Re: Ultimate debunking of Cantor's Theory

In article <1184400850.822238.6310@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
WM <mueckenh@xxxxxxxxxxxxxxxxx> wrote:

On 14 Jul., 01:22, Rotwang <sg...@xxxxxxxxxxxxx> wrote:

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...

This seems irrelevant to the discussion, which was about the diagonal
argument as used to prove that the reals between 0 and 1 are
uncountable.

Cantor used this argument as a simplified proof of his theorem that
the real numbers are uncountable.

And it certainly proves that the set of binary strings is not countable.

Since the set of numbers with dual representation IS countable, an
incountallby infinite set with a countable subset deleted is still
uncountably infinite.

Obviously nonsense is the assumption that a diagonal could be formed
in a list which contains at most numbers which are half as long as the
diagonal.

As any listed number can be extended arbitrarily with zeros without
affecting its value, that dog don't hunt.

And if it is, what do you think that proves? Unless you show some
actual error along the way, doesn't it just show that the mere
existence of decimal expansions of real numbers leads to something
that is nonsense? One can't discard a mathematical proof merely
because one objects to its conclusion.

If the width of the list is at most half of the lengths (or width) of
the diagonal, then the diagonal is not a diagonal, is it?

Extend the "width of the list" with appended zeros. No values are
changed and the problem disappears.

And if the width of the list is sufficient to cover the diagonal, then
the diagonal (minus one digit) is an entry of the list.

That only proves that some rules do not work, but if one's rule takes
binary digits two at a time and replaces the pair '01' by '01, and
everything else by '01' one gets a diagonal NOT in the list.

WM seems to think that if one rule does not work, there can be no rule
which does, but that turns out to be false.
.

Relevant Pages

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