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Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 9:41 AM


On Monday, December 3, 2012 6:21:05 AM UTC8, jaakov wrote: > Dear all: > > > > Given a set X and a cardinal k, is there a set Y such that card(Y)=k and > > X is disjoint from Y? > > > > Is there a proof of this fact that works without the axiom of regularity > > (= axiom of foundation) and does not assume purity of sets? > > > > Thanks in advance > > > > Jaakov. > > > >  news://freenews.netfront.net/  complaints: news@netfront.net 
Is this homework?
Consider these two sets: {1,2,3}, {cat, dog, cow} Then consider the set of even numbers vs the set of odd numbers. I had to look up "purity of sets". Consider the set of real numbers in the range [0,1) and the set in the range [1,2).
You might get into a bit of trouble if you tried to find two disjoint sets with equal cardinality both in the range [0,1]. You'd have to handle sequenses ending in zeros and those ending in all nines. Otherwise you could choose a digit and select based upon it being odd or even.



