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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?
> But there is no surjection from S to P(S). Therefore there is no
> surjection from T to P(S).
I understand (given that there is a surjection from S onto T).
> Therefore P(S)\(P(S) intersect T) is
> nonempty,
I think PS\(PS /\ T) = PS\T, right?
I don't see how to show PS\T not= 0 without again using the axiom of
choice. I figure it this way:
If PS\T = 0, then PS subsetof T, but then PS injects into T, but then
the axiom of choice provides that there is a surjection from T onto PS,
contradicting that there is no surjection from T onto PS. But I don't
see how to show PS\T not= 0 without the axiom of choice.
> choose y in this set. Now S X {y} is a set and is disjoint
> from S.
I understand.
I am slow at this kind of thing, but my question about how to derive
those two lines without the axiom of choice is sincere.
Thanks,
MoeBlee
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