|
|
Re: Bijection Between Complex Numbers and Real Numbers
Posted:
Feb 20, 2012 9:34 PM
|
|
"Richard Tobin" <richard@cogsci.ed.ac.uk> wrote in message
news:jhtdqe$163e$1@matchbox.inf.ed.ac.uk...
> In article
> <513f89be-feb4-4e36-8210-8f2130314b70@x19g2000yqh.googlegroups.com>,
> Michael Ejercito <mejercit@hotmail.com> wrote:
>
>>> Though this has the usual problem with multiple representations. For
>>> example, both 0.1 and 0.00909090... map to (0.1, 0).
>
>> That is correct. This would imply that each complex number maps
>>onto MULTIPLE real numbers, and certainly the reals can not have a
>>GREATER cardinality than the complexes.
>
> Given a surjection both ways, the dual Schroder Bernstein theorem
> shows that there is a bijection.
But doesn't provide an obvious bijection. And this is using a comparatively
advanced result to prove that a solution to a simple problem exists.
> This however relies on the axiom
> of choice.
>
No. Cantor Bernstein Schroder does not require AoC. It is constructive.
> If we take the interleaved digits as a base-11 number we have an injection
> both ways, and standard Schroder Bernstein gives us a bijection.
>
> -- Richard
|
|
