# totally bounded set and pairwise disjoint balls

Let $(X,d)$ be a metric space and let $K\subset X$ be totally bounded. We are given some arbitrary $\epsilon >0$. Let $B\epsilon$ be a family of $\epsilon$ sized balls, such that any 2 balls from this family are disjoint. Assume $B\epsilon$ has infinite cardinality. Is it true to say that $K$ intersects a finite number of balls from the family $B\epsilon$?

• Yes . By definition of totally boundedness. – user118494 Sep 8 '15 at 17:07
• This is not the definition of totally boundedness. The definition is that a totally bounded set can be covered by finitely many $\epsilon$ balls. Here they do not even intersect. And clearly you do not know they cover all of K. – User666x Sep 8 '15 at 17:15
• @user118494 if you have an explanation to your answer, please ellaborate – User666x Sep 8 '15 at 17:42
• I may be wrong since I see an answer posted negating it with an example but this is what I thought : Let $K$ intersect infinitely many balls. And if they do not cover all of $K$ let $K_{0}$ be the subset of $K$ that that is equal to $K\cap \{B_{\epsilon}\}$ . Now subset of a totally bounded set is totally bounded so $K_{0}$ is totally bounded and hence only finitely many of the $B_{\epsilon}$'s are required to cover $K_{0}$ . Since the $B_{\epsilon}$'s do not intersect , the number of such balls intersecting $K$ should be finite . – user118494 Sep 8 '15 at 18:53
• @user118494 The total boundedness of a set $K_0$ means that it can be covered by a finite number of $\epsilon$ sized balls, and not means that any cover of $K_0$ by $\epsilon$ sized balls has a finite subcover. – Alex Ravsky Sep 9 '15 at 0:04

NO.Let $N$ be the positive integers Let $X$ be the set of real sequences $(x_n)_{n\in N}$ for which there is at most one $n>2$ with $x_n \ne 0$. Let $d((x_n)_{n \in N},(y_n)_{n\in N})=\sum_{n \in N}|x_n-y_n|.$ (Of course at most $3$ terms in the summation are non-zero.) Let $$K=\{(x_n)_{n \in N} \in X : 0\le x_1 \le 1 \text{ and } x_n=0\text{ for } n>1\}.$$ $K$ is compact so it is totally bounded.Take real sequences $(A_j)_{j \in N}$ and $(B_j)_{j\in N}$ where $$0<B_{j+1}<A_j<B_j<1\text{ for }j\in N.$$ Now for each $j$ let $D_j$ be the open disc of radius $1$ centered at $$C_j=(\sigma_{j,n})_{n\in N}$$ where $$\sigma_{j,1}=(A_j+B_j)/2 \text{ and } \sigma_{j,j}=1-(B_j-A_j)/2.$$ (All the other co-ordinates of $C_j$ are $0$.). Now I leave it to you to confirm that $\{D_j\}_{j\in N}$ is a pair-wise disjoint family and every D_j intersects K.

• I do not need the centers to be disjoint, I need the balls $D_j$ for $j\in N$ to be pairwise disjoint – User666x Sep 9 '15 at 8:58
• the balls are disjoint. i made a typo. – DanielWainfleet Sep 9 '15 at 9:17
• What happens when j=1? Your 2 definition for sigma do not coincide in this case – User666x Sep 9 '15 at 16:02

It seems the following.

In general, the answer is negative. Let $X=\ell^1$ and $\{e_n\}$ be the standard orts of the space $\ell^1$. Put $K=\{e_1/n: n\in\Bbb N\}$, $\epsilon=1$ and the set of centers of the closed balls of the family $B\epsilon$ is $\{e_n+e_1/n: n\in\Bbb N \}$.

• Interesting... You gave me food for thought that is for sure... – User666x Sep 8 '15 at 19:16
• The same idea as mine but a lot easier to describe and to verify. – DanielWainfleet Sep 8 '15 at 19:20
• @Alex Ravsky is it necessary that the balls be closed? – User666x Sep 8 '15 at 22:14
• @AvivEshed In user254665's counterexample the balls are open. – Alex Ravsky Sep 9 '15 at 0:31