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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.
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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