Re: bijection of R: R <--> Rx.....xR
- From: William Elliot <marsh@xxxxxxxxxxxxxxxxxx>
- Date: Wed, 7 Sep 2005 06:05:18 -0700
On Wed, 7 Sep 2005, Timothy Golden http://www.BandTechnology.com wrote:
> Dirk Once wrote:
> >there is also a bijection
> > R <--> RxRxR
> >and in fact between R and the product set of any countable
> >number of R's:
> > R <--> Rx.....xR
>
> Could someone please expound on this?
> How does R map to RxR?
>
There isn't any explicit map. The existence of a map is shown by
|RxR| = 2^|N| * 2^|N| = 2^(|N| + |N|) = 2^|N| = |R|
tho there's a more general theorem that for any infinite set A
|AxA| = |A| * |A| = |A|
which has a set theory proof that eludes me.
.
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- From: Timothy Golden http://www.BandTechnology.com
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