Re: Ultimate debunking of Cantor's Theory

On 13 Jul., 21:38, Rotwang <sg...@xxxxxxxxxxxxx> wrote:
WM wrote:

[Peter Webb]





When you do the Cantor trick in base 10, you can prove to yourself that it
always produces a number not on the list. Even if you have 0.500.. somewhere
on the list, you can be certain that you will never get the same number in a
different form, such as 0.49999.. as a result of the construction.

If you could find a single example where the Cantor construction failed to
produce a different number - if for example it generated 0.4999.. when 0.5
was on the list - then you can no longer claim that the Cantor construction
ALWAYS produces a new number.

Such an example is easy to find. 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.

The guy to whom you are replying has already given an example where
the Cantor construction fails in base 2. However he was writing about
base 10 above, and with a sensible definition of the construction it
is impossible to find an example where the it fails to find a new
number.

Cantor himself was writing about base 2, using the symbols w and m.
But he did not clearly spell out that he meant real numbers. He only
talked about sequences. And for that case his proof was correct,
because w.mmm... is not the same as m.www... while binary 0.111... =
1.000...

For example one can define the construction so that the
decimal expansion it gives contains only the digits 4 and 7. The only
way that two different decimal expansions can define the same number
is if one of them contains an infinite string of 9's and the other
contains an infinite string of 0's

No. The example I gave shows also that the diagonal 0.111... is
contained in the list:

0.0
0.1
0.11
0.111
....

Of course you can state that the diagonal is an infinite sequence of
1's while every list entry is a finite sequence of 1's. But that does
not help. Every finite number multiplied by 2 is also finite.
Therefore any sequence of 1's of the list numbers is less than half as
long as the sequence of 1's of the diagonal. Therefore the diagonal is
at least twice as long as any sequence of 1's contained in the list,
which is obviously nonsense.

Regards, WM

.

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