Results 1 to 3 of 3

Math Help - Cartesian Product

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    25

    Axiom of Foundation and Cartesian Product

    Can anyone help with this question:

    Use the Axiom of Foundation to show that if A is a non-empty set, then A is not equal to A A.
    [Hint: consider the set a = A ∪ (∪A)]?

    Many thanks.
    Last edited by KSM08; February 5th 2010 at 02:52 AM. Reason: Unhelpful title
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Dec 2009
    Posts
    36
    See here: Axiomatic set theory - Google Books (Suppes, Axiomatic Set Theory, Theorem 107). There you get that A=0, which contradicts it being non-empty.

    Hope this helps!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Feb 2010
    Posts
    466
    Thanks
    4
    I'll give you a proof but I'll ask "WHY?" for certain details you should be able to supply as an easy exercise. If you can't answer the "WHY?"s, then you need to review your introductory set theory and/or logic material.

    'x/\y' will stand for binary intersection.

    Proof by contradiction: Suppose A not empty and A = AxA.

    A not empty, so any m in A is in AxA (WHY?).
    So, if m in A, then there is a j and k such that j in A and k in A and m = {{j} {j k}} (WHY?).
    A not empty, perforce AuUA not empty (WHY?).
    No m in AuUA is empty (WHY?).
    Since AuUA not empty, by the axiom of regularity, there is an m such that m in AuUA and m/\(AuUA) empty. And m not empty, since, as we observed, no member of AuUA is empty.
    Suppose m in A.
    So m subset of UA (WHY?).
    So, since m not empty, we have m/\UA not empty (WHY?), which contradicts m/\(AuUA) empty (WHY?).
    So m not in A.
    So m in UA (WHY?).
    So there is a j and k such that j in A and k in A and m = {j} or m = {j k} (WHY?).
    So m/\A not empty (WHY?), which contradicts m/\(AuUA) is empty (WHY?).
    So we are done (WHY?).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cartesian product of A*A*A*A
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 14th 2011, 07:29 AM
  2. Cartesian product.
    Posted in the Discrete Math Forum
    Replies: 9
    Last Post: November 4th 2010, 02:59 PM
  3. Cartesian product
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: February 26th 2010, 12:30 PM
  4. Cartesian Product
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: November 15th 2009, 09:23 AM
  5. Cartesian product
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: August 27th 2009, 09:51 AM

Search Tags


/mathhelpforum @mathhelpforum