Re: Cantor Confusion
- From: Virgil <virgil@xxxxxxxxxxx>
- Date: Mon, 21 May 2007 13:09:17 -0600
In article <1179756663.678085.237680@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
WM <mueckenh@xxxxxxxxxxxxxxxxx> wrote:
On 21 Mai, 13:47, stevendaryl3...@xxxxxxxxx (Daryl McCullough) wrote:
WM says...
On 20 Mai, 17:57, stevendaryl3...@xxxxxxxxx (Daryl McCullough) wrote:
WM says...
> It is. Set theory is simply biased. Consider the list
> 0.666...
> 0.3666...
> 0.33666...
> 0.333666...
> ...
> If the diagonal number is defined by "replace 6 by 3", then we
> have
> two answers none of which can be preferred by logic, but the
> second of
> which is suppressed by convention.
The diagonal number is 0.333... which is not on the list.
The diagonal number cannot have more 3's than every list number has
3's.
Of course it can.
Consider the following infinite matrix M:
xooo...
xxooo...
xxxooo...
...
You claim the (unchanged) diagonal can have more x's than any line?
Each line contains a finite number of x's.
The diagonal contains more that any finite number of x's.
So "yes"!
Than is wrong because the diagonal consists of line elements.
WM assumes, contrary to fact, the there is a last line and a last
character in each line in each line and a last character in the diagonal.
With your rule for the diagonal: replace 3 by 6, then
if we start with the following list:
0.6000
0.06000
0.006000
0.0006000
0.00006000
...
then the diagonal will be the same: 0.333...
There is one "3" for each number in the list.
What do you think why I chose the special list? In that list the
diagonal number cannot have more 3's than at least one line of the
list (well, one more).
x0 = 0.666...
x1 = 0.3666...
x2 = 0.33666...
x4 = 0.333666...
xn = 1/3 + 1/(3*10^n) for n in N u {0}
y = 1/3
So that xn - y = 1/(3*10^n) =\= 0 for all n in N u {0}
If the diagonal number is complete in the sense that at every
position indexed by a natural number there is a 3, then there must be
a complete sequence of 3's among the list entries too.
No, that's false. It's *provably* false.
It is also provably right. See matrix M above.
WM has yet to construct a valid proof of anything.
I say in this case because not every sequence has a limit. For example,
the sequence
0.600...
0.330...
0.6660...
0.33300...
0.666600...
0.3333000...
has no limit. But it still has a diagonal, namely
0.363636...
But if this diagonal is to be considered a real number, then we need
the factors 10^-i.
Factoring decimals is a muggs game.
Forall n e N we have 0 < 1 - 0.999...9 (with n 9's).
Yes, that's true. 1 does not appear on the list
0.9
0.99
0.999
0.9999
0.99999
etc.
and 0.33333 does not appear on the list
0.6666...
0.36666
0.336666
0.3336666
etc.
So r does not appear on the list.
So the problem of replacing 0 by 9 is not existing.
What are you talking about? Of course the number 0.9999...
(all 9s) exists. And it is equal to 1. But it is not on the
list
0.9
0.99
0.999
0.9999
0.99999
etc.
But we have in the limit j --> oo: D_j = 0.
Yes. That's true, but irrelevant. For the diagonal to be
on the list, it must be that D_j = 0 for some finite j.
That is only required for a finite list.
WRONG!
It's not good enough that limit of D_j = 0.
Consider Dave L. Renfro's example:
0.49999999...
0.49111111...
0.44911111...
0.44491111...
0.44449111...
. . . . . . .
Replacing 4 by 5 and 9 by 0 gives the diagonal number
0.5000...0999... =/= 0.5 for any finite j. Only in the limit j --> oo
the diagonal number is 0.5000... = 0.499..., the first entry. Here it
is good enough to consider the limit to make the proof fail?
That one can construct a rule that fails does not mean all rules fail.
It is enough that there is one rule which does NOT fail. Such as:
If the nth digit of the nth number in the list is not 2,
put a 2 in that place in the diagonal, otherwise put a 3
in that place in the diagonal.
This rule works for all decimals.
.
- References:
- Re: Cantor Confusion
- From: WM
- Re: Cantor Confusion
- From: WM
- Re: Cantor Confusion
- From: Daryl McCullough
- Re: Cantor Confusion
- From: WM
- Re: Cantor Confusion
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