On Tue, 4 Dec 2012, email@example.com wrote: > On Monday, December 3, 2012 10:29:36 PM UTC-8, William Elliot wrote: > > On Mon, 3 Dec 2012, firstname.lastname@example.org wrote: > > > On Monday, December 3, 2012 6:21:05 AM UTC-8, 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 statement: "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.
> So, what is the cardinality of the universal set?
In ZPG and NBG set theory, there no universal set.
> As was noted no set can be a member of itself.
Regularity implies that and the problem asks to not use regularity.
> (Ex ~xeU) <=> ~(Ax xeU)
What does xeU mean? Thatyoulikereadingthis?
> Can a subset of the universal set have the same > cardinality as the universal set?
As noted above, there is no universal set unless you're using NF set theory, which isn't usually used.
> Since neither X nor Y are identified, suppose X is the universal set.
Your supposition is totally unwarranted.
> Can there be a disjoint set Y let alone one with cardinality > k?
What's a disjoint set? That's nonsense. Can there be a set Y with cardinality k that's disjoint to X? That's what the problem is - you're going in circles.
> Maybe I saw the answer in other replies but didn't notice it.
> The syntax being used is a bit strange but seems to be often > used by many.
Your leaps to unwarranted assumptions is more that strange