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Connected Sets.

  1. Apr 19, 2008 #1
    Problem Statement:
    Let Y be a subset of X, and X and Y are connected, show that if A and B form a seperation of X-Y then YUA and YUB are connected.

    Attempt at solution:
    Well I'm not sure where the fact that X is connected comes to play (perhaps it gurantees us the possibility of X-Y not to be connected).
    anyway if we look at: seperations of YUB=CUD and YUA=C'UD'
    then Y=(YUB)-B=(C-B)U(D-B), Y=(YUA)-A=(C'-A)U(D'-A)
    which are both seperations of Y which is a contradiction to Y being connected, am I correct here or yet again wrong?

    any input?

    thanks in advance.
     
  2. jcsd
  3. Apr 20, 2008 #2

    HallsofIvy

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    Looks to me like "proof by contradiction": assume YUA is NOT connected and show that X is not connected.
     
  4. Apr 20, 2008 #3
    Yes I assumed that YUA isn't connected, but got that then Y isn't connected.
    cause if YUA=BUC
    then Y=(YUA)-A=(B-A)U(C-A) which is a seperation of Y.
     
  5. Apr 21, 2008 #4
    Suppose YUA is not connected, with separation CUD. Since Y is connected, then Y must be a subset of C or D, say Y is a subset of C. Now show that X is not connected.

    In the end, you will find that BUC and D are closed subspaces, and thus X = (BUC)UD is a separation of X.
     
    Last edited: Apr 21, 2008
  6. Apr 22, 2008 #5
    You mean Y should be D-A or C-A, let's say C-A, then because X-Y=AUB
    then X=XUX-Y=AU(BUC-A)
    which is a separation of X, something like this?
     
  7. Apr 22, 2008 #6

    HallsofIvy

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    then what's your problem? You assumed YUA isn't connected and arrived at a contradiction.
     
  8. Apr 22, 2008 #7
    Are you sure you separated Y correctly?

    You will need to use closures in your proof. Make use of the fact that if two sets form a separation, then each set does not meet the closure of the other, i.e. ClA is in X-B, ClB is in X-A, ClC is in X-D, ClD is in X-D.

    Suppose YUA is not connected, with separation CUD. Since Y is connected, then Y must be a subset of C or D, say Y is a subset of C. Show that BUC and D are closed subspaces by showing that they equal to their own closure. Then X = (BUC)UD is a separation of X, a contradiction.
     
    Last edited: Apr 22, 2008
  9. Apr 22, 2008 #8
    I just want to see if it's correct, so?
     
  10. Apr 22, 2008 #9
    Your solution is not correct because you didn't separate Y properly (read the precise definition of separation). The correct solution is more complicated than that. The connectedness of X is needed.
     
  11. Apr 24, 2008 #10
    So let me see if I get it A and B form a sparation of X-Y=AUB i.e they are open in X-Y and disjoint.
    now assume YUA=CUD where C and D are open in YUA and disjoint, now
    Y=(YUA)-A=(C-A)U(D-A), but C-A and D-A arent open in Y, then how do i reconcile it.
    I understnad that i need to show that BUC is closed in X and D is closed in X, but how?
     
  12. Apr 26, 2008 #11
    Ok X=(BUC)UD
    and BUC and D are disjoint this I know.
    how to show that cl(BUC) equals BUC?
    we know that
    cl(BUC)=cl(B)Ucl(C)
    and B is closed in X-Y thus B=(X-Y)^cl(B)=cl(B)
    and C is closed in YUA, then C is closed in A and thus also in X-Y, and again we get that B=cl(B)
    for D, we know the same that D is contained in A and thus D=cl(D)
    so both these sets are closed.

    is this correct?
     
  13. Apr 26, 2008 #12
  14. Apr 26, 2008 #13
    so the fact that cl(A) is a subset of X-B comes evidently from the fact that:
    A and B are disjoint, and (X-Y)-B equals A, then because clA contains A, it must be contained in X-B which contains (X-Y)-B.

    Ok I got it, sorry for my thickness, three exams in four days can make me anxious.

    btw, great site, wish I knew it a week ago.
     
    Last edited: Apr 26, 2008
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