Jim Burns schrieb am Samstag, 5. November 2022 um 16:07:51 UTC+1:
> On 11/5/2022 8:10 AM, WM wrote:

> >> If you stop, then (your) ℕ is not ℕ
> >
> > I can continue without end
> > but will never remove all successors.
> Yes.
> Each FISON is almost to another FISON.
> Because they're FISONs.
> Each FISON is not the end of FISONs.

That means FISONs cannot exhaust ℕ.

> > But I can remove them collectively.
> If referring to the collectively makes them "dark",
> then there is nothing not-dark.

Referring is possible *only* collectively for dark numbers.
Referring individually to FISONs or natnumbers will never exhaust ℕ..
Referring collectively to natnumbers will exhaust ℕ.
That is the difference.

Regards, WM

After serious thinking Fritz Feldhase wrote :
> On Sunday, November 6, 2022 at 9:50:39 AM UTC+1, FromTheRafters wrote:
>
>> This page includes a good visual representation of both Zermelo and von
>> Neumann constructions, no Frege though.
>>
>> https://www.bookofproofs.org/branches/set-theoretic-definitions-of-natural-numbers-ernst-zermelo-1908-john-von-neumann-1923/
>
> The first formula on that page is nonsense though. (A typo I guess.)

Right, zero should be the emptyset and the set containing the emptyset
is one (its cardinality and deepest nesting depth).

On 11/6/2022 6:03 AM, WM wrote:
> Jim Burns schrieb am Samstag,
> 5. November 2022 um 16:07:51 UTC+1:
>> On 11/5/2022 8:10 AM, WM wrote:

>>> But I can remove them collectively.
>>
>> If referring to the collectively makes them
>> "dark", then there is nothing not-dark.
>
> Referring is possible *only* collectively
> for dark numbers.
> Referring individually to FISONs or natnumbers
> will never exhaust ℕ.
> Referring collectively to natnumbers
> will exhaust ℕ.
> That is the difference.

Describe a FISON, dark or not-dark.
Reason from the description.
Arrive at a new claim.

There is a FISON.
Dark or not-dark, the new claim is true about it.

That's the part I find useful, the new claims.
And nothing you say affects that part.

But you have made me curious.
Are there dark FISONs in WM-world?

>> Each FISON is almost to another FISON.
>> Because they're FISONs.
>> Each FISON is not the end of FISONs.
>
> That means FISONs cannot exhaust ℕ.

A FISON begins at 0 and ends somewhere.
Each split of a FISON has a successor-step
across that split. Dark or not-dark.

That description _exhausts_ the FISONs.
There are no FISONs for which it's false.
Dark or not-dark.

Can you object to any of that?
What would your objection be?
Do you want to call something else a FISON?

But there are plenty of names left over.
I can call things-which-begin-at-0-and-
-end-somewhere-and-each-split-of-which-
-has-a-successor-step FISONs, or counting
sets, or flying rainbow sparkle ponies,
and they will still be things-which-begin-
-at-0-and-end-somewhere-and-each-split-of-
-which-has-a-successor-step.

It seems perfectly pointless to object
to _what we call_ things.
But then again, dark FISONs?
So, who knows?

On 10/8/2022 1:35 PM, Chris M. Thomasson wrote:
> On 10/4/2022 4:15 PM, Fritz Feldhase wrote:
>> On Wednesday, October 5, 2022 at 12:57:43 AM UTC+2, Chris M. Thomasson
>> wrote:
>>> On 10/4/2022 3:49 PM, Fritz Feldhase wrote:
>>>> On Wednesday, October 5, 2022 at 12:39:33 AM UTC+2, Chris M.
>>>> Thomasson wrote:
>>>>>
>>>>> Well, that means there is are no gaps from 0...1, wrt natural
>>>>> numbers, right?
>>>>>
>>>> Define /gap/. :-P
>>>>
>>>> How about: There's a gap between two natural numbers n, m iff there
>>>> is a natural number k such that n < k < m or m < k < n. ?
>>>>
>>>> If there's no gap between two different natural numbers n and m we
>>>> may call them /consecutive/.
>>>>
>>> I see. Lets start with two different natural numbers. So, basically
>>> there is no gap if[f] the absolute value of the difference between two
>>> [...] natural numbers is equal to one?
>>>
>>> 7 and 6, abs(6-7) = 1, no gap.
>>>
>>> Fair enough?
>>
>> Sure - if we define /gap/ [in the present context] this way. :-P
> [...]
>
> Indeed. :^)
>
> One is atomic in the sense that we are artificially restricting
> ourselves to the natural numbers. Well, within this strict context, the
> number one cannot be broken apart for it is the the smallest unit, so to
> speak. ;^)
>

I forgot to explicitly mention that there is another condition where
there is no gap, using the strict natural numbers only... Well, there is
also no gap when the absolute value of the difference between two
natural numbers is zero. This implies that the two natural numbers are
identical. Anything other than zero, implies that the two natural
numbers are different.

