|
|
Rupert wrote:
> 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.
Unless S is nonempty and T is empty, but in that case the result that
there is no surjection from T to P(S) will still hold.
> 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.
|
|