On 7/2/2022 11:34 AM, WM wrote:
> zelos...@gmail.com schrieb am Samstag, 2. Juli 2022 um 17:34:28 UTC+2:
>> torsdag 30 juni 2022 kl. 14:54:29 UTC+1 skrev WM:
>>> William schrieb am Donnerstag, 30. Juni 2022 um 01:25:26 UTC+2:
>>>> |N_def is a set. A set does not change.
>>> ℕ_def is many sets, depending on what elements are defined. It is a collection.
>> no, it is A set, a singular set that never changes. And N is a subset of it.
>>
>>> Compare the set of known prime numbers. It is a set now but another set next year. It is a potentially infinite collection.
>>
>> Set P_Y2022 and set P_Y2023 may very well be different sets, but they are DIFFERENT SETS STILL!
>>
>> It is not A set that changes!
>
> It is what we call a potentially infinite collection.
>>
>>>
>>> Further, if ℕ was endless, then ∩{E(k) : k ∈ ℕ} could not be empty, because every endsegment which is not the last one cannot make the intersection empty because there is always another endsegment reducing the intersection further. This would not be possible if the intersection was empty already.
>>
>> this is false on its own in so many ways.
>
> It is fact.
>>
>> There is no step by step process. The intersection asks a simple question, which element is in all the sets given?
>
> For all definable numbers k we can obtain the process step by step and find the correct result:
> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
>
>> Because NO element is that, the intersection has NO elements.
>
> Correct.
> ∩{E(k) : k ∈ ℕ} = { }
> But for all definable elements we obtain
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀

wrong, there is no "_def", and WM keeps changing its meaning.

>
> If the rest is no step by step process, then the only reason can be dark numbers.

wrong.

you want your idea of dark numbers to be true, but it is not, you just cannot do real math.

>
> Regards, WM

On 7/2/2022 11:36 AM, WM wrote:
> sergi o schrieb am Samstag, 2. Juli 2022 um 18:01:13 UTC+2:
>
>> if it is impossible to prove a number is a dark number, there are no dark numbers.
>
> It is possible to prove their existence. Here is a proof (one of many):
> For all definable numbers we have the intersection of their endsegments
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀

ℕ_def is undefined.

>
> For all natural numbers, we have the empty intersection of their endsegments:
> ∩{E(k) : k ∈ ℕ} = { }

for k = 1 to oo, then yes

>
> The endsegments can only shrink by on number per endsegment
> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}

sets do not shrink, sets are always fixed size.

the intersection is not an endsegment!

you will never get there by deleting elements from an endsegment to a non-endsegment!

>

>
> Regards, WM

On 7/2/2022 12:55 PM, WM wrote:
> FromTheRafters schrieb am Samstag, 2. Juli 2022 um 19:48:56 UTC+2:
>> WM submitted this idea :
>
>>> For all definable numbers we have the intersection of their endsegments
>>>> ∩{E(k) : k ∈ ℕ_def}| = ℵ₀
>> So, definable just means finite?
>
> When applied to natural numbers, of course. It means that a FISON {1, 2, 3, ..., n} exists.
>
> Regards, WM

so ℕ_def is a FISON ?

On 7/2/2022 2:06 PM, Gus Gassmann wrote:
> On Saturday, 2 July 2022 at 11:49:49 UTC-3, WM wrote:
>> Gus Gassmann schrieb am Freitag, 1. Juli 2022 um 15:54:35 UTC+2:
>>> On Friday, 1 July 2022 at 09:49:34 UTC-3, FromTheRafters wrote:
>>>> WM wrote on 7/1/2022 :
>>> [...]
>>>>> Every k∈|N has ℵo not yet distinguished successors.
>>>> Try proving that without using induction.
>>> Even more to the point, try proving it if n is a "dark number".
>> Impossible.
>
> Good. So now we have
>
> 1. There are infinitely many natural numbers.
> 2. The natural numbers under the standard order are well-ordered.
> 3. There are infinitely many dark numbers and finitely many "distinguished natural numbers".
> 4. Every natural number has ℵo not yet distinguished successors.
> 5. Dark numbers are natural numbers, too.
> 6. Every dark natural number has ℵo not yet distinguished successors.
> 7. Statement 6 can not be proven (at least not by the great perfosser).
>
> Please get lost, you confused, lying piece of shit.

[IAW WM, the above need to be "identified", then "accessed" before use}

On 7/2/2022 12:27 PM, WM wrote:
> William schrieb am Samstag, 2. Juli 2022 um 19:06:25 UTC+2:
>> On Saturday, July 2, 2022 at 1:25:15 PM UTC-3, WM wrote:
>
>>> ... natural numbers reduce the intersection from ℵ₀ to zero?
>>
>> Any infinite subset.
>
> No, only endsegments can do it, one by one:
> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
> This rule must not be violated.

it is not a "rule", it is an "equivalence". Google =

>>
>>> They cannot be distinguished.
>>
>> So what? The fact than they cannot be written down does not mean they do not have other properties,
>
> It means that the endsegments reducing
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> to
> ∩{E(k) : k ∈ ℕ} = { }
> are dark.

your "darkness" is because you are in the dark, when it comes to math.

>
> Regards, WM

this is now well below stupid.

On 7/2/2022 10:49 AM, WM wrote:
> Gus Gassmann schrieb
> am Freitag, 1. Juli 2022 um 15:54:35 UTC+2:
>> On Friday, 1 July 2022 at 09:49:34 UTC-3,
>> FromTheRafters wrote:
>>> WM wrote on 7/1/2022 :

>>>> Every k∈|N has ℵo not yet distinguished successors.
>>>
>>> Try proving that without using induction.
>>
>> Even more to the point, try proving it if n is
>> a "dark number".
>
> Impossible.

Impossible by design.
You want to pontificate.
You don't want to work.
You invent "dark numbers" defined to be
things you can pontificate about with no work.
QED

Occasionally, you let drop a crumb of information
from which we can draw conclusions.
Those are the times you will exert yourself,
in order to sweep away those crumbs.
Otherwise, you only need to know how to type "Impossible".

On 7/2/2022 12:42 PM, WM wrote:
> Jim Burns schrieb
> am Freitag, 1. Juli 2022 um 18:16:23 UTC+2:

>> The claims we make about k ∈ ℕ are as unaffected by
>> the status, distinguished-or-not, of any k ∈ ℕ
>
> For all definable numbers

We can make descriptive claims of k ∈ ℕ.
They will be true of each k ∈ ℕ, definable or not.

We can extend the claims we make, allowing only
claims that must be true when earlier claims are true.
Because we start with claims that are true of each
k ∈ ℕ, definable or not, later claims are
true of each k ∈ ℕ, definable or not.

The claims we make about k ∈ ℕ are as unaffected by
the status, definable-or-not, of any k ∈ ℕ
as they are unaffected by whether any k ∈ ℕ tastes
like chocolate or collects bubble-gum cards.

I don't know how to answer, but it doesn't matter.

> For all definable numbers we obtain
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀

No, you don't obtain that.

That definition is ill-formed.
If we correct your definition to what you say,
then you obtain |∩{E(k) : k ∈ ℕ_def}| = 0

> If the step by step process is denied,
> then the only reason can be dark numbers.

For each split of ℕ,
a step i,i+1 exists which
starts before the split and ends after the split.

As a consequence,
each element of ℕ can be reached by
a terminating process of '+1' steps.

However,
each element i of ℕ is followed by i+1.

As a consequence,
no terminating process of '+1' steps,
no matter where it terminates,
reaches each element of ℕ