true ?

false ?

current theory ?

abstract theory ?

what kind of IDIOT are YOU?
ANSWER NOW!

Graham Cooper <grahamcooper7@gmail.com> writes:

(from the subject) Who really believes ------ EXIST(X) X>INFINITY ?

> true ?

Sometimes. For example with S = {0,INFINITY} and > = {(0,INFINITY)}.

> false ?

Sometimes. For example with S = {0,INFINITY} and > = {(INFINITY,0)}.

> current theory ?

Except in the Ladybird Book of Sets, the term INFINITY is rarely used,
and a symbol other than > is preferred when defining an order relation
between cardinals.

Since the invention of the WWW, learning is easy to come by:
https://www.ucl.ac.uk/~ucahcjm/ast/ast_notes_4.pdf

> abstract theory ?

Are there any mathematical theories that are not abstract?

> what kind of IDIOT are YOU?
> ANSWER NOW!

The kind that answers cranks on Usenet.

--
Ben.

On Tuesday, February 14, 2023 at 11:29:30 PM UTC-5, Graham Cooper wrote:
> true ?
>
> false ?
>
> current theory ?
>
> abstract theory ?
>
>
> what kind of IDIOT are YOU?
> ANSWER NOW!

Guessing at your meaning, are suggesting that if X and Y are infinite sets, then one cannot reasonably be larger than the other? Are you suggesting, for example, that here exists bijection mapping the set of natural numbers N to the power set of N?

Dan

On Tuesday, February 14, 2023 at 11:29:30 PM UTC-5, Graham Cooper wrote:
> true ?
>
> false ?
>
> current theory ?
>
> abstract theory ?
>
>
> what kind of IDIOT are YOU?
> ANSWER NOW!
On Tuesday, February 14, 2023 at 11:29:30 PM UTC-5, Graham Cooper wrote:
> true ?
>
> false ?
>
> current theory ?
>
> abstract theory ?
>
>
> what kind of IDIOT are YOU?
> ANSWER NOW!

Guessing at your meaning here, are suggesting that if X and Y are infinite sets, then one of them cannot reasonably be larger than the other? Are you suggesting, for example, that there exists a bijection mapping the set of natural numbers N to the power set of N?

Dan

On Thursday, February 16, 2023 at 5:18:23 AM UTC+11, Dan Christensen wrote:
> On Tuesday, February 14, 2023 at 11:29:30 PM UTC-5, Graham Cooper wrote:
> > true ?
> >
> > false ?
> >
> > current theory ?
> >
> > abstract theory ?
> >
> >
> > what kind of IDIOT are YOU?
> > ANSWER NOW!
> On Tuesday, February 14, 2023 at 11:29:30 PM UTC-5, Graham Cooper wrote:
> > true ?
> >
> > false ?
> >
> > current theory ?
> >
> > abstract theory ?
> >
> >
> > what kind of IDIOT are YOU?
> > ANSWER NOW!
> Guessing at your meaning here, are suggesting that if X and Y are infinite sets, then one of them cannot reasonably be larger than the other? Are you suggesting, for example, that there exists a bijection mapping the set of natural numbers N to the power set of N?
>
> Dan

There's no bijection from

{0,1}^N to N

but there is a bijection from

{0,1,?}^N to N

ALL INFINITE STRINGS WITH WILDCARD CHARACTER ARE COUNTABLE

technically its

{ e , !e , ?e} ^ N

or

{ TRUE FALSE MAYBE } ^ N

If X is anything but INFINITY and INFINITY is defined to be greater than anything but INFINITY, EXIST(X) X>INFINITY is false (if rather tautological)

If you are (as I assume) talking about the oft heard informal statement ("There are more reals that integers") ,then if we put the statement in a more formal way , "there is no bijection between the integers and the reals," the statement is true.

--
William Hughes

On Wednesday, February 15, 2023 at 5:56:37 PM UTC-4, Graham Cooper wrote:

> There's no bijection from
>
> {0,1}^N to N
>
> but there is a bijection from
>
> {0,1,?}^N to N

Nope, We have ? not the same thing as 1 (We know for sure that 1=1 we do not know for sure that ?=1 so the two things are different).
So in the standard proof we put a 1 whenever we see a ?.

