general topology exercise Consider $\mathbb R^n$ with the usual metric. Let $U \subset \mathbb R^n$ be an open set and $K \subset U$ a compact set. 
Is the following affirmation true? 

There exists an open set $U^{'}$ in $\mathbb R^n$ with $\overline{U^{'}} \subset U $ and $K \subset U^{'}$ "

Can someone please give me some help?
I believe that the affirmation is true...
 A: For each $x\in K$, we have $x\in U$, so since $U$ is open, then there exists $r_x>0$ such that the open ball about $x$ of radius $r_x$ is a subset of $U$. Then the ball about $x$ of radius $\frac{r_x}2$ is certainly a subset of $U$, as is its closure. We can cover $K$ by a set of the latter sort of open ball--in fact, by finitely-many of them, since $K$ is compact. Let $U'$ be the union of that finite collection of open balls, so $U'$ is an open set, and $U'\supseteq K$. Since the closure of a finite union is the union of the closures, and since the closure of each of the open balls is contained in $U$, then $\overline{U'}\subseteq U$.
A: Since $K$ is compact, the distance from $K$ to $R^n \setminus U$ is a positive number, say $d$.
Then take a neighborhood ball with radius $d/2$ of each point of $K$.
Let $U'$ be the union of these balls. It satisfies $K \subset \bar U' \subset U$.
A: $\mathbb R^n$ is a Hausdorff space and $K$ is a compact subspace thereof. Thus $K$ is closed in $\mathbb R^n$.
Since $\mathbb R^n$ is also a normal space, your result follows.
