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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. 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).
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