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- Path: sparky!uunet!munnari.oz.au!bruce.cs.monash.edu.au!monu6!vaxc.cc.monash.edu.au!kevin
- From: kevin@vaxc.cc.monash.edu.au
- Newsgroups: sci.math
- Subject: Partitioning of uncountable sets
- Message-ID: <1992Sep8.182706.90039@vaxc.cc.monash.edu.au>
- Date: 8 Sep 92 18:27:06 +1000
- Organization: Computer Centre, Monash University, Australia
- Lines: 12
-
- 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.
-
-