Re: Ultimate debunking of Cantor's Theory

On Jul 17, 8:01 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:
On 16 Jul., 19:39, Randy Poe <poespam-t...@xxxxxxxxx> wrote:



On Jul 16, 1:22 pm, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

On 16 Jul., 18:16, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:

On Jul 13, 10:14 am, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

Consider the list
0.0
0.1
0.11
0.111
...
and switch 0 to 1 on the diagonal. Then you have at the diagonal the
number 0.111..., but only if this number (with one digit less) is also
in the list.

And that does not refute that the set of real number is uncountable.

Let's first clear the following question: Do you agree that the
diagonal in this special case is in the list?

The diagonal of this list (which is a non-terminating sequence
of 1's) is not in the list (all of the list elements have
a terminating sequence of 1's).

But every digits of the diagonal is adjacent to the last 1 of a
sequence of 1's of the list.

I'd phrase this, "for every n, there is an element which
terminates at digit n". So?

Can a terminating sequence consist of one
elemnt less than a non terminating sequence?

No. But that isn't the case here. The non-terminating
sequence has infinitely more nonzero digits than any
of the terminating numbers.

Why do you ask that question? What is the relevance?

- Randy

.

Relevant Pages

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  • Re: Ultimate debunking of Cantors Theory
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