Existence of a $r>0$ such that $\{y\in \Bbb R^n:||y-x||\le r \text{for some $x\in K$}\}$ is compact 
Let $\Omega\subset\Bbb R^n$ be an open set and $K\subset \Omega$ compact.
Prove that there exists $r>0$ such that $\{y\in \Bbb R^n:||y-x||\le r \text{for some $x\in K$}\}$ is a compact subset of $\Omega$

If $y\in \Omega$ and $\Omega$ is open so $\exists r>0$ such that $B(y,r)\subset \Omega$.
But I need to find such an $r>0$ first which works here for all $y$ .
But I am unable to find such a $r$.
Please give me some hints on how to proceed.
As this question has been asked in an exam,Can someone please tell me what kind of books should I read so I can find these problems and practice them
 A: Let $F=\mathbb{R}^n\setminus\Omega$.  Since $\Omega$ is open, $F$ is closed.  Let us denote $d(x,y)=\|x-y\|$, for all $x,y\in\mathbb{R}^n$.  Then we have the definition of distance between two sets $A,B\subseteq\mathbb{R}^n$ as $d(A,B)=\inf\{d(a,b):a\in A,b\in B\}$.  Use the following standard result (easy to prove!).
$\textbf{Result.}$  Let $(X,d)$ be a metric space, $E\subseteq X$ be closed and $K\subseteq X$ be compact such that $F\cap K=\emptyset$.  Then $d(F,K)>0$.
So in the given problem, since $F\cap K=\emptyset$, let $\delta=d(F,K)>0$.  Then $r=\delta/2$ will be the required '$r$' value (again easy to check!).
Also useful will be the fact that a set $K\subseteq\mathbb{R}^n$ is compact if and only if it is closed and bounded.
A: For every $x\in K$, there is a radius $r_x$ such that $B(x;r_x)\subset \Omega$. Now the key idea is to divide the radii by 2 to give us more of a cushion. You'll see why this is important in a moment. The sets $B(x;r_x/2)$ cover $K$, so we can choose finitely many of these that give a finite subcover. Let the points be $x_i$ with the corresponding radius being $r_i=r_{x_i}/2$ for $0\le i\le n$. Let $r=\min_i r_i$. Now if $\|y-x\| \le r$ for $x\in K$, since the balls $B(x_i; r_i)$ cover $K$, there is some $x_i$ with $\|x_i-x\| < r_i$. Then by the triangle inequality we have $\|x_i-y\| < r_i+r \le 2r_{x_i}/2 = r_{x_i}$. Thus $y\in B(x_i; r_{x_i})\subset \Omega$. Thus the given set is a subset of $\Omega$ for our choice of $r$.
It seems like you're asking about how to find $r$, so I won't address showing that the set is closed and bounded. (Also these are relatively a lot easier I think.)
Also I'm not sure about references, but typically you'll find this sort of problem in analysis or topology books.
