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BxB, sets question, need to verify if this is right.

  1. Oct 7, 2006 #1
    Hello everyone, i was wondering if this is correct or not.
    I circled the 2 I was iffy about. Thanks!
    Note: P stands for power sets, and X stands for Cartesian Product

    [​IMG]
     
  2. jcsd
  3. Oct 7, 2006 #2

    Hurkyl

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    The two you circled are, in fact, wrong.
     
  4. Oct 7, 2006 #3

    HallsofIvy

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    Remember that AxA, AxB, BxB, etc. are ordered sets.
     
  5. Oct 7, 2006 #4

    Hurkyl

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    :confused:
     
  6. Oct 8, 2006 #5
    I went over the chapter again and I still can't see what I did wrong, especially in the above one where i'm finding the power set. Can you give me some feedback on why its wrong or how I can correct it?

    For the BxB would it just be B? {a,b}?
     
  7. Oct 8, 2006 #6

    0rthodontist

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    You want to take the power set of
    [tex]\{\phi, \{\phi\}\}[/tex]
    It may be a little less confusing if you instead try to find the power set of {x, y} where x = [tex]\phi[/tex] and y = [tex]\{\phi\}[/tex], then substitute back.
    in the second one you want to find {a, b} x {a, b}. Can you find {a, b} x {c, d} and then let c = a, d = b?
    In these one thing to check is the size of the set in your answer. If A and B are finite, |P(A)| = 2|A| and |A x B| = |A| x |B|
     
    Last edited: Oct 8, 2006
  8. Oct 8, 2006 #7

    HallsofIvy

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    What a remarkably stupid thing for me to say! It must have been past my bedtime. AxA, AxB, BxB, etc. are sets of ordered pairs. In other words, if a, b are both in B, BxB will contain both (a,b) and (b,a).
     
  9. Oct 8, 2006 #8
    Thanks for the responces...
    In other words, if a, b are both in B, BxB will contain both (a,b) and (b,a).
    So it would just be BxB = { (a,b), (b,a) } ?
    Why wouldn't it include any of the others? like
    BxB = {(a,a),(a,b),(b,a),(b,b) }
    If BXB is an ordered pair that means AXB is, and when you do AXB in this example you do alot more grouping like for instance:

    A = {x,y,z,w}
    B = {a,b}
    AxB = { (x,a), (y,a),(z,a),(w,a),(x,b),(y,b),(z,b),(w,b) }
    You group up each A element with a B.
    So why wouldn't u also do that with BxB? group each B element with itself?
    (a,a),(b,b),(a,b),(b,a)

    Ortho with ur hint,
    to let c = a, d = b. i get the following:
    {a,b} x {c, d} = { (a, c), (a, d), (b, c), (b, d) }
    now sub back in
    BxB = { (a, a), (a, b), (b, a), (b, b) }
     
    Last edited: Oct 8, 2006
  10. Oct 8, 2006 #9
    Also for the power sets i took ur advice ortho and got the following:

    P({null,{null}})
    let x = null
    let y = {null}
    P({x,y}) = {null,{x},{y},{x,y}}
    now sub in for x and y and i got:
    P({x,y}) = {null,{null}, { {null} }, { null,{null} } }
     
  11. Oct 8, 2006 #10

    0rthodontist

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    Those are both right.
     
  12. Oct 8, 2006 #11
    o yeah! thanks for the help again!
    :biggrin:
     
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