In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> Am Dienstag, 3. Dezember 2013 11:21:46 UTC+1 schrieb christian.bau: > > > Shouldn't really replying at all > > Yes, if you prefer to stay blind, you should only talk to other blinds.
Being blind to what is not really there is not a drawback, but WM's seeing things which are not there is. > > >"Zeit Geist"s fine post proved that the real numbers are uncountable by > >starting with an assumption that there is a sequence containing them all > >(which is _by definition_ the opposite of the reals being uncountable) and > >went on to demonstrate a contradiction, which means the assumption of the > >existence of such a sequence must be false. > > It seems you have not yet understood the main point. To contradict a > nonsensical definition does not prove anything: There is no sequence > containing "all" rational numbers.
Yes, there are such sequences, I have even constructed some of them myself.
Consider this well-ordering of |Q, where each rational in |Q is represented by m/n, with m an integer, n a natural. and with m and n having no common factor greater than 1, and ordered as follows: if |m'| + n' < |m"| + n" then m'/n' precedes m"/n" if |m'| + n' = |m"| + n" and m' < m" then m'/n' precedes m"/n"
So 0/1 comes first, then -1/1, and 1/1, then -2/1, -1/2, 1/2 and 2/1, and so on.
If WM claims that this ordering is NOT a well-ordering of all the rationals with a unique non-successor element or that it is NOT order-isomorphic to the naturally ordered naturals, let him try to prove it.
He will fail.
A little thought, if WM is capable of it, should show him that every rational thus ordered has an immediate successor and everyone other than 0/1 has an immediate predecessor.
> > This *would* be the point if such sequence could exist at all. But that is > impossible.
In what respect is the sequencing of rationals described above "impossible"?
Please try at least to understand that you are mistaken if you > think that you can conclude from >
> > The full text of the first line reads > > "every rational number can be enumerated if and only if beyond that > rational number infinitely many are following." > > Can you understand that it is impossible to enumerate any rational number > violating this requirement?
I do not see that there is any rational number which that rule prevents from being correctly ordered. When one has a finite rule, like the one I posted above, it can be applied in infinitely many instances without getting worn out or used up.
> But if countability *would* be a sensible notion, then there is a simple > proof that all definable reals belong to a countable set. Since all > "uncountability proofs" prove the existence of a further definable number, > all are trash.
Only if one assumes up front that that is the case.
Absent WM's a priori assumption, there is no proof that any such assumption is warranted. --