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Question about sets and its closure.

  1. Sep 22, 2011 #1
    I was wondering, is S a subset of S-bar its closure? For example, if p belongs to S, does p belong to S-bar too? Does it go the other way, S-bar is a subset of S?

    If it is true that S is a subset of S-bar does this automatically mean that S is closed?

  2. jcsd
  3. Sep 22, 2011 #2
    Well, what does closure mean? That is, how does your book define it (there are many ways to define it.) For example, an open set does not include all of its limit points (or cluster points) but a closed set does. The closure of a set can be defined as all of the cluster points of a set. If your book doesn't mention this, try to see if you can prove it from your book's definition. Now, what does this say about the question that you asked? Does this help?

    To answer the second question, you seem to at least implicitly know that S-bar is closed. But, why would this mean that a subset of S-bar is closed?
  4. Sep 22, 2011 #3
    This is what my text says about closure:
    Definition 1.76. Let (X,d) be a metric space and let S be a subset of X. Then the
    closure of S, denoted S-bar is the intersection of all closed sets containing S. Thus, S-bar is the smallest closed set containing S. Proposition 1.77. S is closed and if C is any closed set with S as a subset C, then S is a subset C.Theorem 1.78. Let p be an element of X, then p is an element of S-bar if and only if for every r > 0, B(p; r) intersection S is non-empty. Corollary 1.79. Let p be an element of X, then p is an element of S-bar if and only if there is a sequence {pn} a subset of S with lim pn =p.
  5. Sep 22, 2011 #4


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    Okay, it is the intersection of all closed sets containing S. So it is the intersection of some collection of sets, each containing S. Now, is S in that intersection?
  6. Sep 23, 2011 #5
    yes, i think it is. I was just wanting to make sure I was correct. I saw of wikipedia that S-bar is a closed super set of S and that S is closed if and only if S=S-bar. I just wanted to make sure that what I had concluded was correct. Thanks
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