Constructing the open cover from given open cover with the desired property Let $\{U_1,U_2,\ldots\}$ be a cover of $\mathbb{R}^n$ by open sets. Prove that there is an open cover $\{V_1,V_2,\ldots\}$ such that


*

*$V_j \subset U_j$ for each $j$, 

*each compact subset of $\mathbb{R}^n$ is disjoint from all but finitely many of the $V_j$.


My attempt: 
First thing we noticed that any compact set $K\subset \mathbb{R}^n$ is contained in some closed ball $\mathbb{B}[0,n]=B_n.$ Now since, each of the $B_n$ are compact so for each $n\in \mathbb{N}$ there exists finitely many $U_J$'s which will cover the ball $B_n$. Now assume that $j_n$ be the smallest index such that $B_n\subset \bigcup_{j=1}^{j_n} U_j.$ Now I stuck how to construct the sets $V_j$ from these things. 
 A: I shall denote by $B'(p,r)$ the closed ball centered at $p$ with radius $r$.
We know that $B'(0,1)\subset\bigcup_{n\in\mathbb N}U_n$. Since $B'(0,1)$ is compact, it is contained in finitely many of them. Let $F_1\subset\mathbb N$ be a finite subset of natural numbers such that $B'(0,1)\subset\bigcup_{n\in F_1}U_n$.
Now, we start again with $B'(0,2)$. It is contained in some finite union $\bigcup_{b\in F_2}U_n$, with $F_2\supset F_1$. And we can start again: $B'(0,3)\subset\bigcup_{n\in F_3}U_n$ with $F_3$ finite and containing $F_2$. And so on…
Now, if $n\in F_1$, let $V_n=U_n$. If $n\in F_2\setminus F_1$, let $V_n=U_n\setminus B'(0,1)$. If $n\in F_3\setminus F_2$, let $V_n=U_n\setminus B'(0,2)$. And so on… Note that, with this definition, $V_n$ is open and a subset of $U_n$.
Finally, for each $m\in\mathbb{N}\setminus\bigcup_{n\in\mathbb N}F_n$, let $V_m=\emptyset$.
If $K$ is a compact subset of $\mathbb{R}^n$, then $K\subset B'(0,N)$, for some $N\in\mathbb N$. And therefore only finitely many $V_m$'s cover $K$: those such that $m\in F_N$. In other words, $K\subset\bigcup_{k\in F_N}V_k$. All other $V_k$'s are disjoint form $B'(0,N)$ and therefore disjoint from $K$.