On Thursday, February 16, 2023 at 9:18:46 AM UTC+11, William wrote:
> On Wednesday, February 15, 2023 at 5:56:37 PM UTC-4, Graham Cooper wrote:
>
>
> > There's no bijection from
> >
> > {0,1}^N to N
> >
> > but there is a bijection from
> >
> > {0,1,?}^N to N
> Nope, We have ? not the same thing as 1 (We know for sure that 1=1 we do not know for sure that ?=1 so the two things are different).
> So in the standard proof we put a 1 whenever we see a ?.

CONTRADICTION
WILL - 'do not know ?=1'
WILL - 'change ? to 1'

{true false maybe} ^ N
<=BIJECTS=>
N

FUZZY LOGIC SETS

F = { 1? 2 3 5 7 ... }

is a more accurate description than conventional prime definition

On Thursday, February 16, 2023 at 3:01:03 AM UTC-4, Graham Cooper wrote:
> On Thursday, February 16, 2023 at 9:18:46 AM UTC+11, William wrote:
> > On Wednesday, February 15, 2023 at 5:56:37 PM UTC-4, Graham Cooper wrote:
> >
> >
> > > There's no bijection from
> > >
> > > {0,1}^N to N
> > >
> > > but there is a bijection from
> > >
> > > {0,1,?}^N to N
> > Nope, We have ? not the same thing as 1 (We know for sure that 1=1 we do not know for sure that ?=1 so the two things are different).
> > So in the standard proof we put a 1 whenever we see a ?.
> CONTRADICTION
> WILL - 'do not know ?=1'
> WILL - 'change ? to 1'

Nope. "put a '1' " . The "1" that is put is not the "?" changed We *see* the "?" we do not change it..

On Thursday, February 16, 2023 at 6:19:22 PM UTC+11, William wrote:
> On Thursday, February 16, 2023 at 3:01:03 AM UTC-4, Graham Cooper wrote:
> > On Thursday, February 16, 2023 at 9:18:46 AM UTC+11, William wrote:
> > > On Wednesday, February 15, 2023 at 5:56:37 PM UTC-4, Graham Cooper wrote:
> > >
> > >
> > > > There's no bijection from
> > > >
> > > > {0,1}^N to N
> > > >
> > > > but there is a bijection from
> > > >
> > > > {0,1,?}^N to N
> > > Nope, We have ? not the same thing as 1 (We know for sure that 1=1 we do not know for sure that ?=1 so the two things are different).
> > > So in the standard proof we put a 1 whenever we see a ?.
> > CONTRADICTION
> > WILL - 'do not know ?=1'
> > WILL - 'change ? to 1'
> Nope. "put a '1' " . The "1" that is put is not the "?" changed We *see* the "?" we do not change it..

Lets test your theory

ALL SUPERREALS
0.1000..
0.0?00..
0.1111..
0.0101..
....

DIAGONAL = 0.1?11..
WILLS ANTIDIAGONAL = 0.0100..

WILLS ANTIDIAGONAL IS LISTED

ROW 2 = 0.0?00..

------------------------------

Theres a numerous ways to misinterpret the VARIABLE [?]
its pointless going through them when the correct interpretation simply works

{true false maybe}
<=BIJECTS=>
N

> {true false maybe}
> <=BIJECTS=>
> N

should read

{true false maybe} ^ N
<=BIJECTS=>
N

On Thursday, February 16, 2023 at 3:25:58 AM UTC-4, Graham Cooper wrote:
> On Thursday, February 16, 2023 at 6:19:22 PM UTC+11, William wrote:
> > On Thursday, February 16, 2023 at 3:01:03 AM UTC-4, Graham Cooper wrote:
> > > On Thursday, February 16, 2023 at 9:18:46 AM UTC+11, William wrote:
> > > > On Wednesday, February 15, 2023 at 5:56:37 PM UTC-4, Graham Cooper wrote:
> > > >
> > > >
> > > > > There's no bijection from
> > > > >
> > > > > {0,1}^N to N
> > > > >
> > > > > but there is a bijection from
> > > > >
> > > > > {0,1,?}^N to N
> > > > Nope, We have ? not the same thing as 1 (We know for sure that 1=1 we do not know for sure that ?=1 so the two things are different).
> > > > So in the standard proof we put a 1 whenever we see a ?.
> > > CONTRADICTION
> > > WILL - 'do not know ?=1'
> > > WILL - 'change ? to 1'
> > Nope. "put a '1' " . The "1" that is put is not the "?" changed We *see* the "?" we do not change it..
> Lets test your theory
>
> ALL SUPERREALS
> 0.1000..
> 0.0?00..
> 0.1111..
> 0.0101..
> ...
>
> DIAGONAL = 0.1?11..
> WILLS ANTIDIAGONAL = 0.0100..
>
> WILLS ANTIDIAGONAL IS LISTED

