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Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 4, 2012 5:49 AM
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To correct the previoius proof and offer a simpler proof.
On Tue, 4 Dec 2012, William Elliot wrote:
> On Mon, 3 Dec 2012, jaakov wrote:
>
> > Given a set X and a cardinal k, is there a set Y such that card(Y)=k and X is
> > disjoint from Y?
>
> Case |X| < k. Let Y be a set with |Y| = k.
> |Y\X| = k; Y\X and X are disjoint.
> Case |X| <= k. Let A be a set with k < |A|. k < |A| = |A\X|.
> Take Y as any subset of A\X with |Y| = k.
Case k <= |X|; Let A be an infinite set with |X| < |A|. k < |A| = |A\X|.
Take Y as any subset of A\X with |Y| = k.
Better yet is to mere the two cases.
Let A be an infinite set with max{ |X|, k } < |A|.
Since k < |A| = |A\X|, take Y to be any subset of
A\X with |Y| = k.
> > Is there a proof of this fact that works without the axiom of regularity (=
> > axiom of foundation) and does not assume purity of sets?
>
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