Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
sci.math.*
»
sci.math
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
Given a set , is there a disjoint set with an arbitrary cardinality?
Replies:
28
Last Post:
Dec 4, 2012 5:50 PM



jaakov
Posts:
11
Registered:
12/3/12


Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 5:28 PM


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 onetoone. 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.



