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jaakov
Posts:
11
Registered:
12/3/12
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Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 5:28 PM
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On 03.12.2012 17:51, Aatu Koskensilta wrote:
> jaakov<removeit_jaakov@deleteit_ro.ru> writes:
>
>> 1. k is not related to the cardinality of X.
>
> Yes, I misread your original question.
>
>> 2. Your lambda need not exist. To show that lambda exists, one has to
>> show that a set may not contain arbitrarily large ordinals. Is it a
>> known fact or simply false?
>
> For any set A, lambda = U{ alpha in A | alpha is an ordinal} is an
> upper bound on ordinals in A. Take the next k ordinals after lambda and
> you have your desired Y.
>
Thank you, that works. Let me just restate the proof.
Let lambda = sup { alpha in X | alpha is an ordinal}.
External theorem: lambda exists and is an ordinal.
Hope: its proof does not assume the axiom of regularity or set purity.
Let l be any ordinal greater than lambda.
Let Y = { l + a | a is an ordinal and a < k}.
We show:
1. card(Y) = k.
Consider f: k -> Y, a |-> l+a.
By definition of Y, f is onto. Since a1<a2 implies l+a1<l+a2,
f is one-to-one. Thus f is a bijection, so card(Y)=k.
2. Y is disjoint from X.
Assume some x in X and x in Y. By definition of Y there is
a<k such that x = l+a. Since x in X, l+a is an ordinal in X.
Thus l+a <= lambda. But lambda<l<=l+a. By transitivity lambda<lambda.
Contradiction!
Jaakov.
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