Proof of Heine-Borel Theorem; Bartle

Question

I'm reading through the proof for the Heine-Borel Theorem in Bartle's Elements of Real Analysis and getting caught on one point:

We assume that $K$ is a compact subset of $\mathbb{R}^p$ and let $x$ be an element in the complement of $K$. Then we let $G_m=\{y\in\mathbb{R}^p:|x-y|>\frac{1}{m}\}$, $m\in \mathbb{N}$, so that for some $M\in \mathbb{N}$, such that $K$ is contained in $G_M$ and the union of all the $G_m$'s contain all of $\mathbb{R}^p$ except $x$.

I'm good up to here. Then he says, the neighborhood $\{z\in\mathbb{R}^p:|z-x|<\frac{1}{M}\}$ doesn't intersect with $K$, therefore the complement of $K$ is open. Why does this follow? Where are the elements exactly $1/M$ distance from $x$?

Answer 1

Bartle proved that, if $x\in K^\complement$, there is a neighborhood of $x$ that doesn't intersect $K$. In other words, $K^\complement$ is an open set. And this assertion is equivalent to the assertion “$K$ is a closed set”.

Answer 2

Note that $\{G_m : m \in \Bbb{N}\}$ is an open cover of $K$. By compactness of $K$, there must be a maximum $M \in \Bbb{N}$ such that only finite number of $G_m$ with $m\leq M$ cover $K$. But since those are nested subsets, then $G_M$ alone must cover $K$ and does not contain $x$. Because $G_M$ and $ \{z : |z-x| \leq 1/M\}$ are complement of each other, then $\{z : |z-x| < 1/M\}$ is an open neighbourhood of $x$ that does not intersect $K$.

Since this is true for every $x \in K^c$, then $K^c$ is open.