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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
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