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Math Help - Interior and Closure of set proof

  1. #1
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    Interior and Closure of set proof

    Hi There,

    I'm currently studying for a deferred final, and understand the general concept of this question, but was hoping I could get a clear answer from someone knowledgable in order to reinforce my knowledge.

    The Question: Let S be a set in R^n. Explain what is meant by the interior S^o of S. If the closure of S^o is T = [S^o with a bar above it, I'm horrible with LaTeX], and S is a closed set, prove that T is a subset of S. Give an example of a closed set in S where the inclusion is strict

    My attempted solution: Well, I don't totally have a solution. Here's what I know though. The interior of a set is all points in a set that are not on the boundary. That is, all points in S that for any delta > 0 there exists a neighborhood N such that N is an open set in S. I feel my definition is weak, and I don't totally understand what a closure is and thus cannot go much further with this

    If someone could help me with this question I'd greatly appreciate it.
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  2. #2
    Senior Member Tinyboss's Avatar
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    The closure of S is the smallest closed set containing S, equivalently, the intersection of all closed sets containing S. Another equivalent definition is the union of S and the limit points of S. So, since a closed set contains all its limit points, you have the first result.

    As for the strict inclusion, there are closed sets that have no interior. What happens with such a set?
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  3. #3
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    Thanks tinyboss! A closed set which has no interior...wouldn't that set consist entirely of boundary points? I apologize for my absent mindedness but I don't totally see the relevance ><
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  4. #4
    Senior Member Tinyboss's Avatar
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    Doesn't the second part of the question ask you to find a closed set that properly contains the closure of its interior?
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  5. #5
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    Hi Tinyboss, thanks for your patience and help. I'm still having difficulty with this question. Here is my answer so far. A nudge in the right direction from this would be sufficient enough, as I realize I've taken up more time of yours than I should have. I am struggling a lot with the concept of closure, open and closed sets. Anyways, here it is:

    The interior of a set S is the set of all points P such that for some (delta) > 0 there exists an open neighborhood N of radius (delta) centered at P. Suppose S is a closed set, and denote it's interior by I, and it's interior's closure by CL(I). By definition, CL(I) is the smallest closed set containing I, and I is an open subset of S. Since the question defines CL(I) as the complement of I, and CL(I) is a closed set, then CL(I) must subset S. The inclusion is is strict when S is a closed set whose interior is empty.

    I don't think my proof is rigorous enough nor do I really understand it, it's mostly from what you've said and my textbook. Intuitively, wouldn't the union of CL(I) and I be equal to S? And in which case, wouldn't it's inclusion be strict iff I is non-empty?
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  6. #6
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    a good set to test your intuition of this is the following set in R^2:

    S = {(x,y) in R^2: x^2 + y^2 = 1}.

    compare this with the following set:

    T = {(x,y) in R^2: x^2 + y^2 ≤ 1}

    and the set

    U = {(x,y) in R^2: x^2 + y^2 < 1}.

    two of these sets are closed (in the usual metric topology for R^2). one of them has the property that cl(int(A)) ⊂ A. which one?
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  7. #7
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    Deveno, thanks for your help and useful example, it really helped me. Tinyboss, thanks for your help as well.

    Managed to solve this to my own satisfaction.
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