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MoeBlee wrote:
> Rupert wrote:
> > I'd be interested to see your proof that uses the axiom of choice but
> > doesn't use regularity.
>
> Tell me whether you think this is correct (maybe I overlooked
> something):
>
> Show ~x=0 -> Ey(card(x) = card(y) & x and y are disjoint).
>
> If x is finite, then by an easy inductive proof on card(x), we're done.
>
> Suppose x is infinite.
>
> If we have card(x) =< card(Px\x), then we we're done.
>
> Lemma: card(Px) =< card(Px\x) + card(x).
>
> So, if card(Px\x) < card(x), then, by the rule that the sum of two
> infinite cardinals is the greater of the cardinals, card(Px) =<
> card(x), which is impossible.
>
> Axiom of choice is used in assuming trichotomy and in the rule for
> cardinal addition that was used.
>
> MoeBlee
That looks good to me. David's proof looks good to me as well.
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