0

At first I thought that only the empty set could possibly fit the requirement. Since if there is any element in a metric space then the set containing even only one of the element, say $a$, would have only one Cauchy sequence with its limit being $a$. Then the ball is complete. But for the empty set, there is a closed ball, namely, itself. However there is no Cauchy sequence, so nothing to check. Is it complete?　
I also thought about the discrete metric. I can make a discrete metric space with only one point, then according to the definition there is not even a ball in it. Since for a ball $B(p,r)$, $r>0$, but in this case $r$ can only be $0$. Is this going to work?

Ricc
  • 49
  • Suppose the empty set were not complete. Then it would contain a Cauchy sequence that does not converge. Contradiction, since the empty set doesn't contain anything. – silvascientist Sep 12 '15 at 07:05
  • 4
    Try taking $\mathbb Q$ as your metric space. – silvascientist Sep 12 '15 at 07:10
  • THe empty set is always a bit hairy. Most of the times, definitions explicitly exclude the empty set to avoid having to think about pathologies. I could well imagine that many definitions of metric spaces require the underlying set to be nonempty. – PhoemueX Sep 12 '15 at 08:17
  • A closed ball of radius $r$ about $p$ with respect to metric $d$, is $D_d(p,r)= {x : d(x,p)\le r$. For a one point space $ {p}$ we have $D_d(p,r)={p}$ for any $ r>0$. Silvascientist's comment is an example answer for your main Q. – DanielWainfleet Sep 12 '15 at 09:56

0 Answers0