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- Path: sparky!uunet!pipex!unipalm!uknet!gdt!mapsj
- From: mapsj@gdr.bath.ac.uk (Simon Juden)
- Newsgroups: sci.math
- Subject: Re: Is Card(R)=Card(R^2)?
- Message-ID: <1992Aug13.154422.9362@gdr.bath.ac.uk>
- Date: 13 Aug 92 15:44:22 GMT
- References: <1992Aug13.000928.12631@unidus.rz.uni-duesseldorf.de> <1992Aug13.011522.11161@informix.com> <1992Aug13.131518.2982@gdr.bath.ac.uk>
- Organization: School of Mathematics, University of Bath, UK
- Lines: 36
-
- In article <1992Aug13.131518.2982@gdr.bath.ac.uk> mapsj@gdr.bath.ac.uk (Simon Juden) writes:
- >In article <1992Aug13.011522.11161@informix.com> proberts@informix.com (Paul Roberts) writes:
- >>In article <1992Aug13.000928.12631@unidus.rz.uni-duesseldorf.de> heisen@convex.rz.uni-duesseldorf.de (Henner Eisen) writes:
- >>>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. ...
- >>>
- >>>Such mappings exist! To construct a mapping f(x,y) = z choose a
- >>>unique decimal representation of x and y and then merge the digits.
- >>>
- >>>i.E.:
- >>>x = 1 0 0 2. 7 1 8 2 8 2 . . . . . .
- >>>y = 3 0 3 .1 4 1 5 9 3 . . . . . .
- >>>
- >>>z = 1300032.174118529832............
- >>>
- >>>This will be bijective for non-negative real numbers. In order to obtain a
- >>>bijective mapping R <-> R^2 this has to be combined with a bijection
- >>>which maps real numbers to non-negative real numbers (right now, I
- >>>don't remeber by heart how this is done, but it is possible).
- >>
- >>I believe that there is even an everywhere-continuous
- >>bijective mapping from the unit line to the unit square.
- >
- >
- >Indeed so - an elementary version of Peano's construction of such may be found
- >on page 283 of the American Mathematical Monthly, April 1983.
- >
- >Simon
-
- Indeed not - I misread you. Peano's map is a cts surjection but of course as
- has been pointed out an elementary argument yields that there can be no
- bijection. Apologies for the inaccuracy, due to my reading too fast!
-
- Simon
-