So, a = 42, b = 42:

a = abs(b - a) = 0

Well, there is no gap, and a and b are equal because their absolute
difference is zero. The other condition where there is no gap:

a = abs(0 - 1) = 1

There is no gap because the difference is one, aka, the unit granularity
of the natural numbers. Humm... How about:

a = abs(3 - 6) = 3

3 is greater than one, so there is a gap that requires multiple natural
units of 1 to fill in. 3...6: Basically, take 3 to six.

3 + 1 = 4
4 + 1 = 5
5 + 1 = 6

Three units take 3 to 6 because 3 + 3 = 6

Sounding okay so far?

On 9/4/2022 2:46 PM, FromTheRafters wrote:
> Sergio explained :
>> On 9/4/2022 1:10 PM, WM wrote:
>>> Gus Gassmann schrieb am Sonntag, 4. September 2022 um 16:20:32 UTC+2:
>>>> On Sunday, 4 September 2022 at 10:55:28 UTC-3, WM wrote:
>>>> [...]
>>>>> I have this understanding of logic: If two symbols are exchanged,
>>>>> then none of them is deleted.
>>>> This is correct, but insufficient.
>>>
>>> It is all that is needed.
>>
>> wrong,  you are incompetent.
>>
>>>
>>>>>> A In order to prove an inconsistency in ZFC
>>>>>
>>>>> it issufficient to prove that no O can leave the matrix.
>>>> Ah, no. The fact that the reciprocal 1/n is positive for every
>>>> natural number n is insufficient to conclude that the limit of the
>>>> sequence {1, 1/2, 1/3, ..} cannot be 0.
>>>
>>> The limit of a sequence is a number which is approached.
>>
>> insufficient. what do YOU mean by "approached" ?
>
> When an ant becomes a roach, it has been approached.

Indeed. Humm... It looks like a large carpenter ant from 5 feet away.
Approach, now one foot away... Oh shit, its a little roach! Raid... Gas,
nuke the place.

torsdag 3 november 2022 kl. 11:57:56 UTC+1 skrev WM:
> Gus Gassmann schrieb am Mittwoch, 2. November 2022 um 12:16:37 UTC+1:
> > On Wednesday, 2 November 2022 at 06:55:50 UTC-3, WM wrote:
> > > Jim Burns schrieb am Dienstag, 1. November 2022 um 19:15:26 UTC+1:
> > [...]
> > > > A counter-example to
> > > > | All successors have successors.
> > > is [*NOT*] this. ℕ \ 1 \ 2 \ 3 \ ... = { }.
> >
> > You are constantly using unjustified quantifier swaps
> No, I use only justified steps. Try to understand them.
>
> The set ℕ of all natural numbers can be exhausted by subtracting collectively all its elements ℕ \ {1, 2, 3, ...} = Ø.
> But it cannot be exhausted by subtracting only individual elements that have names (like 7 or any fixed substitute for n).
> ∀n ∈ ℕ: ℕ \ {1, 2, 3, ..., n} =/= Ø.
>
> Almost all natural numbers are undefinable.
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> |{1, 2, 3, ..., n}| < ℵo.
>
> Regards, WM

which means your N_def=N!

Jim Burns schrieb am Sonntag, 6. November 2022 um 19:24:51 UTC+1:

> Reason from the description.
> Arrive at a new claim.
>
> There is a FISON.
> Dark or not-dark, the new claim is true about it.
>
> That's the part I find useful, the new claims.
> And nothing you say affects that part.
>
> But you have made me curious.
> Are there dark FISONs in WM-world?

A FISON is not dark because it contains a finite sequence of natnumbers. Every such sequence can be written out, in principle.

> > That means FISONs cannot exhaust ℕ.
> A FISON begins at 0 and ends somewhere.
> Each split of a FISON has a successor-step
> across that split. Dark or not-dark.

That is the hard part to understand. Yes, with every defined n there is n+1, 2n, n^n^n and so on defined. Nevertheless all these numbers will never, never, never complete actual infinity. The reason is that each of such FISONs has ℵo successors |ℕ \ {1, 2, 3, ..., n}| = ℵo. The intersection of a decreasing sequence of infinite inclusion-monotonic sets is not empty.

> That description _exhausts_ the FISONs.

But not the natural numbers.
> There are no FISONs for which it's false.

Yes.
> Dark or not-dark.
>
> Can you object to any of that?
> What would your objection be?
> Do you want to call something else a FISON?

