home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.math
- Path: sparky!uunet!brunix!brunix!dzk
- From: dzk@cs.brown.edu (Danny Keren)
- Subject: Re: Is Card(R)=Card(R^2)? (yes, and a question)
- Message-ID: <1992Aug13.044351.7123@cs.brown.edu>
- Sender: news@cs.brown.edu
- Organization: Brown University Department of Computer Science
- References: <1992Aug12.102140.5231@nntp.hut.fi>
- Date: Thu, 13 Aug 1992 04:43:51 GMT
- Lines: 30
-
- In article <1992Aug12.102140.5231@nntp.hut.fi> samu@lammio.hut.fi writes:
-
- #I have been wondering... It seems first that there are more pairs of
- #real numbers than real numbers, but is it really so? And if not, I would
- #like to see a bijective mapping f:R->R^2. And furthermore, is there some
- #relation between Card(R) and Card(R^n)?
-
- The cardinalities are the same, since Card(R)=2^aleph0
- (aleph0=Card(integers)). Hence:
-
- Card(R^2)=Card(R) X Card(R) = 2^(aleph0+aleph0)=2^aleph0=Card(R).
-
- (This extends immediately to show Card(R)=Card(R^n) for every integer n).
-
- To build a bijection R-->R^2, take every real number, write it in its
- decimal expression, and map it to the real pair of numbers consisting
- of the odd and even digits in the expansion. Actually, there is even a
- *continuous function* from the unit interval onto the unit square! Peano
- gave the first such fuction, which stunned many mathematicians...
-
- 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?).
-
-
- -Danny Keren.
-
-
-
-