# Proving limit points

• Oct 17th 2010, 07:16 PM
zebra2147
Proving limit points
Let $m>0$ be given. If for all $0, then for all $r>0$, $D\bigcap B'_{r}(c) \not= \emptyset$

I understand the definition of limit point and neighborhoods but I'm not sure how to use them to prove this. Any help would be appreciated.
• Oct 18th 2010, 03:51 AM
Math Major
A limit point for a set D in some metric space is a point p such that, every neighborhood of p contains points of D distinct from p, which is pretty much a rewording of what you have written up there. In order to show that a point is a limit point for a set, you just have to verify this property - take an arbitrary positive number $\epsilon$ and show that $\exists q \ne p$ in D such that $q \in B_\epsilon (p)$.

Do you have a specific problem you're working on?
• Oct 18th 2010, 04:49 AM
zebra2147
Thanks for that explanation. I'm trying to prove the statement that I typed above. I think my problem is that $D\bigcap B'_{s}(c) \not= \emptyset$ implies that there is at least one point in D that is also in $B'_{s}(c)$ such that c is not is not included. So if c is not included then how do I show that TWO other points exist to make the statement that you posted to be true?
• Oct 18th 2010, 05:08 AM
zebra2147
I may have just answered my own question...
Since $D\bigcap B'_{s}(c) \not= \emptyset$ then that means that $D\not= \emptyset$ when $0. Therefore, D contains elements. Thus, $D\bigcap B'_{r}(c) \not= \emptyset$ is true for $r>0$ since we just showed that D cointans elements on the interval $0 and $r>0$ contains all these points?????