I can show you the existence of dark natural numbers by the essay attached below. There dark numbers or dark limit processes are proved with no doubt. But before I will give you a short outline of the idea:

1) I assume that all natnumbers are existing in the first column of the well-known matrix.
1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ...
...
2) Then I distribute the natnumbers according to Cantor. In no definable step the numbers of X or O change.
3) That means it is not possible in definable steps to do what Cantor thought possible. (If "in the limit" all enumerations happen, then this is a very dark procedure, not a definable bijection.)
4) Now I conclude from the lack of visible matrix positions for the O that there are dark positions.
5) This implies that also the first column has dark parts.
6) I conclude that the natnumbers initially sitting in the first column contain dark elements.

Essay

If all natural numbers k are existing, then they can be used as indices to index the integer fractions m/1 of the first column. Denoting indexed fractions by X and not indexed fractions by O, we obtain the following matrix:

XOOO...
XOOO...
XOOO...
XOOO...
....

Cantor claimed that all natural numbers k are existing and can be applied to index all positive fractions m/n. They are distributed according to k = (m + n - 1)(m + n - 2)/2 + m. The result is a sequence of fractions 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ...

This sequence is modelled here in the language of matrices. The indices are taken from their initial positions in the first column and are distributed in the given order.

Index 1 remains at the first term 1/1. The next term 1/2 is indexed with 2 which is taken from position 2/1.

XXOO...
OOOO...
XOOO...
XOOO...
....

Then the index 3 it taken from 3/1 and attached to 2/1:

XXOO...
XOOO...
OOOO...
XOOO...
....

Then the index 4 it taken from 4/1 and attached to 1/3:

XXXO...
XOOO...
OOOO...
OOOO...
....

And so on. When finally all exchanges of X and O have been carried through and, according to Cantor, all fractions have been enumerated, we get the matrix

XXXX...
XXXX...
XXXX...
XXXX...
....

It turns out that no fraction without index is visible any longer, but by the technique of lossless exchange of X and O no O can have left the matrix. Therefore there are not less fractions without index than at the beginning.. We know that all O and as many fractions without index are remaining, but we cannot find any one.

Where are they? Fractions that cannot be found we call dark. The O are at dark positions. This is the only explanation.

This proof shows that every column has dark positions and therefore also the integer fractions and also the natural numbers have dark elements. Cantor's enumerating concerns only the visible fractions, never all fractions.

Footnote: Some proponents of set theory claim that Cantor's sequence goes without limit but that in case of the matrix-sequence the set theoretical limit has to be applied. But even if this was justified, we come to the conclusion that all the O remain present in all definable matrices until "in the limit" these infinitely many O have to leave in trouble such that no individual leaves in a definable way. This implies that "in the limit" infinitely many fractions have to become indexed such that none of them can be checked - contrary to Cantor's opinion.