nope
>
> ROW 2 = 0.0?00..
which is not the same 0.0100... In the first case the second digit after the decimal may be 1 but we do not know this. In the second case the second digit after the decimal is 1 and we know this.

On Tuesday, February 14, 2023 at 11:29:30 PM UTC-5, Graham Cooper wrote:
> true ?
>
> false ?
>
> current theory ?
>
> abstract theory ?
>
>
> what kind of IDIOT are YOU?
> ANSWER NOW!

The problem is that you imply something by the term 'INFINITY' without defining it.
Let me try a fresh definition for you.
Let's state explicitly that we are working in radix ten values.
This generally need not be explicitly stated; for it is an assumption of convention.
Still, this detail exposes the fact that we will never garner a large radix ten value without explicitly stating its lower digits.
Let's set them to threes:
x = 333...33
Let's add one:
x + 1 = 333...34
Now we can say that x+1 > x by reason alone, but here we have some actual infinite values that are computable in this way. X was infinite already, and x+1 is infinite, too. That we can take the product
(x)(x+1) = 111...11222...22
and because products require this many digits we can see consistency is being maintained. That these two aleph forms are monstrously larger than the one aleph forms is clear. That they are computable (some of them anyways) is quite a good thing. Their structure is consistent. The only problem really is the usage of the ellipses here. I accept that their usage places these values on a branch of mathematics, but as to what else lays on that branch is a question of what else uses ellipses. Using this condition as a curricular branch puts rather a lot of mathematics on this same branch. For instance, back when we were taught that:
1/3 = 0.333...
did we trouble over those unending digits? Well, a little bit; yes. Fascinating just how many take this form. That they trail off to zero rapidly somewhat alleviates the trouble I suppose, but as we attempt to engage infinity, and we can do so simply by dropping that decimal point, and the function of the decimal point as an appendage to the natural value ought to come into this analysis just about here; well; there you have it.

the Aleph of old versus these digital alephs are different things. The radix analysis awaits this thread, but it has been posted plenty elsewhere here in sci.math.

On Friday, February 17, 2023 at 2:58:56 AM UTC+11, William wrote:
> On Thursday, February 16, 2023 at 3:25:58 AM UTC-4, Graham Cooper wrote:
> > On Thursday, February 16, 2023 at 6:19:22 PM UTC+11, William wrote:
> > > On Thursday, February 16, 2023 at 3:01:03 AM UTC-4, Graham Cooper wrote:
> > > > On Thursday, February 16, 2023 at 9:18:46 AM UTC+11, William wrote:
> > > > > On Wednesday, February 15, 2023 at 5:56:37 PM UTC-4, Graham Cooper wrote:
> > > > >
> > > > >
> > > > > > There's no bijection from
> > > > > >
> > > > > > {0,1}^N to N
> > > > > >
> > > > > > but there is a bijection from
> > > > > >
> > > > > > {0,1,?}^N to N
> > > > > Nope, We have ? not the same thing as 1 (We know for sure that 1=1 we do not know for sure that ?=1 so the two things are different).
> > > > > So in the standard proof we put a 1 whenever we see a ?.
> > > > CONTRADICTION
> > > > WILL - 'do not know ?=1'
> > > > WILL - 'change ? to 1'
> > > Nope. "put a '1' " . The "1" that is put is not the "?" changed We *see* the "?" we do not change it..
> > Lets test your theory
> >
> > ALL SUPERREALS
> > 0.1000..
> > 0.0?00..
> > 0.1111..
> > 0.0101..
> > ...
> >
> > DIAGONAL = 0.1?11..
> > WILLS ANTIDIAGONAL = 0.0100..
> >
> > WILLS ANTIDIAGONAL IS LISTED
> nope
> >
> > ROW 2 = 0.0?00..
> which is not the same 0.0100...

