Re: ** says: Definition: sum{i in N} i = 0

On 9 Jul., 15:05, hagman <goo...@xxxxxxxxxxxxx> wrote:
On 9 Jul., 09:52, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:





On 9 Jul., 07:32, hagman <goo...@xxxxxxxxxxxxx> wrote:

On 8 Jul., 22:50, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

On the contrary it is simple because Cantor himself considered his
proof an existence proof for trancendental numbers.

Well, but that does not make transcendental number the *cause* for
the uncountability of the reals.
For example, the existence of transcendentals is also possible
by approximation theory: sum 10^(-n!) is transcendental.
Thus transcendental numbers exist. But this tiny fact doesn't
prove uncountabilty of the reals.

Of course only the (asserted) existence of *all* transcendentals makes
R uncountable. Those few transcendentals which were constructed by
Liouville himself and the handful being found later on by Hermite, v.
Lindemann, Schneider, Gelfand and others is certainly not sufficient
to make anything uncountable. (By the way, it was Liouville's aim
already to prove e transcendental.)

Let a_0 be a sequence of ones and twos.

You meant (a_n)?

Yep.



Then
sum a_n*10^(-n!)
is transcendental of Liouville type and the set of these is
uncountable,
of course by the same diagonal argument as used for the reals
themselves.

The set of all infinite sequences (a_n) of this kind is uncountable.
For this sake you need not append any "*10(-n!)".

I need the factor 10^(-n!) in order to produce a convergent series
and indeed a *rapidly* converging series such that the limit
is transcendental and no two sequences (a_n) produce
the same limit.
Note that without the factor 10^(-1), sum a_n is not defined

But sequences are defined. And what I told you is
1) that you need not transcendental numbers in order to show that
there are uncountably many infinite sequences, and
2) that Liouville did not create uncountably many transcendental
numbers.

Regards, WM

.