Regards, WM

måndag 7 november 2022 kl. 10:58:36 UTC+1 skrev WM:
> Jim Burns schrieb am Sonntag, 6. November 2022 um 19:24:51 UTC+1:
>
> > Reason from the description.
> > Arrive at a new claim.
> >
> > There is a FISON.
> > Dark or not-dark, the new claim is true about it.
> >
> > That's the part I find useful, the new claims.
> > And nothing you say affects that part.
> >
> > But you have made me curious.
> > Are there dark FISONs in WM-world?
> A FISON is not dark because it contains a finite sequence of natnumbers. Every such sequence can be written out, in principle.
> > > That means FISONs cannot exhaust ℕ.
> > A FISON begins at 0 and ends somewhere.
> > Each split of a FISON has a successor-step
> > across that split. Dark or not-dark.
> That is the hard part to understand. Yes, with every defined n there is n+1, 2n, n^n^n and so on defined. Nevertheless all these numbers will never, never, never complete actual infinity. The reason is that each of such FISONs has ℵo successors |ℕ \ {1, 2, 3, ..., n}| = ℵo. The intersection of a decreasing sequence of infinite inclusion-monotonic sets is not empty.
> > That description _exhausts_ the FISONs.
> But not the natural numbers.
> > There are no FISONs for which it's false.
> Yes.
> > Dark or not-dark.
> >
> > Can you object to any of that?
> > What would your objection be?
> > Do you want to call something else a FISON?
> I can show you the existence of dark natural numbers by the essay attached below. There dark numbers or dark limit processes are proved with no doubt. But before I will give you a short outline of the idea:
>
> 1) I assume that all natnumbers are existing in the first column of the well-known matrix.
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...
> 2) Then I distribute the natnumbers according to Cantor. In no definable step the numbers of X or O change.
> 3) That means it is not possible in definable steps to do what Cantor thought possible. (If "in the limit" all enumerations happen, then this is a very dark procedure, not a definable bijection.)
> 4) Now I conclude from the lack of visible matrix positions for the O that there are dark positions.
> 5) This implies that also the first column has dark parts.
> 6) I conclude that the natnumbers initially sitting in the first column contain dark elements.
>
> Essay
>
> If all natural numbers k are existing, then they can be used as indices to index the integer fractions m/1 of the first column. Denoting indexed fractions by X and not indexed fractions by O, we obtain the following matrix:
>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
>
> Cantor claimed that all natural numbers k are existing and can be applied to index all positive fractions m/n. They are distributed according to k = (m + n - 1)(m + n - 2)/2 + m. The result is a sequence of fractions 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ...
>
> This sequence is modelled here in the language of matrices. The indices are taken from their initial positions in the first column and are distributed in the given order.
>
> Index 1 remains at the first term 1/1. The next term 1/2 is indexed with 2 which is taken from position 2/1.
>
> XXOO...
> OOOO...
> XOOO...
> XOOO...
> ...
>
> Then the index 3 it taken from 3/1 and attached to 2/1:
>
> XXOO...
> XOOO...
> OOOO...
> XOOO...
> ...
>
> Then the index 4 it taken from 4/1 and attached to 1/3:
>
> XXXO...
> XOOO...
> OOOO...
> OOOO...
> ...
>
>
> And so on. When finally all exchanges of X and O have been carried through and, according to Cantor, all fractions have been enumerated, we get the matrix
>
> XXXX...
> XXXX...
> XXXX...
> XXXX...
> ...
>
> It turns out that no fraction without index is visible any longer, but by the technique of lossless exchange of X and O no O can have left the matrix. Therefore there are not less fractions without index than at the beginning. We know that all O and as many fractions without index are remaining, but we cannot find any one.
>
> Where are they? Fractions that cannot be found we call dark. The O are at dark positions. This is the only explanation.
>
> This proof shows that every column has dark positions and therefore also the integer fractions and also the natural numbers have dark elements. Cantor's enumerating concerns only the visible fractions, never all fractions.
>
> Footnote: Some proponents of set theory claim that Cantor's sequence goes without limit but that in case of the matrix-sequence the set theoretical limit has to be applied. But even if this was justified, we come to the conclusion that all the O remain present in all definable matrices until "in the limit" these infinitely many O have to leave in trouble such that no individual leaves in a definable way. This implies that "in the limit" infinitely many fractions have to become indexed such that none of them can be checked - contrary to Cantor's opinion.
>
> Regards, WM

None of that proves anythign "dark"

On Monday, 7 November 2022 at 07:37:16 UTC-4, zelos...@gmail.com wrote:
> [...]
> None of that proves anythign "dark"

Oh, I'd say that this is ample proof for one dark thing: Muckenheim's mind. Thought is no longer possible for him.

WM formulated on Monday :
> Jim Burns schrieb am Sonntag, 6. November 2022 um 19:24:51 UTC+1:
>
>> Reason from the description.
>> Arrive at a new claim.
>>
>> There is a FISON.
>> Dark or not-dark, the new claim is true about it.
>>
>> That's the part I find useful, the new claims.
>> And nothing you say affects that part.
>>
>> But you have made me curious.
>> Are there dark FISONs in WM-world?
>
> A FISON is not dark because it contains a finite sequence of natnumbers.
> Every such sequence can be written out, in principle.
>
>>> That means FISONs cannot exhaust ℕ.
>> A FISON begins at 0 and ends somewhere.
>> Each split of a FISON has a successor-step
>> across that split. Dark or not-dark.
>
> That is the hard part to understand. Yes, with every defined n there is n+1,
> 2n, n^n^n and so on defined.

With a 'defined n' and when there is n+1 and induction, there is no
need to define 2n, n^n^n and so on because there are already defined.

That's why your various schemes always end up meaning N_def = |N

On 11/6/2022 2:25 PM, Chris M. Thomasson wrote:
> On 9/4/2022 2:46 PM, FromTheRafters wrote:
>> Sergio explained :
>>> On 9/4/2022 1:10 PM, WM wrote:
>>>> Gus Gassmann schrieb am Sonntag, 4. September 2022 um 16:20:32 UTC+2:
>>>>> On Sunday, 4 September 2022 at 10:55:28 UTC-3, WM wrote:
>>>>> [...]
>>>>>> I have this understanding of logic: If two symbols are exchanged, then none of them is deleted.
>>>>> This is correct, but insufficient.
>>>>
>>>> It is all that is needed.
>>>
>>> wrong,  you are incompetent.
>>>
>>>>
>>>>>>> A In order to prove an inconsistency in ZFC
>>>>>>
>>>>>> it issufficient to prove that no O can leave the matrix.
>>>>> Ah, no. The fact that the reciprocal 1/n is positive for every natural number n is insufficient to conclude that the limit of the sequence {1, 1/2,
>>>>> 1/3, ..} cannot be 0.
>>>>
>>>> The limit of a sequence is a number which is approached.
>>>
>>> insufficient. what do YOU mean by "approached" ?
>>
>> When an ant becomes a roach, it has been approached.
>
> Indeed. Humm... It looks like a large carpenter ant from 5 feet away. Approach, now one foot away... Oh shit, its a little roach! Raid... Gas, nuke the
> place.

