^n$ or $\mathbb{H}^n$ if it is true).
Thanks for your help!
 A: To find such a $U$, note that, because $M$ is a manifold, each $x \in K$ has an open neighbourhood, $N_x$ say, that is homeomorphic to the open $n$-dimensional unit ball via a homeomorphism that extends to the closed $n$-dimensional unit ball. The union of the $N_x$ cover $K$, hence as $K$ is compact, a finite subset $N_{x_1}, N_{x_2}, \ldots N_{x_k}$ of the $N_x$ also covers $K$. Take $U$ to be the union of this finite subset. Then, $U$ is open, $K \subseteq U$ and the closure $\bar U$ is compact, because it is the (finite) union of the closures of the $N_{x_i}$ each of which is compact (because it is homeomorphic to the closed $n$-dimensional unit ball).
It is not true that every closed subset of a manifold is compact. E.g. $\mathbb{R}^n$ is a closed but not compact subset of itself.
A: Take a compact neighbourhood $C_y$ of  each point  $y\in K$ and then extract a finite covering from the open covering $(\mathring C_y)_y$ .
Finally take  the union  of the finitely many selected $C_y$ 's.  
[This construction works for any locally compact topological space :  the manifold structure is irrelevant]
