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Help with a topology proof

  1. Jan 22, 2008 #1
    Hello

    I have a proof that I need to try to work out but I'm not really getting too far and need help if you could at all. The question is

    Let A and B be two subsets of a metric space X. Prove that:
    Int(A)[tex]\bigcup[/tex]Int(B)[tex]\subseteq[/tex]Int(A[tex]\bigcup[/tex]B) and Int(A)[tex]\bigcap[/tex]Int(B) = Int(A[tex]\bigcap[/tex]B)
    I also have to give an example of two subsets A and B such that Int(A) [tex]\bigcup[/tex] Int (B) ≠ Int (A[tex]\bigcup[/tex] B)

    Any help at all would be so great
    Cheers
     
  2. jcsd
  3. Jan 22, 2008 #2
    Let x be in intA U int B. Then, WLOG, assume x is in int A. Then there exist a basis element U containing x and contained in A and therefore contained in A U B. The second question requires doing this in both directions.
     
  4. Jan 22, 2008 #3
    Topology

    Hi andytoh

    Thank you for taking the time to reply to my question. Just one thing, I'm new to this so could you tell me what WLOG means?

    Thanks again



     
  5. Jan 22, 2008 #4
    "Without loss of generality", since the roles of A and B are symmetric.

    Incidentally, #1 can be generalized to any collection (finite or infinite) of subsets, and #2 can be generalized to any finite collection of subsets.
     
    Last edited: Jan 22, 2008
  6. Jan 22, 2008 #5
    Counterexample of Int(A) U Int (B) ≠ Int (A U B):
    A=[-1,0], B=[0,1]

    Counterexample of #2 with an infinite collection of subsets:
    Use {(-1,1), (-1/2,1/2), (-1/3,1/3), (-1/4,1/4), ... }
    Left side = {0}, but right side = empty.
     
    Last edited: Jan 22, 2008
  7. Jan 23, 2008 #6
    topology

    Thank you so much, once again, for your replies. I really appreciate your help

    Cheers
     
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