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Math Help - Proving limit points

  1. #1
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    Proving limit points

    Let m>0 be given. If for all 0<s<m, D\bigcap B'_{s}(c) \not= \emptyset, 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.
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  2. #2
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    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?
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  3. #3
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    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?
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  4. #4
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    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<s<m. 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<s<m and r>0 contains all these points?????
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