|
|
Re: Matheology § 038
Posted:
Jun 15, 2012 2:08 PM
|
|
On 15 Jun., 19:37, PotatoSauce <kiwisqu...@gmail.com> wrote: > On Friday, June 15, 2012 12:42:39 PM UTC-4, WM wrote: > > Of course there is a limit involved. Otherwise you could not biject > > all numbers. > > Bijection between Natural numbers and the Even Natural numbers. > > f: N ----> 2N, > > f(n) = 2n. > > What limit? > > > Without limit you cannot include all numbers unless you > > find a last one. > > What "last one"? Tell me exactly why I need to find a "last one" to define the map > > f: N ---> 2N > > f(n) = 2n. >
You define nothing else, but a bijection between finite initial segments.
> > Do you think there was a last natural number in an > > infinite set? > > 1) Your question makes no sense.
You are wrong. What does not make sense is to talk about "all" natural numbers as if there was a last one. What does absolutely not make sense is finished infinity. > > 2) It has absolutely nothing to do with bijections. > Look here: Every initial segment of the even positive numbers (2, 4, ..., 2n) has cardinality n, i.e., less than 2n. That can be proved for every finite initial segment and every natural number n. |{2, 4, ..., 2n}| = n < 2n You need the limit in order to breack that rule.
Unless we go to the limit, there is no chance to have a cardinality aleph_0 of a set that contains only finite numbers 2n.
Regards, WM
|
|