Results 1 to 5 of 5

Math Help - Axiom of Choice equivalents

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    2

    Axiom of Choice equivalents

    So apparently for all pairwise disjoint sets S not containing the empty set

    \exists f [\exists Y (f:S \rightarrow Y) and \forall x  (x \in S \implies f(x) \in x)] \iff \exists t [\forall z z \in S \implies \exists! w (w \in z \cap t)]

    I had no problems proving the first version from the second, and assuming the first one and taking f(S) (image of the whole set S under f) for t, then obviously for all z in S f(z) is in z and f(S). The reason I'm posting is because I can't for the life of me prove that it's the only element of z intersection f(S)!
    Last edited by kzxrr; October 26th 2010 at 09:11 AM. Reason: LaTeX
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,965
    Thanks
    1785
    Awards
    1
    Here is a useful website.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Feb 2010
    Posts
    466
    Thanks
    4
    Suppose w in z/\range(f), and v in z/\range(f).

    Show w=v:

    w=f(x) for some x in S, and f(x) in x. But S is pairwise disjoint, so z=x, so w = f(z).

    v=f(p) for some p in S, and f(p) in p. But S is pairwise disjoint, so z=p, so v = f(z).

    So w=v.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2010
    Posts
    2
    I wasn't seeing that z and x were in S which was disjoint so their intersection must be empty or they must be equal. Ty so much it's so clear now.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Feb 2010
    Posts
    466
    Thanks
    4
    Yeah, in trying to prove stuff, if you don't see an intuitive path, it often helps just to start pulling out all the "information" or "implications" you can from the given premises. Extract everything you can from the premises, then lay it out in front of you to then see whether there might be a way to combine all that stuff back into a proof.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Axiom of choice.
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: January 11th 2010, 09:50 AM
  2. With or without Choice-axiom?
    Posted in the Discrete Math Forum
    Replies: 11
    Last Post: December 19th 2009, 08:08 PM
  3. axiom of choice
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 13th 2009, 03:01 PM
  4. Axiom of choice
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: February 18th 2009, 03:27 AM
  5. Axiom of Choice
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: January 18th 2006, 02:52 PM

/mathhelpforum @mathhelpforum