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On 25 Aug 2006 17:04:52 -0700, "MoeBlee" <jazzmobe@xxxxxxxxxxx> wrote:
>Rupert wrote:
>> MoeBlee wrote:
>> > For ZC (Z set theory with the axiom of choice), I'm pretty sure I see
>> > how to prove:
>> >
>> > AxEy(x equinumerous with y & x/\y = 0)
>> >
>> > where '/\' stands for binary intersection.
>> >
>> > But can this be proven without the axiom of choice? If so, how to do
>> > it?
>> >
>>
>> Yes, consider x X {x}.
>
>Of course I thought of that. But that doesn't the disjointedness
>require the axiom of regularity? I should have been clear to exclude
>use of regularity also. So the axioms are
>
>extensionality
>separation
>pair
>union
>power
>infinity (though, I'd like to exclude it also)
We're willing to use power sets, great.
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}.
>Thanks,
>
>MoeBlee
************************
David C. Ullrich
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