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Topic:
Given a set , is there a disjoint set with an arbitrary cardinality?
Replies:
28
Last Post:
Dec 4, 2012 5:50 PM




Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 4, 2012 1:29 AM


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?



