Jim Burns schrieb am Samstag, 14. August 2021 um 20:58:59 UTC+2:
> On 8/14/2021 10:20 AM, WM wrote:

> > Therefore all endsegments satisfying
> > ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
> > can be combined to
> > |∩{E(k) : k ∈ ℕ_def}| = ℵo .

> The empty set is the intersection of all infinite end segments.

Nionsense. Endsegments are inclusion monotonic. A set of infinite endsegments has an infinite intersection.

> > |∩{E(k) : k ∈ ℕ_def}| = ℵo .
> No.

You may be qualified for mathematical paralympics.

The intersection depends only on the endsegments. If those endsegments which satisfy (*) can form an infinite set, their intersection is infinite. If they can't, then ℕ_def =/= ℕ.

Regards, WM

William schrieb am Samstag, 14. August 2021 um 20:25:42 UTC+2:
> On Saturday, August 14, 2021 at 1:28:08 PM UTC-4, WM wrote:
> > William schrieb am Samstag, 14. August 2021 um 17:09:13 UTC+2:
>
> > > Piffle. If every element of |N_F is used then no element of |N_F is not used;
> > Wrong.
> ?!

If every element of |N_F is used most elements are not used, because every element belongs to a finite initial segment followed by aleph_0 elements.

All elements are not followed by aleph_0 elements. If every element is used, then aleph_0 elements remain. If all elements are used, then none remains.

Regards, WM

On 8/11/2021 12:55 PM, Sergio wrote:
> On 8/11/2021 2:46 PM, WM wrote:
>> William schrieb am Mittwoch, 11. August 2021 um 16:50:30 UTC+2:
>>> On Wednesday, August 11, 2021 at 9:57:42 AM UTC-4, WM wrote:
>>>> William schrieb am Dienstag, 10. August 2021 um 15:14:35 UTC+2:
>>>>> On Tuesday, August 10, 2021 at 5:35:20 AM UTC-4, WM wrote:
>>>>>> William schrieb am Montag, 9. August 2021 um 17:55:55 UTC+2:
>>>>>
>>>>>>> (Note, every element of |N_F is finite and followed by almost all elements of |N_F.)
>>>>>> Every definable index, yes
>>>>> Indeed, every index you can write down, and every index you cannot write down too.
>>>>>> But all indexes are not followed by indexes:
>>>>> Another use of the ambiguous "all".
>>>> It isn't ambiguous
>>> He said staring at two different meanings of the phrase " All indexes are not followed by indexes"
>>
>> There is only one meaning here. All indexes are used up in the bijection of indexes and endsegments. No additional natural numbers are available as contents of endsegments.
>
>
> you ran out of natural numbers ? you can "re-use" them you know.
>
> In case you run out, here are some extras 1's 1 1 1 1 1 1 1 1 1 1 1
> 2's 2 2 2 2 2 2 2 2 2 2 2 2
> 3's 3 3 3 3 3 3 3 3 3

LOL!

How can one run out of an infinite source?

