# Compact set covered by closed balls with positive radii [closed]

Let $$K\subseteq\mathbb{R}^2$$ be a compact subset. Suppose that it is covered by a collection of closed disks with positive radii. Is there a finite subcover?

• What have you tried so far? Commented Jun 27 at 4:53
• @CyclotomicField I thought about increasing the radii of the closed disks by $\epsilon$ to get finite subcover, but I couldn't go any further... Commented Jun 27 at 4:57
• The difference between an open cover and a closed cover is in the boundary points. See if you can find a counterexample using this idea. Commented Jun 27 at 5:01
• @CyclotomicField I can't... Can you give me more hint? Commented Jun 27 at 5:22

Here is a counter example. Let $$K = \{x\in R^2 \ |\ ||x||_2 \leq 1\}$$, denoted by $$B(0,1)$$, where $$B(x,r)$$ has center $$x$$ and radius $$r$$ and is a closed disk. $$K$$ has a cover $$\{B(0,r) | 0 < r < 1 \} \cup \{ B(x,1) | ||x||_2 = 2\}$$.
Note that the boundary of $$K$$ is covered point by point, so it does not have a finite subcover.
• I suppose you can take $K=\{x:\|x\|_2=1\}$ to avoid the balls $B(0,r),$0<r<1$. Commented Jun 27 at 5:51 • @geetha290krm, you are right! Commented Jun 27 at 6:07 • Do you know if there is a counter-example in the case of$\mathbb R$? Commented Jun 27 at 6:39 • @geetha290krm, a closed disk in$\mathbb{R}$is a closed interval$B(x,r)$. Assume$K =\{0\} \cup \{1/n, n \in \mathbb{N}_+ \} \subset \mathbb{R}$, then it is compact (closed and bounded).$K$is covered by$[-1, 0] \cup \{ [1/(n+1), 1/n], n \in \mathbb{N}_+ \}\$, which can not have a finite cover. Commented Jun 27 at 6:52