home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.math
- Path: sparky!uunet!comp.vuw.ac.nz!cc-server4.massey.ac.nz!TMoore@massey.ac.nz
- From: news@massey.ac.nz (USENET News System)
- Subject: Re: Is Card(R)=Card(R^2)?
- Message-ID: <1992Aug13.000715.21464@massey.ac.nz>
- Organization: Massey University
- References: <1992Aug12.102140.5231@nntp.hut.fi>
- Date: Thu, 13 Aug 92 00:07:15 GMT
- Lines: 25
-
- In article <1992Aug12.102140.5231@nntp.hut.fi>, samu@lammio.hut.fi (Samuli Siltanen) 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)?
- >
- > Hoping that this is not a trivial one
- >
- It is not trivial, but not too hard either. This is the one which slowed
- acceptance of set theory - many mathematicians were reluctant to
- accept the result and therefore felt that set theory must be wrong.
-
- There is a bijective mapping.
-
- First attempt: use assign alternate digits of the decimal expansion
- of z to x and y. (To keep it simple let 0 <= z <= 1). This does not
- quite work because 0.52909090... maps to (0.5999..., 0.2000...)
- and so does 0.62000... . We have, however, shown that
- #([0,1]) >= #([0,1] x [0,1]).
- Cantor laboured a great deal to get an explicit bijection, but now
- we can just use the Schroeder-Bernstein theorem:
- #A >= #B and #A <= #B implies #A = #B.
-
- A similar argument shows that #R = #R^n.
-