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Prove that the closure is the following set.

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data

    Suppose (X,d) is a metric space. For a point in X and a non empty set S (as a subset of X), define d(p,S) = inf({(d(p,x):x belongs to S}). Prove that the closure of S is equal to the set {p belongs to S : d(p,S) =0}

    2. Relevant equations

    Closure of S = S U S' , where S' is the set of limit points of S.

    3. The attempt at a solution

    Any hint would be useful. My line of reasoning is the following.
    I started with a picture of all the metric space. I took cases like p being part of S and then p not being part of S. But, if p is not part of S, then the intersection of Neighborhood of p with some small radius , r with the set S could be null as d(p,x) > 0. Where as the least distance of any such point "p" and x is zero. Also, d(p,S) belongs to R. Hence, there shall be a real number between any nonzero d(p,S) and zero. As a result, all elements of inf{d(p,S)} = 0.
  2. jcsd
  3. Oct 2, 2007 #2

    matt grime

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    p is a limit point of S if and only if there is a sequence s_i in S tending to p. Just writing the definition pretty much solves the problem.
  4. Oct 2, 2007 #3
    Does not make any sense to me

    Edit: Yup, it makes sense now. For some reason, the class on sequence is chapter 3 in Rudin but the homework problem given to me is after Chapter 2. so, it was not not easy understanding the subsequence part. However, I solved it in a different way.
    Last edited: Oct 2, 2007
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