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Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 4, 2012 3:20 AM
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On Monday, December 3, 2012 10:29:36 PM UTC-8, William Elliot wrote:
> On Mon, 3 Dec 2012, forbisgaryg@gmail.com wrote:
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> > On Monday, December 3, 2012 6:21:05 AM UTC-8, jaakov wrote:
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> > > Given a set X and a cardinal k, is there a set Y such that card(Y)=k and
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> > >
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> > > X is disjoint from Y?
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> > >
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> > > Is there a proof of this fact that works without the axiom of regularity
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> > >
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> > > (= axiom of foundation) and does not assume purity of sets?
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> >
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> > Based upon some of your replies I have a question.
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> > Given your statment: "Given a set X and a cardinal k"
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> > 1. Are you referring to any set X or specifically a set of real numbers?
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> Any set or course.
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> > 2. Are you saying saying k isn't necessarily the cardinality of X?
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> That's right.
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> All that was said about the given cardinal k was card Y = k.
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> > Given your question "> Is there a proof of this fact that works without
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> > the axiom of regularity (= axiom of foundation) and does not assume
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> > purity of sets?
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> >
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> > 1. Are you referring to any set or specifical sets of real numbers?
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> Any set of course. No where was it stated or even suggested
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> that X or Y, is a subset of the reals. Where do you get this
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> 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.
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