when you find a "dark Ant in the spaghetti, you gotta throw the whole bowl out"

On 11/7/2022 4:56 AM, WM wrote:
> Jim Burns schrieb am Sonntag,
> 6. November 2022 um 19:24:51 UTC+1:
>> On 11/6/2022 6:03 AM, WM wrote:
>>> Jim Burns schrieb am Samstag,
>>> 5. November 2022 um 16:07:51 UTC+1:
>>>> On 11/5/2022 8:10 AM, WM wrote:

>>>>> But I can remove them collectively.
>>>>
>>>> If referring to the collectively makes them
>>>> "dark", then there is nothing not-dark.
>>>
>>> Referring is possible *only* collectively
>>> for dark numbers.
>>> Referring individually to FISONs or natnumbers
>>> will never exhaust ℕ.
>>> Referring collectively to natnumbers
>>> will exhaust ℕ.
>>> That is the difference.

>> But you have made me curious.
>> Are there dark FISONs in WM-world?
>
> A FISON is not dark because
> it contains a finite sequence of natnumbers.
> Every such sequence can be written out,
> in principle.

Irrelevant.
You:
>>> Referring is possible *only* collectively
>>> for dark numbers.
>>> That is the difference.

It is unclear
how "dark" applies to _collections_ of
_individuals_ but you give an example.

>>> Referring individually to FISONs or natnumbers
>>> will never exhaust ℕ.
>>> Referring collectively to natnumbers
>>> will exhaust ℕ.

Who knows why you didn't mention that
referring collectively to FISONs
will ALSO exhaust ℕ ?

FISONs collectively exhaust the correct ℕ
the ℕ to which our theorems refer,
the ℕ which supersets each FISON and
which subsets anything else which
supersets each FISON.
∀𝐹 : 𝐹₀⋮⌝ ⟹ 𝐹 ⊆ ℕ
(∀𝐹 : 𝐹₀⋮⌝ ⟹ 𝐹 ⊆ 𝐵) ⟹ ℕ ⊆ 𝐵

𝐹₀⋮⌝ == "𝐹 is a FISON"

You have something else, ℕᴹᵂ, which FISONs
do not exhaust individually or collectively.
The key to darkness, though, is
>>> Referring is possible *only* collectively
>>> for dark numbers.

Our ℕ is enough to prove there are dark FISONs
using the method you use to prove there are
dark natural numbers.

The existence of dark FISONs changes no part
of what we're doing with our ℕ

Describe a FISON, dark or not-dark.
Reason from the description.
Arrive at a new claim.

There is a FISON, dark or not-dark.
Dark or not-dark, the new claim is true of it.

That's the part we find useful, the new claims.
And nothing you say affects that part.

> The intersection of a decreasing sequence of
> infinite inclusion-monotonic sets is not empty.

No,
consider an infinite sequence ⟨𝑆⟩ of
disjoint non-∅ sets
𝑆 ≠ 𝑆₂ ⟹ 𝑆∩𝑆₂ = ∅

One example of ⟨𝑆⟩ would be
⟨ {0} {1} {2} {3} ⋯ ⟩

⟨𝑆⟩ is a _sequence_
There is a first set 𝑆₀ in ⟨𝑆⟩
For each 𝑆≠𝑆₀ in ⟨𝑆⟩ a predecessor 𝑆⁻⁻ in ⟨𝑆⟩
For each 𝑆 in ⟨𝑆⟩ a successor 𝑆⁺⁺ in ⟨𝑆⟩

Next, consider the sequence ⟨𝐸(𝑆)⟩ of
end segments of ⟨𝑆⟩
𝐸(𝑆) = ⋃[𝑆≼𝑆₂] 𝑆₂

∀𝑆₂ ∈ ⟨𝑆⟩ : 𝑆≼𝑆₂ ⟹ 𝑆₂ ⊆ 𝐸(𝑆)
(∀𝑆₂ ∈ ⟨𝑆⟩ : 𝑆≼𝑆₂ ⟹ 𝑆₂ ⊆ 𝐵) ⟹ 𝐸(𝑆) ⊆ 𝐵

Each 𝐸(𝑆) in ⟨𝐸(𝑆)⟩ is infinite.
It has at least one element from
each of infinitely-many 𝑆₂, 𝑆≼𝑆₂

The sequence ⟨𝐸(𝑆)⟩ is inclusion-monotonic.

