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Powerset proof

  1. Nov 6, 2003 #1
    P for powerset, n for intersection

    show that P(AnB)=P(A) n P(B)

    Studying for a midterm, seen this question in our textbook and on an old midterm. No idea how to do it. Anyone know?
     
    Last edited: Nov 6, 2003
  2. jcsd
  3. Nov 7, 2003 #2

    NateTG

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    It's pretty basic, just apply definitions, and you should be ok.
    Just prove that if x is in P(A ∩ B) then x is also in P(A) ∩ P(B), and
    that if x is in P(A) ∩ P(B) then x is in P(A ∩ B).

    It should just be applying definitions.
     
  4. Nov 7, 2003 #3
    I know the idea behind doing a proof heh, I think my textbook is somewhat lacking though. What definitions shoudl I be attempting to make use of. Only info ive found on powersets in textbook is what the powerset actually is (the set containing all the subsets). Cant think of any helpful way to apply that to a general case though.
     
  5. Nov 7, 2003 #4

    NateTG

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    Ok:

    x is in P(A ∩ B) is equivalent to saying that
    x is a subset of A ∩ B
    so each element χ of x is in A ∩ B
    so each element χ of x is in A and in B
    so x is a subset of A and x is a subset of B
    so x is in P(A) and x is in P(B)

    you should have no problem filling in the holes, and going in the other direction from there.
     
  6. Nov 7, 2003 #5

    jcsd

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    Shouldn't that be x is an element of....
     
  7. Nov 8, 2003 #6

    HallsofIvy

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    No, x is a subset of A∩B is correct.

    x is an element of P(A∩B) which is the collection of all subsets of A∩B.
     
  8. Nov 8, 2003 #7

    jcsd

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    I see, set theory is quite new to me, it was taken off our curriclum at school and isn't much used in undergraduate physics (well not in the first year anyway).
     
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