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Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 4, 2012 3:20 AM


On Monday, December 3, 2012 10:29:36 PM UTC8, William Elliot wrote: > On Mon, 3 Dec 2012, forbisgaryg@gmail.com wrote: > > > On Monday, December 3, 2012 6:21:05 AM UTC8, 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? > > > > > > > > Is there a proof of this fact that works without the axiom of regularity > > > > > > > > (= axiom of foundation) and does not assume purity of sets? > > > > > > Based upon some of your replies I have a question. > > > > > > Given your statment: "Given a set X and a cardinal k" > > > > > > 1. Are you referring to any set X or specifically a set of real numbers? > > > > Any set or course. > > > > > 2. Are you saying saying k isn't necessarily the cardinality of X? > > > > That's right. > > All that was said about the given cardinal k was card Y = k. > > > > > Given your question "> Is there a proof of this fact that works without > > > the axiom of regularity (= axiom of foundation) and does not assume > > > purity of sets? > > > > > > 1. Are you referring to any set or specifical sets of real numbers? > > > > Any set of course. No where was it stated or even suggested > > that X or Y, is a subset of the reals. Where do you get this > > notion that the discussion is about set of real numbers?
So, what is the cardinality of the universal set?
As was noted no set can be a member of itself.
(Ex ~xeU) <=> ~(Ax xeU)
Can a subset of the universal set have the same cardinality as the universal set? Since neither X nor Y are identified, suppose X is the universal set. Can there be a disjoint set Y let alone one with cardinality k?
Maybe I saw the asnwer in other replies but didn't notice it. The syntax being used is a bit strange but seems to be often used by many.