On Saturday, August 14, 2021 at 1:23:39 PM UTC-7, Ross A. Finlayson wrote:
> On Saturday, August 14, 2021 at 12:46:56 PM UTC-7, Jim Burns wrote:
> > On 8/14/2021 1:23 PM, WM wrote:
> > > Jim Burns schrieb
> > > am Samstag, 14. August 2021 um 19:02:48 UTC+2:
> > >> On 8/14/2021 10:29 AM, WM wrote:
> >
> > >>> There is only one important point:
> > >>> You must not confuse pot. and act.
> > >>> as set theorists deplorably deliberately do.
> > >>
> > >> Set theorists do not confuse [changing] last endsegments
> > >> with [unchanging] last end segments.
> > >
> > > They do.
> > You do.
> > <WM<JB>>
> > >>
> > >> WM rejects No Last End Segment.
> > >
> > > No, but I reject that last endsegments can yield
> > > a set with no last endsegment -
> > > other than potentially infinite.
> > > But that has always a last one.
> > >
> > </WM<JB>>
> > > Every endsegment which can be last is insufficient for
> > > an actual infinity.
> > Each end segment can be described with one description..
> > Having been described, it can be reasoned about.
> >
> > Depending upon your mood, I suppose, you call this either
> > potential infinity or actual infinity.
> > Whatever you call it, we can reason about _all_ of
> > those infinitely many individuals.
> > > Set theorists however claim that there are no others and
> > > that actual infinit exists. That is wrong.
> > I claim that we can reason about each of some collection
> > of described individuals, and that we can do this whether
> > or not there is a last one of these individuals, or whether
> > or not it is possible to step to one-by-one each of these
> > individuals, and that we can reason about them WITHOUT our
> > claims applying to other individuals we have NOT described.
> I like a theory where "ubiquitous ordinals" make for
> order type is powerset is successor.
>
> When we reason about 0..oo it's in a sense same
> as oo..0 with the notion what "it's difference what
> implements counting".
>
> The usual "sputnik of quantification" like "Russell's
> element or Burali-Forti's ordinal and so on", point to
> why it's necessary for a rapprochement in foundations,
> basically to include the "extra-standard".
>
> Then as for whether a standard or ground model of
> the integers even exists, is that bounded fragments
> and extensions do.
>
> I spent a lot of time studying foundations
> mostly because it's illuminating. Also because
> it's difficult in comprehension and strong exercise.
>
> ... That then it becomes ingrained and free ...,
> that foundations is foundations is foundations.
>
> Mark my words: mark my words.
>
> Line-drawing: marked by word.

Also because it's simple then to establish mathematics usually
the definition and derivation, good throughout
"Hilbert's Infinite Living Museum, of mathematics".

On Saturday, August 14, 2021 at 5:45:59 PM UTC-4, WM wrote:
> William schrieb am Samstag, 14. August 2021 um 20:25:42 UTC+2:
> > On Saturday, August 14, 2021 at 1:28:08 PM UTC-4, WM wrote:
> > > William schrieb am Samstag, 14. August 2021 um 17:09:13 UTC+2:
> >
> > > > Piffle. If every element of |N_F is used then no element of |N_F is not used;
> > > Wrong.
> > ?!
> If every element of |N_F is used most elements are not used,
?!

--
William Hughes

On 8/14/2021 4:48 PM, Chris M. Thomasson wrote:
> On 8/11/2021 12:55 PM, Sergio wrote:
>> On 8/11/2021 2:46 PM, WM wrote:
>>> William schrieb am Mittwoch, 11. August 2021 um 16:50:30 UTC+2:
>>>> On Wednesday, August 11, 2021 at 9:57:42 AM UTC-4, WM wrote:
>>>>> William schrieb am Dienstag, 10. August 2021 um 15:14:35 UTC+2:
>>>>>> On Tuesday, August 10, 2021 at 5:35:20 AM UTC-4, WM wrote:
>>>>>>> William schrieb am Montag, 9. August 2021 um 17:55:55 UTC+2:
>>>>>>
>>>>>>>> (Note, every element of |N_F is finite and followed by almost
>>>>>>>> all elements of |N_F.)
>>>>>>> Every definable index, yes
>>>>>> Indeed, every index you can write down, and every index you cannot
>>>>>> write down too.
>>>>>>> But all indexes are not followed by indexes:
>>>>>> Another use of the ambiguous "all".
>>>>> It isn't ambiguous
>>>> He said staring at two different meanings of the phrase " All
>>>> indexes are not followed by indexes"
>>>
>>> There is only one meaning here. All indexes are used up in the
>>> bijection of indexes and endsegments. No additional natural numbers
>>> are available as contents of endsegments.
>>
>>
>> you ran out of natural numbers ?  you can "re-use" them you know.
>>
>> In case you run out, here are some extras 1's  1 1 1 1  1 1  1 1  1 1 1
>> 2's  2 2  2 2 2  2 2  2 2 2  2 2
>> 3's  3 3 3  3 3  3 3 3 3
>
> LOL!
>
> How can one run out of an infinite source?

I want to try running out, but I need to find that infinite source of
money first ...

On Saturday, August 14, 2021 at 4:16:15 PM UTC+2, WM wrote:

> > > Important however is that all elements are used. [WM]
> > >
> > He said staring at an example where every element of |N_F is used, [William]

Look, idiot, here William means _each and every_ == _all_.

Hint: "Every x e IN is ..." <=> "All x in IN are .."
See: https://en.wikipedia.org/wiki/Universal_quantification

> Every is not enough. Every has almost all as successors.

