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On 29 Aug 2006 15:48:16 -0700, MoeBlee <jazzmobe@xxxxxxxxxxx> said:
> 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.
Doh! Of course that's right. Obviously 0 is obviously not a function
on S. I was too focused on banging out the interesting case:
>> 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.
Cool.
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