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MoeBlee wrote:
> Rupert wrote:
> > No, David's proof is correct and doesn't use choice.
>
> Then I don't understand the proof, since I can't see how it goes
> through without choice. Would you kindly convey it your own manner? I
> am very much interested in seeing the theorem proven without choice,
> regularity, or replacement.
>
> Thanks,
>
> MoeBlee
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. But there is no surjection from S to P(S). Therefore there is no
surjection from T to P(S). Therefore P(S)\(P(S) intersect T) is
nonempty, choose y in this set. Now S X {y} is a set and is disjoint
from S.
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