home *** CD-ROM | disk | FTP | other *** search
- Path: sparky!uunet!ogicse!das-news.harvard.edu!cantaloupe.srv.cs.cmu.edu!GS6.SP.CS.CMU.EDU!jmount
- From: jmount+@CS.CMU.EDU (John Mount)
- Newsgroups: sci.math
- Subject: Re: Partitioning of uncountable sets
- Message-ID: <Bu9Ksv.G04.2@cs.cmu.edu>
- Date: 8 Sep 92 14:41:18 GMT
- Article-I.D.: cs.Bu9Ksv.G04.2
- References: <1992Sep8.182706.90039@vaxc.cc.monash.edu.au>
- Sender: news@cs.cmu.edu (Usenet News System)
- Organization: Carnegie Mellon University
- Lines: 24
- Nntp-Posting-Host: gs6.sp.cs.cmu.edu
-
- 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.
-
- This seems correct, and like a good way to prove the theorem, but
- are you sure this avoids the axiom of choice? How do you show
- card(X)=card(Y) without using AC? (In general cardinal comparability
- is equivalent to AC. And the proof I know for the absorption law of
- cardinal arithmetic uses AC pretty strongly.)
-
- --
- --- It is kind of strange being in CS theory, given computers really do exist.
- John Mount: jmount+@cs.cmu.edu (412)268-6247
- School of Computer Science, Carnegie Mellon University,
- 5000 Forbes Ave., Pittsburgh PA 15213-3891
-