The intersection ⋂⟨𝐸(𝑆)⟩ is empty.
| | Assume otherwise.
| Assume k ∈ ⋂⟨𝐸(𝑆)⟩
| | k ∈ 𝐸(𝑆) for some 𝑆 ∈ ⟨𝑆⟩
| 𝐸(𝑆) = ⋃[𝑆≼𝑆₂] 𝑆₂
| k ∈ 𝑆₂ for some 𝑆₂ ≽ 𝑆
| The sets in ⟨𝑆⟩ are disjoint.
| For each 𝑆₃ ≽ 𝑆₂⁺⁺, k ∉ 𝑆₃
| 𝐸(𝑆₂⁺⁺) = ⋃[𝑆₂⁺⁺≼𝑆₃] 𝑆₃
| k ∉ 𝐸(𝑆₂⁺⁺)
| k ∉ ⋂⟨𝐸(𝑆)⟩
| Contradiction.

Therefore, ⋂⟨𝐸(𝑆)⟩ = ∅

>> That description _exhausts_ the FISONs.
>
> But not the natural numbers.

No.
The natural numbers are the FISON-ends.

For each FISON, there is a unique FISON-end.
For each FISON-end, there is a unique FISON.

Exhausting the FISONs exhausts the FISON-ends.

On 11/7/2022 8:43 AM, Sergi o wrote:
> On 11/6/2022 2:25 PM, Chris M. Thomasson wrote:
>> On 9/4/2022 2:46 PM, FromTheRafters wrote:
>>> Sergio explained :
>>>> On 9/4/2022 1:10 PM, WM wrote:
>>>>> Gus Gassmann schrieb am Sonntag, 4. September 2022 um 16:20:32 UTC+2:
>>>>>> On Sunday, 4 September 2022 at 10:55:28 UTC-3, WM wrote:
>>>>>> [...]
>>>>>>> I have this understanding of logic: If two symbols are exchanged,
>>>>>>> then none of them is deleted.
>>>>>> This is correct, but insufficient.
>>>>>
>>>>> It is all that is needed.
>>>>
>>>> wrong,  you are incompetent.
>>>>
>>>>>
>>>>>>>> A In order to prove an inconsistency in ZFC
>>>>>>>
>>>>>>> it issufficient to prove that no O can leave the matrix.
>>>>>> Ah, no. The fact that the reciprocal 1/n is positive for every
>>>>>> natural number n is insufficient to conclude that the limit of the
>>>>>> sequence {1, 1/2, 1/3, ..} cannot be 0.
>>>>>
>>>>> The limit of a sequence is a number which is approached.
>>>>
>>>> insufficient. what do YOU mean by "approached" ?
>>>
>>> When an ant becomes a roach, it has been approached.
>>
>> Indeed. Humm... It looks like a large carpenter ant from 5 feet away.
>> Approach, now one foot away... Oh shit, its a little roach! Raid...
>> Gas, nuke the place.
>
>
>
> when you find a "dark Ant in the spaghetti, you gotta throw the whole
> bowl out"

:^D

Actually, one time I made a nice potato salad for the fourth of July. As
soon as I opened it, a god damn fly landed right in the middle of it.
That little bastard! Grrrrrrr...

On 11/7/2022 3:54 PM, Chris M. Thomasson wrote:
> On 11/7/2022 8:43 AM, Sergi o wrote:
>> On 11/6/2022 2:25 PM, Chris M. Thomasson wrote:
>>> On 9/4/2022 2:46 PM, FromTheRafters wrote:
>>>> Sergio explained :
>>>>> On 9/4/2022 1:10 PM, WM wrote:
>>>>>> Gus Gassmann schrieb am Sonntag, 4. September 2022 um 16:20:32 UTC+2:
>>>>>>> On Sunday, 4 September 2022 at 10:55:28 UTC-3, WM wrote:
>>>>>>> [...]
>>>>>>>> I have this understanding of logic: If two symbols are exchanged, then none of them is deleted.
>>>>>>> This is correct, but insufficient.
>>>>>>
>>>>>> It is all that is needed.
>>>>>
>>>>> wrong,  you are incompetent.
>>>>>
>>>>>>
>>>>>>>>> A In order to prove an inconsistency in ZFC
>>>>>>>>
>>>>>>>> it issufficient to prove that no O can leave the matrix.
>>>>>>> Ah, no. The fact that the reciprocal 1/n is positive for every natural number n is insufficient to conclude that the limit of the sequence {1,
>>>>>>> 1/2, 1/3, ..} cannot be 0.
>>>>>>
>>>>>> The limit of a sequence is a number which is approached.
>>>>>
>>>>> insufficient. what do YOU mean by "approached" ?
>>>>
>>>> When an ant becomes a roach, it has been approached.
>>>
>>> Indeed. Humm... It looks like a large carpenter ant from 5 feet away. Approach, now one foot away... Oh shit, its a little roach! Raid... Gas, nuke
>>> the place.
>>
>>
>>
>> when you find a "dark Ant in the spaghetti, you gotta throw the whole bowl out"
>
> :^D
>
> Actually, one time I made a nice potato salad for the fourth of July. As soon as I opened it, a god damn fly landed right in the middle of it. That
> little bastard! Grrrrrrr...

