NetNews Usenet Archive 1992 #20/NN_1992_20.iso/spool/sci/math/11118

Internet Message Format | 1992-09-08 | 2.0 KB

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From: [e-mail] (Herman Rubin)
Newsgroups: sci.math
Subject: Re: Partitioning of uncountable sets
Message-ID: <[e-mail]>
Date: 8 Sep 92 14:58:25 GMT
Article-I.D.: mentor.Bu9LLE.Av8
References: <[e-mail]>
Sender: [e-mail] (USENET News)
Organization: Purdue University Statistics Department
Lines: 26

In article <[e-mail]> [e-mail] writes:
>A proof that every uncountable set can be partioned into two uncountable
>sets.
>Let X be an uncountable set. Consider the set of ordered pairs, (x,0),(x,1)
>where x is in X. Call this set Y. Then Y is also uncountable, moreover,
>{(x,0) with x in X}=Y(0) and {(x,1) with x in X}=Y(1) are both uncountable.
>But the cardinaltiy of X is the cardinality of Y. Thus, there is a bijection
>f from Y to X. f(Y(0)) and f(Y(1)) form the desired partition of X.
>The proof is of course easier if one assumes A.C, which I have avoided.

If finite sets are not included in countable, the theorem is trivially true.
But otherwise, you have not avoided all of AC, although what you have used
is less.  It is not necessarily the case that the union of two disjoint
uncountable sets of the same cardinality has that cardinality.  In fact,
the theorem itself is false, counterexample as follows.

There are Fraenkel-Mostowski models in which there is an infinite set A with
all subsets finite or co-finite.  This set is already a counterexample.  But
if one insists on transfinite, we adjoin such a set to the integers, and we
obtain a transfinite set such that every partition has a finite or countable
element.
--
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: [phone]
[e-mail] (Internet, bitnet)
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