Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 4, 2012 5:49 AM


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? >



