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Math Help - Topology proof

  1. #1
    Senior Member Danneedshelp's Avatar
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    Topology proof

    Q: Let A be bounded above so that s=supA exists. Show that s\in{A^{'}}, where A^{'} denotes the closure of A.

    Below is my try at it. I am not sure about my use of the monotone convergence theorem or the rest of the proof for that matter. We just started basic topology, so I am still getting use to the definitions. Embedded in the proof is a lemma about open intervals and limit points. Not sure if it is needed, because it is not even a very complete proof.

    A: Suppose A is bounded above and s=supA exists. Since A is bounded and monotone, there exists a sequence (a_{n}) in A such that (a_{n})\rightarrow\\s, by the monotone convergence theorem. To conlude s is a limit point, we need to show that, for any \epsilon-neighborhood V_{\epsilon}(s), V_{\epsilon}(s) \cap A\neq \{s\}. However, s may or may not be contained in A, so we must consider both cases. First, if s\notin{A}, then given \epsilon>0 it follows that (s-\epsilon\\,s+\epsilon) contains infinitly many points of A\ \{s\}; hence, s is a limit point of A, when A is open. A similar arguement shows that, s is a limit point when A is closed. Moreover, when A is closed it must contain its limit points. Thus, s\in{A\cup\\L}, where L denotes the set of all limit points of A (not necessarily in A). Therefore, by definition s\in{A^{'}}.
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    Quote Originally Posted by Danneedshelp View Post
    Q: Let A be bounded above so that s=supA exists. Show that s\in{A^{'}}, where A^{'} denotes the closure of A.
    To say that s \in  A', the closure, means that s \in A it is limit point.
    If  s\in A we are done so suppose that  s\not\in  A, we must show that it is limit point.
    By definition of  s=\sup(A) if c>0 then  s-c is not an upper bound.
    This \left( {\forall c > 0} \right)\left( {\exists x_c  \in A} \right)\left( {s - c < x_c  < s} \right)
    You go to the text material to find out if that is enough to show that it a limit point.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Plato View Post
    To say that s \in  A', the closure, means that s \in A it is limit point.
    If  s\in A we are done so suppose that  s\not\in  A, we must show that it is limit point.
    By definition of  s=\sup(A) if c>0 then  s-c is not an upper bound.
    This \left( {\forall c > 0} \right)\left( {\exists x_c  \in A} \right)\left( {s - c < x_c  < s} \right)
    You go to the text material to find out if that is enough to show that it a limit point.
    So, since the set (s-c,s)\cap{A} is infinite, we can conclude s\in{A^{'}} for both cases.
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