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Closure of A is the smallest closed set containing A

  1. Jan 4, 2008 #1
    Ok, the proof looks simple since by defintion Cl A = intersection of all closed sets containing A. And textbooks give a quick proof that we all understand, but I have a question: Don't we first have to prove that a smallest closed set containing A exists in the first place?

    I'm trying to prove the existence using Zorn's Lemma...
     
  2. jcsd
  3. Jan 4, 2008 #2
    Here is the statement of my question:

    Given a subset A in a topological space X, use Zorn's lemma to prove that there exists a smallest closed set containing A.

    Let S be the collection of all closed subsets of X containing A. Partially order S by set containment, i.e. U < V iff U contains V. Let C = {B_i} be a totally ordered subcollection of S. Let B = intersection of all the B_i. Then B is an upper bound of C (since B is a subset of every B_i), and B is in S since B is a closed set containing A. Thus every totally ordered subcollection of S contains an upper bound in S so by Zorn's Lemma, S contains a maximal element, i.e. there exists a smallest closed set containing A.

    Does that look right?
     
    Last edited: Jan 4, 2008
  4. Jan 4, 2008 #3

    morphism

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    That's unnecessary. The smallest closed set containing A is the intersection of all closed sets containing A. Since A sits in a topological space X, the collection of closed sets containing it is nonempty - it contains X - so the intersection of all the members of this collection makes sense.
     
    Last edited: Jan 4, 2008
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