why compactness of metric space K means it can be covered by finite small open ball $d(x,p)<\delta$

In the process to prove theorem 7.25 of the Principles of Mathematical Analysis, I can't exactly understand why

"Since K is compact, there are finitely many points $p_1,...,p_r$ in K such that to every $x \in K$ corresponds at least one $p_i$ with $d(x,p_i)<\delta$"

is valid,

I understand the definition of compactness in chapther 2 of Rudin, and I can "heuristically" understand above statement. But I want to know the exact proof of above statement. Thank you!

• What is the definition of compactness in chapter 2 of Rudin? Aug 5 '17 at 14:07
• "A subset K of metric space X is said to be compact if every open cover of K contains a finite subcover."
– 백주상
Aug 5 '17 at 14:09

If $K$ is a subset in a metric space and $\delta > 0$, then $$K \subset \bigcup_{x \in K} B(x, \delta)$$ where $B(x,\delta)$ is the open ball centered at $x$ with radius $\delta$. This union is an open cover of $K$. If $K$ is compact, there exists a finite subcover $$K \subset \bigcup_{i = 1,\cdots,n} B(p_i, \delta)$$

• I'm wondering how we can control the size of ball. metric always contain all bounded real number?
– 백주상
Aug 5 '17 at 14:20
• @백주상, size of the balls doesn't matter, look at the indexing of the union: $x\in K$. That means that you've taken every point in $K$ plus something around it. Aug 5 '17 at 14:22
• In many concrete cases, the number of points $p_i$ will grow when $\delta$ approaches 0. For example if $K$ is the closed unit ball in $\mathbb{R}^d$, the number of points $p_i$ will typically be proportional to $\delta^{-d}$ when $\delta$ is small. Aug 5 '17 at 14:24
– 백주상
Aug 5 '17 at 14:26
• Then we assume that the points on metric space surrounding other points without any 'holes'?
– 백주상
Aug 5 '17 at 14:38

K is compact, so for a given open cover $\{N_{\alpha}\}$ that consists of neighborhouds with center $p_i$, there is a finite subcover (by the compactness). This means:

$K \subset N_{p_1} \cup \dots \cup N_{p_n}$ for finitely many points $p_1, \dots, p_n$

so for every $x \in K$, we can find a neighborhood $N_{p_i}$ such that $x \in N_{p_i}$, meaning that $d(x,p_i) < r_i = \delta$

• I'm wondering how we can control the size of ball. metric always contain all bounded real number?
– 백주상
Aug 5 '17 at 14:20
• This answer is very sloppy. So, what are $N_\alpha$'s and what is $p_i$? One would have to read from the end to the beginning to understand what you are doing. Aug 5 '17 at 14:26

Let $K$ be compact. Pick any point $x$. Cover $K$ with $\{ B(x,n) : n \in N \}$.
A finite number of those balls will cover K. Pick the largest.

A subset that is contained in a ball is called bounded.
The ball that contains $K$ may not be small. For example $K = [0,10^{100}]$.