I think that quote above came from OJ Simpson trial, helped the jury let him go, one of his lawyers, Kardician, with "cockroach" in place of "dark ant"
but I could not find it using google, so I think Google is actively reconfiguring the past...

Sergi o formulated the question :
> On 11/7/2022 3:54 PM, Chris M. Thomasson wrote:
>> On 11/7/2022 8:43 AM, Sergi o wrote:
>>> On 11/6/2022 2:25 PM, Chris M. Thomasson wrote:
>>>> On 9/4/2022 2:46 PM, FromTheRafters wrote:
>>>>> Sergio explained :
>>>>>> On 9/4/2022 1:10 PM, WM wrote:
>>>>>>> Gus Gassmann schrieb am Sonntag, 4. September 2022 um 16:20:32 UTC+2:
>>>>>>>> On Sunday, 4 September 2022 at 10:55:28 UTC-3, WM wrote:
>>>>>>>> [...]
>>>>>>>>> I have this understanding of logic: If two symbols are exchanged,
>>>>>>>>> then none of them is deleted.
>>>>>>>> This is correct, but insufficient.
>>>>>>>
>>>>>>> It is all that is needed.
>>>>>>
>>>>>> wrong,  you are incompetent.
>>>>>>
>>>>>>>
>>>>>>>>>> A In order to prove an inconsistency in ZFC
>>>>>>>>>
>>>>>>>>> it issufficient to prove that no O can leave the matrix.
>>>>>>>> Ah, no. The fact that the reciprocal 1/n is positive for every
>>>>>>>> natural number n is insufficient to conclude that the limit of the
>>>>>>>> sequence {1, 1/2, 1/3, ..} cannot be 0.
>>>>>>>
>>>>>>> The limit of a sequence is a number which is approached.
>>>>>>
>>>>>> insufficient. what do YOU mean by "approached" ?
>>>>>
>>>>> When an ant becomes a roach, it has been approached.
>>>>
>>>> Indeed. Humm... It looks like a large carpenter ant from 5 feet away.
>>>> Approach, now one foot away... Oh shit, its a little roach! Raid... Gas,
>>>> nuke the place.
>>>
>>>
>>>
>>> when you find a "dark Ant in the spaghetti, you gotta throw the whole bowl
>>> out"
>>
>> :^D
>>
>> Actually, one time I made a nice potato salad for the fourth of July. As
>> soon as I opened it, a god damn fly landed right in the middle of it. That
>> little bastard! Grrrrrrr...
>
> I think that quote above came from OJ Simpson trial, helped the jury let him
> go, one of his lawyers, Kardician, with "cockroach" in place of "dark ant"
> but I could not find it using google, so I think Google is actively
> reconfiguring the past...

Perhaps looking for something akin to 'if it doesn't fit, you must
acquit' will work better. I think it was in reference to the single
glove they found.

zelos...@gmail.com schrieb am Montag, 7. November 2022 um 12:37:16 UTC+1:

> None of that proves anythign "dark"

Where are the O's?

Regards, WM

Gus Gassmann schrieb am Montag, 7. November 2022 um 12:59:08 UTC+1:
> On Monday, 7 November 2022 at 07:37:16 UTC-4, zelos...@gmail.com wrote:
> > [...]
> > None of that proves anythign "dark"
> Oh, I'd say that this is ample proof for one dark thing:

Any limit desired?

Regards, WM

FromTheRafters schrieb am Montag, 7. November 2022 um 14:49:18 UTC+1:
> WM formulated on Monday :

> > That is the hard part to understand. Yes, with every defined n there is n+1,
> > 2n, n^n^n and so on defined.
> With a 'defined n' and when there is n+1 and induction, there is no
> need to define 2n, n^n^n and so on because there are already defined.
>
> That's why your various schemes always end up meaning N_def = |N

That's why it's not:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ ℕ_def| = ℵo .

Regards, WM

Jim Burns schrieb am Montag, 7. November 2022 um 21:16:05 UTC+1:
> On 11/7/2022 4:56 AM, WM wrote:
> >> That description _exhausts_ the FISONs.
> >
> > But not the natural numbers.
> No.
> The natural numbers are the FISON-ends.

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ ℕ_def| = ℵo .

