property is here
proof is here
My question is :
I don' know why" For any limit point of F must already be a limit point of some finite union". I have not yet learned the concept of locally finite in topological space. If it has to be used , can you explain it in toplogical space in R^n ?. and it is better for me without the concept to understand above question. so I expect to know that only by using definiton of limit point ,if possible
Since $K_k \subseteq E_k$ for all $k$, we know all points $x \in K_k$ have $k-1 \leq |x| < k$. So for any point $y \in \mathbb{R}^n$, we know $y$ has a distance at least 1 away from all $K_k$ sets for which $k$ satisfies $k\geq |y|+2$. Indeed, if $k \geq |y|+2$ and if $x \in K_k$, then $|x-y|\geq |x|-|y| \geq k-1-|y|\geq 1$.
So if $y$ is a limit point of $\cup_{k=1}^{\infty} K_k$, there are only a finite number of $K_k$ sets within a radius $1/2$ of the point $y$. That is, $y$ can only be a limit point of $\cup_{k=1}^N K_k$ for some finite integer $N$.