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David C. Ullrich wrote:
> On 27 Aug 2006 15:59:21 -0700, "MoeBlee" <jazzmobe@xxxxxxxxxxx> wrote:
>
> >David C. Ullrich wrote:
> >> Say S is a set. Some elements of S may be ordered
> >> pairs; let T be the set of y such that there exists
> >> x such that (x,y) is in S. Now card(T) < card(P(S));
> >> choose y in P(S) not equal to any element of T,
> >> and consider S x {y}.
> >
> >Then T = range(S), which is okay (S has a range even if S is not a
> >relation).
> >
> >But how do justify card(T) < card(P(S))?
> >
> >I can justify it this way, but it uses the axiom of choice:
> >
> >Az card(range(z)) =< card(z) < card(Pz).
> >
> >I used the axiom of choice to derive Az card(range(z)) =< card(z).
> >
> >So how did you get card(T) < card(P(S)) without the axiom of choice?
>
> Come now.
>
> Without AC I'm not sure what card even means. So forget that.
> But come on now - all we need here is that there exists y such
> that y is not in T; the fact that card(T) < card(P(S)) was just
> supposed to be a convenient way to see there is such a thing,
> but in fact P(S) is irrelevant.
>
If we have regularity then we can define the cardinal of a set to be
the set of all sets equipollent to that set of least possible rank. But
in any event we can just interpret card(T) < card(P(S)) to mean: there
is an injection from T to P(S), and there is no bijection from T to
P(S). Your argument is fine.
> Theorem: For every T there exists y such that y is not an element of
> T.
>
> Proof: Surely you've seen the proof of this? QED.
>
> >MoeBlee
>
>
> ************************
>
> David C. Ullrich
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