Re: bijection of R: R <--> Rx.....xR


Proginoskes wrote:
> Jiri Lebl wrote:
> > Timothy Golden http://www.BandTechnology.com wrote:
> > > It's amazing how fast results come back on this system these days.
> > > I preferred it when I would come check in the next day.
> > > It gave me some time to think.
> > > Anyhow, it does not look like a clean mechanism to go even from R to
> > > RxR.
> > > It is radix dependent and is not actually a mathematical relationship.
> > > I got into this problem from studying angular coordinate systems versus
> > > cartesian.
> >
> > You are very unlikely to find a simple bijection from R to RxR.
>
> Probably the nicest one can be constructed as follows:
>
> (1) Let f:R -> P(N) be a bijection from the real numbers to the
> subsets of the positive integers. Such a bijection exists.
>
> (2) Let g:P(N) -> P(N) be defined by
> n is in g(A) iff 2*n is in A,
> and let h:P(N) -> P(N) be defined by
> n is in h(A) iff 2*n-1 is in A.
> Clearly g and h are bijections.
>
> (3) Now define F:R -> R^2 by
> F(r) = (F^(-1)(g(f(r))), F^(-1)(h(f(r)))).
> F is a composition of bijections, so F is a bijection.
>
> Cantor saw this proof (in some form) and said, "I see it, but I don't
> believe it."
> --- Christopher Heckman

Here is another one. Well, it's between the unit interval and the unit
square; well-known bijections between those and all of R or RxR can be
used to extend this. I hope this recursive definition has the right
mathematical flavour, and it doesn't need excursions into Set Theory.

A bijection between [0,1] and [0,1]x[0,1]:

First define the following function of one variable:

g(0) = 0
g(x+1) = (g(x)+3)/4 (for x >= 0)
g(1/x) = 1 - g(x) (for x > 1)

Sofar g is defined on Q+ (positive rationals). Extend the domain to R+
(positive reals) by taking limits (and show that this is well-defined).

The resulting function from R+ to R+ is then continuous at irrationals,
although it has discontinuities at all rationals (for example, the left
and right limits at 1 both exist, but are different:
g(1-) = 1/4
g(1+) = g(1) = 3/4

Now define f(x,y) with two variables: [0,1] x [0,1] -> [0,1]

f(x,y) = (4g(x)+8g(y))/9

This function is bijective, as well as bi-monotonic,
i.e. x1 > x2 ==> f(x1,y) > f(x2,y)
y1 > y2 ==> f(x,y1) > f(x,y2)

Michel

(See also "Mächtigkeit des Kontinuums" in de.sci.mathematik, Jan 5)

.

Relevant Pages

  • Re: bijection of R: R <--> Rx.....xR
    ... You are very unlikely to find a simple bijection from R to RxR. ... need to totally ignore any geometry (and topology) of R and RxR. ...
    (sci.math)
  • Re: measurable functions (and hilbert spaces)
    ... I can say if R is a separable metric space with a bijection ... > between R and RxR, ... > Therefore I have an isomorphism, h, (bi-measurable bijection) between ...
    (sci.math)
  • Re: bijection of R: R <--> Rx.....xR
    ... Jiri Lebl wrote: ... > You are very unlikely to find a simple bijection from R to RxR. ... n is in giff 2*n is in A, ...
    (sci.math)
  • Re: measurable functions (and hilbert spaces)
    ... I can say if R is a separable metric space with a bijection ... between R and RxR, ... Therefore I have an isomorphism, h, (bi-measurable bijection) between ...
    (sci.math)
  • Re: An uncountable countable set
    ... (You need only consider the images of 1, these diverge to -oo.) ... As I only consider positive rationals, ... Such a bijection does not exist, because there is no smallest positive ... But it would, if actual infinity existed. ...
    (sci.math)
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