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Cl(A) smallest closed set containing A.

  1. Sep 21, 2010 #1
    My professor proved this result in class, but I don't understand the "simple" direction. He said that the above result is in another words proving that Cl(a) = intersection of all closed sets that contain A.

    So he proved Cl(A) subset of intersection of all of the closed sets containing A. And intersection of closed sets containing A is a subset of Cl(A). However, I don't understand this last direction.

    He simply says:

    Cl(A) = int(A) U Bd(A); by definition.
    => Cl(A) is one of the closed sets containing A.
    => intersection of closed sets containing A is a subset of Cl(A).

    That's what I just don't get. I understand that Cl(A) is closed and it contains A, but I just don't see how we can then say that the intersection of all closed sets containing A has Cl(A) in it. I know its going to be one of those I'll be intersecting all the other closed sets with, but that Cl(A) is actually INSIDE the intersection I just don't see.

    I'm having a massive brain freeze so an explanation would be appreciated.

  2. jcsd
  3. Sep 21, 2010 #2


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    If you intersect two sets, say A and B, then the intersection will be a subset of A. After all, it contains only elements in A (which satisfy a specific property, namely that they are also elements in B).
    By the same argument, it will also be a subset of B.

    Now suppose that you intersect Cl(A) with I, the intersection of all other closed sets containing A. Then Cl(A) intersected with I is a subset of both Cl(A) and of I. It is the former you are interested in here.
  4. Sep 21, 2010 #3
    Ahhh thanks a lot. I get it. Stupid me just wasn't seeing it. Thanks a lot!
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