? = UNKNOWN

You can NOT assert 0.0100.. is MISSING
when 0.0?00.. is LISTED

Why is everyone is SCI.MATH such a clueless idiot ??

On Thursday, February 16, 2023 at 5:50:24 PM UTC-4, Graham Cooper wrote:
> On Friday, February 17, 2023 at 2:58:56 AM UTC+11, William wrote:
> > On Thursday, February 16, 2023 at 3:25:58 AM UTC-4, Graham Cooper wrote:
> > > On Thursday, February 16, 2023 at 6:19:22 PM UTC+11, William wrote:
> > > > On Thursday, February 16, 2023 at 3:01:03 AM UTC-4, Graham Cooper wrote:
> > > > > On Thursday, February 16, 2023 at 9:18:46 AM UTC+11, William wrote:
> > > > > > On Wednesday, February 15, 2023 at 5:56:37 PM UTC-4, Graham Cooper wrote:
> > > > > >
> > > > > >
> > > > > > > There's no bijection from
> > > > > > >
> > > > > > > {0,1}^N to N
> > > > > > >
> > > > > > > but there is a bijection from
> > > > > > >
> > > > > > > {0,1,?}^N to N
> > > > > > Nope, We have ? not the same thing as 1 (We know for sure that 1=1 we do not know for sure that ?=1 so the two things are different).
> > > > > > So in the standard proof we put a 1 whenever we see a ?.
> > > > > CONTRADICTION
> > > > > WILL - 'do not know ?=1'
> > > > > WILL - 'change ? to 1'
> > > > Nope. "put a '1' " . The "1" that is put is not the "?" changed We *see* the "?" we do not change it..
> > > Lets test your theory
> > >
> > > ALL SUPERREALS
> > > 0.1000..
> > > 0.0?00..
> > > 0.1111..
> > > 0.0101..
> > > ...
> > >
> > > DIAGONAL = 0.1?11..
> > > WILLS ANTIDIAGONAL = 0.0100..
> > >
> > > WILLS ANTIDIAGONAL IS LISTED
> > nope
> > >
> > > ROW 2 = 0.0?00..
> > which is not the same 0.0100...
> ? = UNKNOWN
>
> You can NOT assert 0.0100.. is MISSING
> when 0.0?00.. is LISTED
0.0?00... is listed but it is not the same as 0.0100... which is not listed.

> > > > ALL SUPERREALS
> > > > 0.1000..
> > > > 0.0?00..
> > > > 0.1111..
> > > > 0.0101..
> > > > ...
> > > >
> > > > DIAGONAL = 0.1?11..
> > > > WILLS ANTIDIAGONAL = 0.0100..
> > > >
> > > > WILLS ANTIDIAGONAL IS LISTED
> > > nope
> > > >
> > > > ROW 2 = 0.0?00..
> > > which is not the same 0.0100...
> >
> > ? = UNKNOWN
> >
> > You can NOT assert 0.0100.. is MISSING
> > when 0.0?00.. is LISTED
>
> 0.0?00... is listed but it is not the same as 0.0100... which is not listed.

Given the sequence

< 0 X 0 0 >
and
< 0 1 0 0 >

it is possible that the sequences are equal when X=1

|N| = |{ T F ? } ^ N| >= |{T F} ^ N| = |R|

Imagine a SET with 3 membership options

PRIMES = { 1? 2 3 5 7 ... }

M e S
M !e S
M ?e S

Then POWERSET3( N ) =

{
{ 1 2 3 }
{ 1 2 3 4 5 6 ... }
{ 1? 2 3 5 7 ... }
{ 1? 2 3? 4 5? ... }
...
}

That is - you can explicitly define MEMBERS, NON-MEMBERS and POSSIBLE MEMBERS

Then the POWERSET2 proof fails and

|N| = |R|

On Friday, February 17, 2023 at 3:25:58 AM UTC+11, Timothy Golden wrote:
> On Tuesday, February 14, 2023 at 11:29:30 PM UTC-5, Graham Cooper wrote:
> > true ?
> >
> > false ?
> >
> > current theory ?
> >
> > abstract theory ?
> >
> >
> > what kind of IDIOT are YOU?
> > ANSWER NOW!
> The problem is that you imply something by the term 'INFINITY' without defining it.