> referring collectively to FISONs
> will ALSO exhaust ℕ ?

No. Dark means that *only* collectively mentioning is possible.

> FISONs collectively exhaust the correct ℕ
ℕ_def
>the ℕ to which our theorems refer,
>the ℕ which supersets each FISON

That is the ℕ_def with only individually definable elements. But it is not the ℕ of Cantor's.

> For each FISON, there is a unique FISON-end.
> For each FISON-end, there is a unique FISON.
>
> Exhausting the FISONs exhausts the FISON-ends.

Yes, but the natnumbers of Cantor's ℕ are more.

Regards, WM

On Tuesday, November 8, 2022 at 12:35:55 PM UTC+1, WM wrote:
> FromTheRafters schrieb am Montag, 7. November 2022 um 14:49:18 UTC+1:
> >
> > That's why your various schemes always end up meaning N_def = |N
> >
> That's why it's not:
>
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> |ℕ \ ℕ_def| = ℵo .

IN_def is not defined this way. (Hint: Both equations hold for IN_def = { } or IN_def = {4711}, etc. Actually, they hold for ANY set IN_def c IN.)

So let's DEFINE it:

IN_def := {n e IN : |IN \ {1, 2, 3, ..., n}| = ℵo}.

From this we get: IN_def = IN.

tisdag 8 november 2022 kl. 12:32:32 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 7. November 2022 um 12:37:16 UTC+1:
>
> > None of that proves anythign "dark"
> Where are the O's?
>
> Regards, WM
Those do not matter because thati s not how the bijection is constructed!

tisdag 8 november 2022 kl. 12:35:55 UTC+1 skrev WM:
> FromTheRafters schrieb am Montag, 7. November 2022 um 14:49:18 UTC+1:
> > WM formulated on Monday :
> > > That is the hard part to understand. Yes, with every defined n there is n+1,
> > > 2n, n^n^n and so on defined.
> > With a 'defined n' and when there is n+1 and induction, there is no
> > need to define 2n, n^n^n and so on because there are already defined.
> >
> > That's why your various schemes always end up meaning N_def = |N
> That's why it's not:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> |ℕ \ ℕ_def| = ℵo .
>
> Regards, WM
except
|ℕ \ ℕ_def| = ℵo .
Is false, N_def=N

tisdag 8 november 2022 kl. 12:47:14 UTC+1 skrev WM:
> Jim Burns schrieb am Montag, 7. November 2022 um 21:16:05 UTC+1:
> > On 11/7/2022 4:56 AM, WM wrote:
> > >> That description _exhausts_ the FISONs.
> > >
> > > But not the natural numbers.
> > No.
> > The natural numbers are the FISON-ends.
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> |ℕ \ ℕ_def| = ℵo .
> > referring collectively to FISONs
> > will ALSO exhaust ℕ ?
> No. Dark means that *only* collectively mentioning is possible.

which is meaningless.

> > FISONs collectively exhaust the correct ℕ
> ℕ_def

=N

> >the ℕ to which our theorems refer,
> >the ℕ which supersets each FISON
> That is the ℕ_def with only individually definable elements. But it is not the ℕ of Cantor's.
> > For each FISON, there is a unique FISON-end.
> > For each FISON-end, there is a unique FISON.
> >
> > Exhausting the FISONs exhausts the FISON-ends.
> Yes, but the natnumbers of Cantor's ℕ are more.

Nope

>
> Regards, WM

On Tuesday, November 8, 2022 at 12:35:55 PM UTC+1, WM wrote:
> FromTheRafters schrieb am Montag, 7. November 2022 um 14:49:18 UTC+1:
> >
> > That's why your various schemes always end up meaning N_def = |N
> >
> That's why it's not:
>
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> |ℕ \ ℕ_def| = ℵo .

IN_def is not defined this way. (Hint: Both equations hold for IN_def = { } or IN_def = {4711}, etc. Actually, they hold for ANY _finte_ set IN_def c IN.)

So let's DEFINE it:

IN_def := {n e IN : |IN \ {1, 2, 3, ..., n}| = ℵo}.

From this we get: IN_def = IN.

This means: |ℕ \ ℕ_def| = ℵo does not hold.

On Tuesday, 8 November 2022 at 07:33:22 UTC-4, WM wrote:
> Gus Gassmann schrieb am Montag, 7. November 2022 um 12:59:08 UTC+1:
> > On Monday, 7 November 2022 at 07:37:16 UTC-4, zelos...@gmail.com wrote:
> > > [...]
> > > None of that proves anythign "dark"
> > Oh, I'd say that this is ample proof for one dark thing:
> Any limit desired?

No, but it would help if you could read. But when 0 is an upper bound on the number of neurons still firing in your brain, I know that even reading comprehension is too much to ask from you.