If a sequence in a metric space has no convergent subsequence, then there exist a open ball around each point

Question

"If a sequence in a metric space has no convergent subsequence, then there exist a open ball around each point that does not intersect any other point of the sequence "

I think the important idea for this proof is that if we take any point of the sequence and consider all points less than radius 1, we can take the minimum or half of the minimum and the ball around this point would not intersect any other point of the sequence. Now the reason why the neighborhood around any point must only contain a finite number of other points is because the sequence has no convergent subsequence. Otherwise we can create a convergent subsequence.

Is this correct?

The answers to your question here indicate that ${x_n \mid n \in A}$ has no limit points. You should be able to conclude from there. — Aryaman Maithani, Feb 14 '21 at 04:30
I guess if that lemma is already proven, then this is pretty much a corollary — Bill, Feb 14 '21 at 04:35

score 1 · Answer 1 · answered Feb 14 '21 at 04:44

1

If every open ball around point $p$ intersects a point of the sequence other than $p$, then you can construct a subsequence that converges to $p$.

answered Feb 14 '21 at 04:44

Robert Israel

448,999

If a sequence in a metric space has no convergent subsequence, then there exist a open ball around each point

1 Answers1