how about

EXIST( oo ) ALL( n ) oo>n

> Let me try a fresh definition for you.
> Let's state explicitly that we are working in radix ten values.
> This generally need not be explicitly stated; for it is an assumption of convention.
> Still, this detail exposes the fact that we will never garner a large radix ten value without explicitly stating its lower digits.
> Let's set them to threes:
> x = 333...33
> Let's add one:
> x + 1 = 333...34

an infinite string has no ending character
certainly not one that can be specified as a mismatch to the infinite stream

[...] means TO INFINITUM

> > Let's set them to threes:
> > x = 333...33
> > Let's add one:
> > x + 1 = 333...34
> an infinite string has no ending character
> certainly not one that can be specified as a mismatch to the infinite stream
>
> [...] means TO INFINITUM

EXCEPTION

x1,x2...xn

where the number of terms is explicit

On Thursday, February 16, 2023 at 7:09:26 PM UTC-4, Graham Cooper wrote:
> > > > > ALL SUPERREALS
> > > > > 0.1000..
> > > > > 0.0?00..
> > > > > 0.1111..
> > > > > 0.0101..
> > > > > ...
> > > > >
> > > > > DIAGONAL = 0.1?11..
> > > > > WILLS ANTIDIAGONAL = 0.0100..
> > > > >
> > > > > WILLS ANTIDIAGONAL IS LISTED
> > > > nope
> > > > >
> > > > > ROW 2 = 0.0?00..
> > > > which is not the same 0.0100...
> > >
> > > ? = UNKNOWN
> > >
> > > You can NOT assert 0.0100.. is MISSING
> > > when 0.0?00.. is LISTED
> >
> > 0.0?00... is listed but it is not the same as 0.0100... which is not listed.
> Given the sequence
>
> < 0 X 0 0 >
> and
> < 0 1 0 0 >
>
> it is possible that the sequences are equal when X=1

and it is certain that <0 1 0 0> and <0 1 0 0> are equal. So the two sequences are different.

>
>
> |N| = |{ T F ? } ^ N| >= |{T F} ^ N| = |R|

Assume for the sake of argument the incorrect idea that the powerset proof does not apply to (T F ?)^N. This would not imply that |(T F ?)^N| was equal to |N| just that we could not use the powerset proof to show the two are not equal. The powerset proof still applies to the set {T F} so we have |{ T F ? } ^ N| >= |{T F} ^ N| > |N|

On Friday, February 17, 2023 at 11:47:04 AM UTC+11, William wrote:
> On Thursday, February 16, 2023 at 7:09:26 PM UTC-4, Graham Cooper wrote:
> > > > > > ALL SUPERREALS
> > > > > > 0.1000..
> > > > > > 0.0?00..
> > > > > > 0.1111..
> > > > > > 0.0101..
> > > > > > ...
> > > > > >
> > > > > > DIAGONAL = 0.1?11..
> > > > > > WILLS ANTIDIAGONAL = 0.0100..
> > > > > >
> > > > > > WILLS ANTIDIAGONAL IS LISTED
> > > > > nope
> > > > > >
> > > > > > ROW 2 = 0.0?00..
> > > > > which is not the same 0.0100...
> > > >
> > > > ? = UNKNOWN
> > > >
> > > > You can NOT assert 0.0100.. is MISSING
> > > > when 0.0?00.. is LISTED
> > >
> > > 0.0?00... is listed but it is not the same as 0.0100... which is not listed.
> > Given the sequence
> >
> > < 0 X 0 0 >
> > and
> > < 0 1 0 0 >
> >
> > it is possible that the sequences are equal when X=1
> and it is certain that <0 1 0 0> and <0 1 0 0> are equal. So the two sequences are different.

your interpretation of simple variable assignment is bizarre

it is POSSIBLE the sequences are equal
therefore you CANNOT conclude that they are different

different syntax maybe

> >
> >
> > |N| = |{ T F ? } ^ N| >= |{T F} ^ N| = |R|
> Assume for the sake of argument the incorrect idea that the powerset proof does not apply to (T F ?)^N.

OK

> This would not imply that |(T F ?)^N| was equal to |N| just that we could not use the powerset proof to show the two are not equal.

RIGHT

The anti-diagonal argument doesnt apply to infinite strings with variables

> The powerset proof still applies to the set {T F} so we have |{ T F ? } ^ N| >= |{T F} ^ N| > |N|

Its like saying SQRT(-1) is undefined.

