Let be given. If for all , then for all ,

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.

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- Oct 17th 2010, 08:16 PMzebra2147Proving limit points
Let be given. If for all , then for all ,

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, 04:51 AMMath 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 and show that in D such that .

Do you have a specific problem you're working on? - Oct 18th 2010, 05:49 AMzebra2147
Thanks for that explanation. I'm trying to prove the statement that I typed above. I think my problem is that implies that there is at least one point in D that is also in 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, 06:08 AMzebra2147
I may have just answered my own question...

Since then that means that when . Therefore, D contains elements. Thus, is true for since we just showed that D cointans elements on the interval and contains all these points?????