Holy shit!

Look dumbo, if ALL natural numbers are >= 0, then there aren't any left which aren't >= 0.

Remember: _every_ == _all_ (here).

On 8/14/2021 4:45 PM, WM wrote:
> William schrieb am Samstag, 14. August 2021 um 20:25:42 UTC+2:
>> On Saturday, August 14, 2021 at 1:28:08 PM UTC-4, WM wrote:
>>> William schrieb am Samstag, 14. August 2021 um 17:09:13 UTC+2:
>>
>>>> Piffle. If every element of |N_F is used then no element of |N_F is not used;
>>> Wrong.
>> ?!
>
> If every element of |N_F is used most elements are not used, because every element belongs to a finite initial segment followed by aleph_0 elements.
>
> All elements are not followed by aleph_0 elements. If every element is used, then aleph_0 elements remain. If all elements are used, then none remains.
>
> Regards, WM
>

1. All elements are not followed by aleph_0 elements

2. If every element is used, then aleph_0 elements remain.

3. If all elements are used, then none remains.

4. Every element of |N_F is used in E= (E(1), E(2), E(3),...)

5. For every element i of |N_F E(i) has cardinality aleph_0

6. All indexes are not followed by indexes.

7. Every element of |N_F is not followed by indexes . False

8. The set |N_F is not followed by indexes. True.

9. every element of |N_F is finite and followed by almost all elements
of |N_F.

On 8/14/2021 6:24 PM, Greg Cunt wrote:
> On Saturday, August 14, 2021 at 4:16:15 PM UTC+2, WM wrote:
>
>>>> Important however is that all elements are used. [WM]
>>>>
>>> He said staring at an example where every element of |N_F is used, [William]
>
> Look, idiot, here William means _each and every_ == _all_.
>
> Hint: "Every x e IN is ..." <=> "All x in IN are .."
>
> See: https://en.wikipedia.org/wiki/Universal_quantification
>
>> Every is not enough. Every has almost all as successors.
>
> Holy shit!
>
> Look dumbo, if ALL natural numbers are >= 0, then there aren't any left which aren't >= 0.
>
> Remember: _every_ == _all_ (here).
>
>

WM has 'actual infinity' hesitation.

On Sunday, August 15, 2021 at 12:08:11 AM UTC+2, William wrote:
> On Saturday, August 14, 2021 at 5:45:59 PM UTC-4, WM wrote:
> >
> > If every element of |N_F is used most elements are not used,
> ?!

Huh?! You are "discussing" with a deluded crank. What the hell do you expect?

A rational debate? *lol*

On 8/14/2021 6:24 PM, Sergio wrote:
> On 8/14/2021 4:45 PM, WM wrote:
>> William schrieb am Samstag, 14. August 2021 um 20:25:42 UTC+2:
>>> On Saturday, August 14, 2021 at 1:28:08 PM UTC-4, WM wrote:
>>>> William schrieb am Samstag, 14. August 2021 um 17:09:13 UTC+2:
>>>
>>>>> Piffle. If every element of |N_F is used then no element of |N_F is not used;
>>>> Wrong.
>>> ?!
>>
>> If every element of |N_F is used most elements are not used, because every element belongs to a finite initial segment followed by aleph_0 elements.
>>
>> All elements are not followed by aleph_0 elements. If every element is used, then aleph_0 elements remain. If all elements are used, then none remains.
>>
>> Regards, WM
>>
>
> 1. All elements are not followed by aleph_0 elements
>
> 2. If every element is used, then aleph_0 elements remain.
>
> 3. If all elements are used, then none remains.
>
> 4. Every element of |N_F is used in E= (E(1), E(2), E(3),...)
>
> 5. For every element i of |N_F E(i) has cardinality aleph_0
>
> 6. All indexes are not followed by indexes.
>
> 7. Every element of |N_F is not followed by indexes . False
>
> 8. The set |N_F is not followed by indexes. True.
>
> 9. every element of |N_F is finite and followed by almost all elements
> of |N_F.
>

forgot this one;

10. If every element of |N_F is used most elements are not used.

is that potentially used ? or actually used ?

