TITLE: exhaustion by compactness
QUESTION [0 upvotes]: Let $V \subseteq \mathbb R^n$ show that there exists a sequence $(K_m)_{m \in \mathbb N}$ of subsets of $V$ that satisfies :
$(i) K_m \subset K_{m+1}$
$(ii) \cup_{n \in \mathbb N} K_m = V$
First i considered $V = \mathbb R^n$ and i did show part $(i)$ easily. But i do not know how to show part $(ii)$.
For the case $V \subset \mathbb R^n$ i have to use $K_m := \{x\in \mathbb R^n \vert d(x,A) \ge \frac{1}{m}\}$  for some suitable set $A$. But i do not know, how to choose an $A$ so that $K_m$ will be compact.
How should i approach $(i)$ and $(ii)$ for $V \subset \mathbb R^n$ and $V = \mathbb R^n$?

REPLY [0 votes]: It suffices to take $$K_m=\bar{B}(0,m)=\{x\in\mathbb{R}^n| d(0,x)\le m\}.$$
It is clear that $K_m$ is compact and it satisfies both $(i)$ and $(ii)$.
In fact, $K_m$ is a closed ball in $\mathbb R^n$, then it is compact.
On the other hand, let $y\in\mathbb R^n$, it exists $M\in\mathbb N$ large enough that $d(0,y)\le M$. Then $$y\in K_M\subseteq\cup K_m,$$ that is $\mathbb R^n\subseteq\cup K_m$.
If you want to write it as $$K_m=\{x\in\mathbb R^n|d(x,A)\ge\frac{1}{m}\}$$ it suffices to define $A_m$ as the set $B\left(0,m+\frac{1}{m}\right)^{c}$, while I don't understand why you need this description.
{"set_name": "stack_exchange", "score": 0, "question_id": 4187057}