Then you define i

Then you have a LARGER NUMBER SET

That number set INCLUDES the previous number set

Then the previous (non)definition no longer applies

as it doesn't apply to the LARGER number set.

A subset of a complete grammar may contain a contradiction not part of the full grammar

On Thursday, February 16, 2023 at 9:01:14 PM UTC-4, Graham Cooper wrote:
> On Friday, February 17, 2023 at 11:47:04 AM UTC+11, William wrote:
> > On Thursday, February 16, 2023 at 7:09:26 PM UTC-4, Graham Cooper wrote:

> > > |N| = |{ T F ? } ^ N| >= |{T F} ^ N| = |R|
> > Assume for the sake of argument the incorrect idea that the powerset proof does not apply to (T F ?)^N.
> OK
> > This would not imply that |(T F ?)^N| was equal to |N| just that we could not use the powerset proof to show the two are not equal.
> RIGHT
>
> The anti-diagonal argument doesnt apply to infinite strings with variables
> > The powerset proof still applies to the set {T F} so we have |{ T F ? } ^ N| >= |{T F} ^ N| > |N|
> Its like saying SQRT(-1) is undefined.
>
> Then you define i
>
> Then you have a LARGER NUMBER SET

Correct. |{ T F ? } ^ N| >= |{T F} ^ N|

On Friday, February 17, 2023 at 12:42:41 PM UTC+11, William wrote:
> On Thursday, February 16, 2023 at 9:01:14 PM UTC-4, Graham Cooper wrote:
> > On Friday, February 17, 2023 at 11:47:04 AM UTC+11, William wrote:
> > > On Thursday, February 16, 2023 at 7:09:26 PM UTC-4, Graham Cooper wrote:
>
> > > > |N| = |{ T F ? } ^ N| >= |{T F} ^ N| = |R|
> > > Assume for the sake of argument the incorrect idea that the powerset proof does not apply to (T F ?)^N.
> > OK
> > > This would not imply that |(T F ?)^N| was equal to |N| just that we could not use the powerset proof to show the two are not equal.
> > RIGHT
> >
> > The anti-diagonal argument doesnt apply to infinite strings with variables
> > > The powerset proof still applies to the set {T F} so we have |{ T F ? } ^ N| >= |{T F} ^ N| > |N|
> > Its like saying SQRT(-1) is undefined.
> >
> > Then you define i
> >
> > Then you have a LARGER NUMBER SET
> Correct. |{ T F ? } ^ N| >= |{T F} ^ N|

Lets NOT assume |POWERSET2(N)| > |N|

However |POWERSET2(N)| > | ANY LIST OF R |

This is uninteresting as R is incomplete anyway

R c SR
|SR| = |N|
|R| = |N|

On Friday, February 17, 2023 at 12:59:20 PM UTC+11, Graham Cooper wrote:
> On Friday, February 17, 2023 at 12:42:41 PM UTC+11, William wrote:
> > On Thursday, February 16, 2023 at 9:01:14 PM UTC-4, Graham Cooper wrote:
> > > On Friday, February 17, 2023 at 11:47:04 AM UTC+11, William wrote:
> > > > On Thursday, February 16, 2023 at 7:09:26 PM UTC-4, Graham Cooper wrote:
> >
> > > > > |N| = |{ T F ? } ^ N| >= |{T F} ^ N| = |R|
> > > > Assume for the sake of argument the incorrect idea that the powerset proof does not apply to (T F ?)^N.
> > > OK
> > > > This would not imply that |(T F ?)^N| was equal to |N| just that we could not use the powerset proof to show the two are not equal.
> > > RIGHT
> > >
> > > The anti-diagonal argument doesnt apply to infinite strings with variables
> > > > The powerset proof still applies to the set {T F} so we have |{ T F ? } ^ N| >= |{T F} ^ N| > |N|
> > > Its like saying SQRT(-1) is undefined.
> > >
> > > Then you define i
> > >
> > > Then you have a LARGER NUMBER SET
> > Correct. |{ T F ? } ^ N| >= |{T F} ^ N|
> Lets NOT assume |POWERSET2(N)| > |N|
>
> However |POWERSET2(N)| > | ANY LIST OF R |
>
> This is uninteresting as R is incomplete anyway
>
> R c SR
> |SR| = |N|
> |R| = |N|

let LIST(f(x)) be some list of some or all of the elements of f(x)

