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Chris Menzel wrote:
> On 29 Aug 2006 10:37:51 -0700, MoeBlee <jazzmobe@xxxxxxxxxxx> said:
> > Rupert wrote:
> >> Let S be a set. Let T=ran(S), that is, T is the set of y such that
> >> there exists an x such that <x,y> is in S. There is a surjection from S
> >> to T.
> >
> > How do we show there is a surjection from S to T without the axiom of
> > choice?
>
> Either S contains at least one ordered pair or it doesn't. If it
> doesn't, then T is empty and 0 is a surjection from S to T.
Hmm, by my definition, 0 is a surjection from 0 onto 0, but not from a
non-empty S onto 0. But it doesn't matter here, since if T is 0 then
the next step in Rupert's proof works anyway.
> If it does,
> then there is an ordered pair z in S and hence also the surjective
> function f_z : S -> T such that
>
> 2nd(w) if w is an ordered pair
> f_z(w) =
> 2nd(z), otherwise
>
> (where, of course, for ordered pairs z = <x,y>, 2nd(z) = y).
Got it.
And I just figured out to show PS\T not= 0.
So that's two proofs that work to answer my original question.
Thanks,
MoeBlee
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