# 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 self-containing set can not contain only the non-self-containing

sets

a non-self-containing set can not contain all non-self-containing sets

The phrase "all non-self-containing sets", has no meaning until

we specify where they can come from.

We can have non-self-containing sets in Y as subset of P(X)

Then

a set not in Y, can contain all the

non-self-containing 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

.

**Follow-Ups**:**Re: diagonalization***From:*Graham Cooper

**References**:**diagonalization***From:*ubuntuboy

**Re: diagonalization***From:*Aatu Koskensilta

**Re: diagonalization***From:*Daryl McCullough

**Re: diagonalization***From:*Graham Cooper

**Re: diagonalization***From:*William Hughes

**Re: diagonalization***From:*Graham Cooper

**Re: diagonalization***From:*William Hughes

**Re: diagonalization***From:*Graham Cooper

**Re: diagonalization***From:*William Hughes

**Re: diagonalization***From:*Graham Cooper

**Re: diagonalization***From:*William Hughes

**Re: diagonalization***From:*Graham Cooper

**Re: diagonalization***From:*William Hughes

**Re: diagonalization***From:*Graham Cooper

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