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Thread: Topology Problem

  1. #1
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    Topology Problem

    i think i'm just having trouble visualizing this problem...

    they define x in S as an isolated point of S if there exists r>0 such that B(x;r)∩S={x}. the problem asks to show that the closure of subset S of X is the disjoint union of limit pts of S and isolated pts of S.

    how would i go about solving this problem?
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  2. #2
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    Quote Originally Posted by squarerootof2 View Post
    i think i'm just having trouble visualizing this problem...

    they define x in S as an isolated point of S if there exists r>0 such that B(x;r)∩S={x}. the problem asks to show that the closure of subset S of X is the disjoint union of limit pts of S and isolated pts of S.

    how would i go about solving this problem?
    Can you explain what you mean by "disjoint union"?

    Maybe the following will help: The closure of S is the union of S and its limit points.
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  3. #3
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    i think disjoint union just means a collection. i just looked at the solutions in the back and it seems to suggest a proof by contradiction, i.e. assume that x is not a limit point of S and to mess around with an epsilon ball containing only finitely many pts of S. i was just wondering what the best way to approach this problem was. thanks.
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  4. #4
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    Quote Originally Posted by squarerootof2 View Post
    i think i'm just having trouble visualizing this problem...

    they define x in S as an isolated point of S if there exists r>0 such that B(x;r)∩S={x}. the problem asks to show that the closure of subset S of X is the disjoint union of limit pts of S and isolated pts of S.
    It should be clear from the definition that no isolated point is a limit point of S.
    It follows that every non-isolated point is a limit point of S.
    Now let $\displaystyle S'$ denote the set of all limits points of S.
    Then $\displaystyle S\backslash S'$ would be the set of all isolated points of S.
    Observing that the closure of S consists of all points of S union all limits points of S, the proof follows.
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