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Power set proof

  1. Dec 3, 2007 #1
    This seems like a simple proof but i'm not familiar with power set proofs

    If A[tex]\subseteq[/tex]B then P(A) [tex]\subseteq[/tex] P(B)
     
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  3. Dec 4, 2007 #2

    Kurdt

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    A good place to start might be with the definitions of a subset and a power set. So the general set A is a subset of another general set B if every element of A is contained in B.

    The power set P(A) of a set A is defined as [itex] P(A) = \{X:X\subseteq A\} [/itex], that is the set of all the subsets of A.
     
  4. Dec 4, 2007 #3
    i'm more inclined to start with x [tex]\in[/tex] P(a), can i start the proof this way?
     
  5. Dec 4, 2007 #4

    CRGreathouse

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    That's probably a good way.
     
    Last edited: Dec 4, 2007
  6. Dec 5, 2007 #5

    HallsofIvy

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    Why do you say "more inclined"? That was exactly what was suggested.
     
  7. Dec 5, 2007 #6

    Office_Shredder

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    If [tex]x \in P(A)[/tex] what is x? In particular, what set are all of x's elements in?
     
    Last edited: Dec 5, 2007
  8. Dec 5, 2007 #7
    x is just an arbitrary element. And if A [tex]\subseteq[/tex] B then prove P(A) [tex]\subseteq[/tex] P(B). This need to be proved formally as well for my assignment!
     
  9. Dec 5, 2007 #8

    Office_Shredder

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    Sorry, the latex got screwed up. Re-read it now
     
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