1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Naive set theory problem

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that(power set) P(E) U P(F) is a subset of P(E U F)


    2. Relevant equations

    P(E) U P(F) is a subset of P(E U F)

    3. The attempt at a solution
    P(E)U P(F)={x:xεP(E) or xεP(F)}
    but P(E)={X:X is a subset of E} or P(E)={x:xεX→xεE}
    so we get P(E)U P(F)={x:xεX→xεE or xεX→xεF}
    but logical implication p→q⇔non(p) or q
    so we get P(E) U P(F)={x:non(xεX) or xεE or non(xεX) or xεF}
    therefore by the idempotence property and comutativity of the logical operators
    P(E) U P(F)={x:non(xεX) or xεE or xεF}
    and we get P(E) U P(F)={x:non(xεX) or xεE U F}
    which is exactly P(E U F).My question is why is it that only the inclusion stands, and why not equality also.Thank you.
     
  2. jcsd
  3. Nov 26, 2013 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You've gone wrong here. Power sets are collections of sets, so you have to be able to work with sets as elements to do this correctly.

    To show equality does not hold, you could find an example where it doesn't hold.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Naive set theory problem
  1. Set Theory Problem (Replies: 11)

  2. 2 set theory problems (Replies: 3)

Loading...