
Proving limit points
Let $\displaystyle m>0$ be given. If for all $\displaystyle 0<s<m, D\bigcap B'_{s}(c) \not= \emptyset$, then for all $\displaystyle r>0$, $\displaystyle D\bigcap B'_{r}(c) \not= \emptyset$
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.

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 $\displaystyle \epsilon $ and show that $\displaystyle \exists q \ne p $ in D such that $\displaystyle q \in B_\epsilon (p) $.
Do you have a specific problem you're working on?

Thanks for that explanation. I'm trying to prove the statement that I typed above. I think my problem is that $\displaystyle D\bigcap B'_{s}(c) \not= \emptyset$ implies that there is at least one point in D that is also in $\displaystyle B'_{s}(c)$ 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?

I may have just answered my own question...
Since $\displaystyle D\bigcap B'_{s}(c) \not= \emptyset$ then that means that $\displaystyle D\not= \emptyset$ when $\displaystyle 0<s<m$. Therefore, D contains elements. Thus, $\displaystyle D\bigcap B'_{r}(c) \not= \emptyset$ is true for $\displaystyle r>0$ since we just showed that D cointans elements on the interval $\displaystyle 0<s<m$ and $\displaystyle r>0$ contains all these points?????