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Proof: if A is subset of B then closure of A is subset of closure of B

  1. Sep 29, 2011 #1
    I'm thinking proof by contradiction but I can't seem to get anywhere.
     
  2. jcsd
  3. Sep 29, 2011 #2

    pwsnafu

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    Think about limit points. x is in the closure of A, so it is a limit point of A...
     
  4. Sep 30, 2011 #3

    Fredrik

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    You should probably state the definition of "closure" that you would like to use, and maybe also your definition of "closed". I like the definition that says that the closure of E is the smallest closed set that contains E. If we use that definition, the proof is very simple.

    I see that you're new here. If you're wondering why we don't just tell you the complete solution, it's because the forum rules tell us not to do that when someone posts a textbook-style question. We are only allowed to give you hints, and tell you if you're doing something wrong.
     
  5. Sep 30, 2011 #4

    disregardthat

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    It might be easier to prove that if A is contained in a closed set (here the closure of B, which follows since B is contained in the closure of B and A is contained in B), then the closure of A is contained in the same closed set.
     
  6. Sep 30, 2011 #5

    PAllen

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    May I ask why this thread is here? It isn't number theory, and it looks like homework to me.
     
  7. Sep 30, 2011 #6

    HallsofIvy

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    I agree. I have moved the thread to "Homework and Coursework- Calculus and Beyond".

    I also agree with Fredrik that you need to state what definition of closure you are using. One common definition is that the closure of a set, A, is the smallest closed set that contains A. Using that definition this theorem is pretty close to trivial.
     
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