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- From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
- Newsgroups: sci.math
- Subject: Re: Is Card(R)=Card(R^2)? (yes, and a question)
- Message-ID: <a_rubin.713717421@dn66>
- Date: 13 Aug 92 14:50:21 GMT
- References: <1992Aug12.102140.5231@nntp.hut.fi> <1992Aug13.044351.7123@cs.brown.edu>
- Lines: 30
- Nntp-Posting-Host: dn66.dse.beckman.com
-
- In <1992Aug13.044351.7123@cs.brown.edu> dzk@cs.brown.edu (Danny Keren) writes:
-
- >If I recall correctly, the fact that for *every* infinite cardinality
- >alpha it is true that alpha = alpha X alpha is equivalent to the
- >axiom of choice (help, anyone? I can prove that the axiom of choice
- >implies alpha = alpha X alpha, what about the other direction?).
-
- See Rubin & Rubin, _Equivalents of the Axiom of Choice_ for many and proofs
- equivalents. (The Rubin's are my parents). My recollection of the proof:
-
- Let x be a set. (Without loss of generality, x contains no ordinals). Let
- y be H(x) = {al: al an ordinal and al <= x} (Hartog's function). (It
- follows from elementary considerations that ~(y <= x).) Consider x U y.
- If (x U y) X (x U y) = (x U y), then we can select f:x X y <= x U y. If
- (Ez in x)(A(al)<y) (f(z,al) in x), then f(z,.):y <= x, a contradiction.
- Otherwise, define g on x by g(z) = f(z,first al<y such that f(z,al) in y).
- Clearly g: x <= y, so x can be well-ordered. (It seems clear to me that
- the well-ordering principle implies the axiom of choice. Of course, I've
- been working in the field on and off for 20 years.)
-
- I don't have TeX here; with a little work I could convert that proof into
- something that would appear reasonable in TeX, but then _I_ couldn't read
- it.
-
-
- --
- Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
- 216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
- My opinions are my own, and do not represent those of my employer.
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