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On 28 Aug 2006 04:19:41 -0700, "Rupert" <rupertmccallum@xxxxxxxxx>
wrote:
>
>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.
No doubt - I decided not to worry about this, because bringing
P(S) into the argument in the first place was a stupid irrelevancy.
All we need is that for every T there exists y such that y is not
in T, and proving that doesn't require anything about cardinality
or power sets.
>> 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
************************
David C. Ullrich
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