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Math Help - Limit points in a metric space

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    Limit points in a metric space

    Let (X,d) be a metric space and assme that A \subsetX. Prove that x \in Cl(A) iff there exists a sequence in A converging to x.
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    Quote Originally Posted by Andreamet View Post
    Let (X,d) be a metric space and assume that A \subsetX. Prove that x \in Cl(A) iff there exists a sequence in A converging to x.
    Here's a rushed proof. you can streamline it.

    Assume there is a sequence of points in A that converges to x. Then x lies in any closed set containing A. That is, x \in \text{Cl}(A).

    Conversely, assume x \in \text{Cl}(A). Then the "ball" of radius 1/k centered at x will contain points in A. For each k, pick one of these points and call it a_k, then the sequence \{ a_k \} converges to x


    I assume you know how to describe a "ball" in a metric space. we derive this from the metric
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    Quote Originally Posted by Jhevon View Post
    For each k, pick one of these points and call it a_k, then the sequence \{ a_k \} converges to x

    What theorem or axiom in mathematics will allow you to do this picking process and so be able to form the sequence in question??
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xalk View Post
    What theorem or axiom in mathematics will allow you to do this picking process and so be able to form the sequence in question??
    gee, i don't know of a theorem that says we can do this. it just makes sense that we can based on our definitions.

    what does it mean to be in \text{Cl}(A)? Well, it means that if x \in \text{Cl}(A) then any "open ball" centered at x will contain points in A. what does "open ball" mean in a metric space? the "open ball" centered at x is the set \{ y \in A \mid d(x,y) < r , \text{ for some given } r > 0 \}. so being in the closure means that this set is nonempty. if the set is nonempty, what's stopping us from picking elements in the set?
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    Quote Originally Posted by Jhevon View Post
    gee, i don't know of a theorem that says we can do this. it just makes sense that we can based on our definitions.

    what does it mean to be in \text{Cl}(A)? Well, it means that if x \in \text{Cl}(A) then any "open ball" centered at x will contain points in A. what does "open ball" mean in a metric space? the "open ball" centered at x is the set \{ y \in A \mid d(x,y) < r , \text{ for some given } r > 0 \}. so being in the closure means that this set is nonempty. if the set is nonempty, what's stopping us from picking elements in the set?
    Ask ZERMELO
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xalk View Post
    Ask ZERMELO
    i don't think we have to pull out the big guns for this problem. it doesn't seem like the type where you would be expected to go (in depth) into things like the Zermelo-Fraenkel axioms. our definitions have them built in, i think.
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