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- From: holmes@opal.idbsu.edu (Randall Holmes)
- Subject: Re: Partitioning of uncountable sets
- Message-ID: <1992Sep8.164657.17267@guinness.idbsu.edu>
- Sender: usenet@guinness.idbsu.edu (Usenet News mail)
- Nntp-Posting-Host: opal
- Organization: Boise State University Math Dept.
- References: <1992Sep8.182706.90039@vaxc.cc.monash.edu.au>
- Date: Tue, 8 Sep 1992 16:46:57 GMT
- Lines: 23
-
- In article <1992Sep8.182706.90039@vaxc.cc.monash.edu.au> kevin@vaxc.cc.monash.edu.au 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.
- >
- >Love,
- >Kevin Davey.
- >
-
- You have not avoided using AC. k + k = k is not a theorem of ZF. I
- also doubt the truth of the assertion "every uncountable set can be
- partitioned into two uncountable sets" in the absence of AC, but I'm
- not certain of its status.
- --
- The opinions expressed | --Sincerely,
- above are not the "official" | M. Randall Holmes
- opinions of any person | Math. Dept., Boise State Univ.
- or institution. | holmes@opal.idbsu.edu
-