Question regarding metric spaces and union of balls

Question

If $X$ is a compact space and $\epsilon > 0$, I want to show that there exists $n$ point $x_1, x_2, ... x_n$ such that $\bigcup_{i=1}^n B(x_i, \epsilon) = X$

I am not sure how to do that.

Intuitively I understand that since $X$ is compact, there exist a bunch of element of sequences of $X$ that belong in $X$. I think we have to show that these elements are dense enough in $X$ to find an $\epsilon$ such all the ball around those points will intersect and will form a space X.

How do I transform that intuition in a mathematical proof?

Suppose otherwise. What if there did not exist finitely many open $\epsilon$-balls whose union included the compact set $X$. — AJY, May 28 '16 at 06:57
In other words, this problem becomes a lot easier if you think of compactness in terms of its definition involving open covers, rather than its sequential properties. — Greg Martin, May 28 '16 at 07:01
@GregMartin maybe his definition was the sequential definition? — Henno Brandsma, May 28 '16 at 08:14

score 1 · Answer 1 · answered May 28 '16 at 08:28

Suppose not. Then there exists some $\varepsilon>0$ such that for all finite subsets $F \subseteq X$, $\cup_{x \in F} B(x,\varepsilon) \neq X$.

So start by picking $x_0 \in X$. By the above fact about $\varepsilon$, take $F = \{x_0\}$ and so we have that $B(x_0, \varepsilon) \neq X$, so pick $x_1$ with $d(x_1,x_0) \ge \varepsilon$. Use recursion:

Now suppose we have already picked $x_0,\ldots,x_n$ such that $d(x_i, x_j) \ge \varepsilon$ for all $i \neq j$ in $\{0,\ldots,n\}$. Use the property again for $F = \{x_0,\ldots,x_n\}$, and pick $x_{n+1} \in X \setminus \cup_{x \in F} B(x, \varepsilon)$. Then $d(x_{n+1},x_i) \ge \varepsilon $ for all $i \in \{0,\ldots,n\}$ and the other mutual distances are already OK. So $\{x_0,\ldots,x_n,x_{n+1}\}$ is as required.

This defines a sequence $(x_n)$ such that all distances between distinct elements are at least $\varepsilon > 0$. Now show that such a sequence cannot have a convergent subsequence (or even a Cauchy subsequence), because the distances then tend to $0$ on a subsequence. This contradicts (sequential) compactness.

score 0 · Answer 2 · answered May 28 '16 at 07:19

You are trying to show that a compact metric space is totally bounded. (In fact, a metric space is compact iff it is totally bounded and complete.)

For any $\varepsilon>0$, we have $$K=\bigcup_{x\in K} B(x,\varepsilon).$$ So if $K$ is compact, then there exists a finite subset $\{x_1,\ldots,x_n\}\subset K$ such that $$K=\bigcup_{i=1}^n B(x_i,\varepsilon),$$ and hence $K$ is totally bounded.

Question regarding metric spaces and union of balls

2 Answers2