POWERSET PROOF
-----------------------------
ALL L
L = LIST( {T F}^N )
->
|L| < | {T F}^N | .........#

HOWEVER

EXIST L
L = LIST( {T F ?} ^N)
^ |L| = | {T F ?} ^ N |

The Triple Powerset can be listed

Since
| {T F}^N | < | {T F ?}^N |

The Powerset Proof can be safely ignored

# - this is not a cardinality inequality merely a statement that L is incomplete

On Thursday, February 16, 2023 at 9:59:20 PM UTC-4, Graham Cooper wrote:
> On Friday, February 17, 2023 at 12:42:41 PM UTC+11, William wrote:
> > On Thursday, February 16, 2023 at 9:01:14 PM UTC-4, Graham Cooper wrote:
> > > On Friday, February 17, 2023 at 11:47:04 AM UTC+11, William wrote:
> > > > On Thursday, February 16, 2023 at 7:09:26 PM UTC-4, Graham Cooper wrote:
> >
> > > > > |N| = |{ T F ? } ^ N| >= |{T F} ^ N| = |R|
> > > > Assume for the sake of argument the incorrect idea that the powerset proof does not apply to (T F ?)^N.
> > > OK
> > > > This would not imply that |(T F ?)^N| was equal to |N| just that we could not use the powerset proof to show the two are not equal.
> > > RIGHT
> > >
> > > The anti-diagonal argument doesnt apply to infinite strings with variables
> > > > The powerset proof still applies to the set {T F} so we have |{ T F ? } ^ N| >= |{T F} ^ N| > |N|
> > > Its like saying SQRT(-1) is undefined.
> > >
> > > Then you define i
> > >
> > > Then you have a LARGER NUMBER SET
> > Correct. |{ T F ? } ^ N| >= |{T F} ^ N|
> Lets NOT assume |POWERSET2(N)| > |N|

We do not have to assume it. It is true.

On Friday, February 17, 2023 at 4:05:03 PM UTC+11, William wrote:
> On Thursday, February 16, 2023 at 9:59:20 PM UTC-4, Graham Cooper wrote:
> > On Friday, February 17, 2023 at 12:42:41 PM UTC+11, William wrote:
> > > On Thursday, February 16, 2023 at 9:01:14 PM UTC-4, Graham Cooper wrote:
> > > > On Friday, February 17, 2023 at 11:47:04 AM UTC+11, William wrote:
> > > > > On Thursday, February 16, 2023 at 7:09:26 PM UTC-4, Graham Cooper wrote:
> > >
> > > > > > |N| = |{ T F ? } ^ N| >= |{T F} ^ N| = |R|
> > > > > Assume for the sake of argument the incorrect idea that the powerset proof does not apply to (T F ?)^N.
> > > > OK
> > > > > This would not imply that |(T F ?)^N| was equal to |N| just that we could not use the powerset proof to show the two are not equal.
> > > > RIGHT
> > > >
> > > > The anti-diagonal argument doesnt apply to infinite strings with variables
> > > > > The powerset proof still applies to the set {T F} so we have |{ T F ? } ^ N| >= |{T F} ^ N| > |N|
> > > > Its like saying SQRT(-1) is undefined.
> > > >
> > > > Then you define i
> > > >
> > > > Then you have a LARGER NUMBER SET
> > > Correct. |{ T F ? } ^ N| >= |{T F} ^ N|
> > Lets NOT assume |POWERSET2(N)| > |N|
> We do not have to assume it. It is true.

Any enumeration of a powerset(N) is incomplete

| POWERSET(N) - LIST(POWERSET(N)) | > 0

Like
| { numbers-divisible-by-3 } - { numbers-divisible-by-6 } | > 0

There is no >|N| implied

| LIST( {0 1}^N ) | < | {0 1}^N |

| LIST( {0 1 ?}^N | = | {0 1 ?}^N |

There is an incompatibility with POWERSET2 and LIST
But thats because {0 1}^N is just a subset of the TOTAL POWERSET {0 1 ?}^N

So all you have proven is a subset of the TOTAL POWERSET is incompletely listable