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Let $M$ be a metric space, let $K \subset V \subset M$, $K$ compact, $V$ open. Prove that there exist $r>0$ such that $\bigcup_{x \in K} B(x,r) \subset V$

I came up with a proof, but there is a step that I am not sure if it is correct.

Since $K$ is compact, is complete and totally bounded. Therefore, given $\epsilon >0$ there exist $x_1,x_2,...,x_n \in K$ such that $K=B(x_1,\epsilon)\cup ...\cup B(x_n,\epsilon)$.

Define $X_i = \{x \in K : d(x,x_i)<\epsilon\}$. Now I state the following. There exist $m_i\in \mathbb{N}$ such that for all $x \in X_i$, $B(x,1/m_i) \subset B(x_i,\epsilon)$. Suppose that this $m_i$ doesn't exist, then for all $m \in \mathbb{N}$ there exist $x \in X_i$ such that $B(x,1/m) \nsubseteq B(x_i,\epsilon)$ wich contradicts that $B(x_i,\epsilon)$ is open (this is the step I am doubting.)

Now taking $m=min\{m_i : i \in \{1,..n\}\}$ We have that $\bigcup_{x \in K}B(x,1/m) \subset \bigcup_{i = 1}^{n}B(x_i,\epsilon)=K \subset V$

What do you think?

Joaquin Liniado
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1 Answers1

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Yes, you are right to doubt that: there is no such $m_i$ in general. Just consider the interval $(-1,1)$ as an example: although it's open, there is no $m$ such that for every $x\in (-1,1)$ the ball $B(x,1/m)$ is contained in $(-1,1)$. This is evident geometrically: they closer $x$ is to the boundary, the smaller its neighborhood has to be to fit inside.

Observe that your goal is to prove that the distance between compact set $K$ and closed set $V^c$ is positive. This has been asked and answered many times: see A and B disjoint, A compact, and B closed implies there is positive distance between both sets and linked questions.

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