Re: diagonalization
 From: William Hughes <wpihughes@xxxxxxxxx>
 Date: Tue, 15 Feb 2011 14:16:01 0800 (PST)
On Feb 15, 5:33 pm, Graham Cooper <grahamcoop...@xxxxxxxxx> wrote:
On Feb 16, 7:13 am, William Hughes <wpihug...@xxxxxxxxx> wrote:
On Feb 15, 5:07 pm, Graham Cooper <grahamcoop...@xxxxxxxxx> wrote:
On Feb 16, 6:29 am, William Hughes <wpihug...@xxxxxxxxx> wrote:
On Feb 15, 2:20 pm, Graham Cooper <grahamcoop...@xxxxxxxxx> wrote:
On Feb 16, 4:10 am, William Hughes <wpihug...@xxxxxxxxx> wrote:
On Feb 15, 1:59 pm, Graham Cooper <grahamcoop...@xxxxxxxxx> wrote:
On Feb 16, 3:13 am, William Hughes <wpihug...@xxxxxxxxx> wrote:
On Feb 15, 12:19 pm, Graham Cooper <grahamcoop...@xxxxxxxxx> wrote:
On Feb 16, 1:38 am, William Hughes <wpihug...@xxxxxxxxx> wrote:
On Feb 15, 11:08 am, Graham Cooper <grahamcoop...@xxxxxxxxx> wrote:
On Feb 9, 3:17 am, stevendaryl3...@xxxxxxxxx (Daryl McCullough) wrote:
Aatu Koskensilta says...
William Hughes <wpihug...@xxxxxxxxx> writes:
The best way to think of this is probably thatCantor'sdiagonal
method applies only to "square" lists.
The didactically preferable version presents just the mathematical and
logical essence of the argument:
Let f be a function taking elements of a set A to subsets of
A. There is then a subset of A not in the range of f. For consider
the subset D of A defined by
D = {x in A  x not in f(x)}
and suppose there were an a in A such that f(a) = D. We would have,
by definition of D, that a in D iff a not in D, a contradiction.
I've been too brainwashed by the Cantorian establishment; I can't
get in the frame of mind in which I can possibly see an objection
to this.

Daryl McCullough
You can't??
Let me translate:
Let a room full of numbered boxes each contain some plastic number
fridge magnets.
Which box contains all the fridge magnets of the numbers of the boxes
that don't contain a fridge magnet of the number of their own box?
You think this implies higher infinites than the far right of the
number line?
Your PROOF is you need a fresh box to hold those fridge magnets on
FINITE examples.
Here is your OBJECTIONLESS proof!
BOX 1 = 4, 6, 7 [does NOT contain a fridge magnet with its box
number]
BOX 2 = 2, 8, 189 [DOES contain a fridge magnet with it's box number]
BOX 3 = 1, 2, 4, 5 [does NOT contain a fridge magnet with it's box
number]
NEW BOX = [1, 3] [THE HAVE NOTS!]
Look, an algorithm to make a box full of fridge magnets unlike any
other!
And this algorithm will apply to any set of boxes
of fridge magnets.
So we conclude that no matter what set of boxes we have,
we can produce a box not in that set.
And note that this algorithm can be easily generalized
to *any* set (not just a set of boxes of fridge magnets).
So is that your proof Daryl?
You honestly base higher infinities of higher infinities on that proof
On the generalization to any set, yes.
 William Hughes
Well that's a start!
a selfcontaining set can not contain only the nonselfcontaining
sets
a nonselfcontaining set can not contain all nonselfcontaining sets
The phrase "all nonselfcontaining sets", has no meaning until
we specify where they can come from.
We can have nonselfcontaining sets in Y as subset of P(X)
Then
a set not in Y, can contain all the
nonselfcontaining sets in Y.
No paradox.
 William Hughes
So you allow the formerly paradoxical construct, define it over
infinite range, and conclude *the only conlusion* is the existence of
a partition between lists and sets?
No. I note if the assumption is made that there
is a bijection between N and P(N) we get a contradiction.
My conclusion is that the assumption is false and there
is no such bijection. At no time do I talk about
"a partition between lists and sets" whatever that means.
 William Hughes
right! But your contradiction is arrived from 2 assumptions.
1/ a bijection to the powerset
2/ The existence of a variation of Russel's set
No
[Note the original proof was of the form
let f:X>P(X) then there exists y in P(X) not in
the range of f.
We have shifted to the slightly different
let f:X>P(X) then f is not a bijection]
The only assumption made is that a bijection between
X and P(X) exists.
The existence of a variation of Russel's set follows
from this assumption and is the contradiction used to
show the assumption is false.
 William Hughes Hide quoted text 
 Show quoted text 
This line of reasoning will never go anywhere, but you are merely
assuming an infinite Russel set is well defined.
Quite the opposite. The fact that the infinite Russel set is not
well defined is used to show the original assumption is false.
 William Hughes
no. here is my bijection.
What you have given if the first part of a list
of lists. At this point we have no idea if it
is a bijection or not. Assume it is a bijection.
1  1 2 3 4 5 6
2  1 3 5 6 7
3  1 3 5 6 7 8 9
4  2 3 4 5 6 7
5  11 12 113 154
...
You construct an infinite russel set
Correct. This is a contradiction. Thus the
original assumption must be false. What
you have given cannot be a bijection.
and call it an unmappable
element.
No, only in the first proof.
Not in the direct proof by contradiction.
 William Hughes
.
 FollowUps:
 Re: diagonalization
 From: Graham Cooper
 Re: diagonalization
 References:
 diagonalization
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 Re: diagonalization
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 Re: diagonalization
 From: Daryl McCullough
 Re: diagonalization
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 Re: diagonalization
 From: William Hughes
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 Re: diagonalization
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