Re: Ultimate debunking of Cantor's Theory


"Calvin" <crice5@xxxxxxxxxxxxxx> wrote in message news:1184334767.128726.26470@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Jul 13, 2:03 am, "Peter Webb"
<webbfam...@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
What Prog said is quite true. He just picked an example which (whilst
correct) is probably a little bit too clever.

He hasn't claimed that you can write down a list of all Reals in base 2. He
claims that the Cantor diagonal argument cannot be used in base 2 to prove
this.

Here is a somewhat clearer example.

Imagine the list is:

a(1) = 0.011111 ...
a(2) = 0.01
a(3) = 0.001
a(4) = 0.0001
.
.

Form the diagonal by flipping bits. You get

cantor diagonal = 0.1000000...

But 0.0111... is the same number 0.1 (unless you also believe that 0.999...
<> 1), and so the Cantor diagonal does appear on the list.

Now we all know that the list above doesn't contain every Real, but the
Cantor diagonal construction doesn't itself prove this to be true.

(His example was a "bit too clever" because he set a(2) = a(3) = a(4) ....,
which is quite valid but obscures his central argument).

My problem with what he did is not a failure to
understand that 0.0111... is the same number as 0.1;
I accepted that immediately.

My problem with it is that he produced a CONTRIVED list,
flipped the diagonal and said that the new number
thus produced is indeed already in the list. So
what?

That's not the sort of thing Cantor's argument is.
He takes a traditional reductio ad absurdum approach
and 'assumes' that what he wants to prove is not true.
If it were not true, then a countable list of decimal
(or binary) expansions of the reals between 0 and 1
could exist. He didn't say how such a list might be
generated. How could he? Why would he? He wants
to show that the existence of such a complete list
is impossible.

To make a simple analogy, if you are proving that
there is no rational number the square of which
is 2, you assume the contrary and posit integers p
and q in lowest terms, q not 0, for which the square
of p/q is 2. You don't claim to know what p and q
would have to be. That would be outrageously absurd,
since your motive is to prove that they don't exist
as posited.

I could be very much in sympathy with a complaint
against Cantor's diagonal argument that said it is
not acceptable to operate on the assumed impossible
list to the extent that he does in order to 'show'
that such a countable list cannot be complete.
I've always been uneasy about that. But the sort
of approach that Proginoskes uses to 'refute' Cantor
in the case of a list of binary forms of the reals
is an insult to everyone's intelligence, it seems
to me.


I said this last time, I will say it again now.

He is not arguing that you can construct a list of all Reals in base 2. He was correcting somebody who said the Cantor diagonal proof was easy in base 2, which it isn't, which is interesting because base 2 is the only base where the Cantor construction can't be used (in its simplest form).

The problem is that if you just flip bits, you could end up with a number which is already on the list, because (in the example given) the procedure creates 0.100.. when 0.0111.. is already on the list.

Now, you only need to show one example where the Cantor construction fails to produce a number not on the list, and you can no longer claim that the number produced by the Cantor construction is always a number not on the list. So it is possible it is already on the list.

Its easy to get around this problem that (for example) 0.5 = 0.4999... in other base other than base 2, and if you have a look at the Cantor construction the problem doesn't exist in base 10 or any other base (except 2).

The standard workaround in base 2 is to consider pairs of digits, which is equivalent to using base 4. Not that this is needed; if the Reals aren't equinumerous with the integers in base 10, then they aren't in any base.


.

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