On 8/14/2021 4:40 PM, WM wrote:
> Jim Burns schrieb am Samstag, 14. August 2021 um 20:58:59 UTC+2:
>> On 8/14/2021 10:20 AM, WM wrote:
>
>>> Therefore all endsegments satisfying
>>> ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
>>> can be combined to
>>> |∩{E(k) : k ∈ ℕ_def}| = ℵo .
>
>> The empty set is the intersection of all your intervals are diversion..
>
> Nionsense. Endsegments are inclusion monotonic. A set of infinite endsegments has an infinite intersection.

wrong. there are no natural numbers in that intersection.

Proof;
let k be a natural number
E(k+1) does not have k in it
therefore the intersection of all endsegments cannot have k in it.
k can be any natural number
therefore the intersection of all endsegments is empty.

let that sink in.

read it slowly a second and third time.

Your 'invented math' fails.

On Sunday, August 15, 2021 at 2:00:16 AM UTC+2, Sergio wrote:
> On 8/14/2021 4:40 PM, WM wrote:
> >
> > [All] set[s] of infinite endsegments [have] an infinite intersection.
> >
> wrong. there are no natural numbers in [the] intersection [of all endsegments].
>
> Proof;
> let k be a natural number
> E(k+1) does not have k in it
> therefore the intersection of all endsegments cannot have k in it.
> k can be any natural number
> therefore the intersection of all endsegments is empty.
>
> let that sink in. [...]

The problem with WM is that is DOES NOT sink in. NEVER.

You see: "Cranks characteristically dismiss all evidence or arguments which contradict their own unconventional beliefs, making any rational debate a futile task and rendering them impervious to facts, evidence, and rational inference."

In short: "A crank is defined as a man who cannot be turned." (Nature, 8 Nov 1906, 25/2)

Source: https://en.wikipedia.org/wiki/Crank_(person)

Crank Wolfgang Mueckenheim, aka WM wrote:
> Jim Burns schrieb am Samstag, 14. August 2021 um 20:58:59 UTC+2:
>> On 8/14/2021 10:20 AM, WM wrote:
>
>>> Therefore all endsegments satisfying
>>> ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
>>> can be combined to
>>> |∩{E(k) : k ∈ ℕ_def}| = ℵo .
>
>> The empty set is the intersection of all infinite end segments.
>
> Nionsense. Endsegments are inclusion monotonic. A set of infinite endsegments has an infinite intersection.

This is WRONG. PERIOD. (E_k) is an obvious counterexample to this
claim of yours, Crank Wolfgang Mueckenheim from Hochschule Augsburg.

Using an alleged FALSE theorem of yours in front of the most simple
counter-example that proves it wrong is quite a silly position, Crank
Wolfgang Mueckenheim from Hochschule Augsburg.

On 8/14/2021 5:40 PM, WM wrote:
> Jim Burns schrieb
> am Samstag, 14. August 2021 um 20:58:59 UTC+2:
>> On 8/14/2021 10:20 AM, WM wrote:

>>> Therefore all endsegments satisfying
>>> ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
>>> can be combined to
>>> |∩{E(k) : k ∈ ℕ_def}| = ℵo .
>>
>> The empty set is the intersection of all infinite end segments.
>
> Nionsense. Endsegments are inclusion monotonic.

For each infinite end segment E,
there is another infinite end segment E' such that
E' is a proper subset of E.

E is not a subset of E'.

> A set of infinite endsegments has an infinite intersection.

Let IAIES be the intersection of all infinite end segments.
For each infinite end segment E', IAIES is a subset of E'.
Because intersection.

Assume that an infinite end segment E exists such that E = IAIES.
There is another infinite end segment E' such that
IAIES = E is not a subset of E'.
Contradiction.

Therefore,
no infinite end segment E exists such that E = IAIES.

So, no.

>>> |∩{E(k) : k ∈ ℕ_def}| = ℵo .
>>
>> No.
>
> You may be qualified for mathematical paralympics.
> The intersection depends only on the endsegments.
> If those endsegments which satisfy (*) can form an infinite set,
> their intersection is infinite. If they can't, then ℕ_def =/= ℕ.

On 8/14/2021 3:24 PM, Sergio wrote:
> On 8/14/2021 4:48 PM, Chris M. Thomasson wrote:
>> On 8/11/2021 12:55 PM, Sergio wrote:
>>> On 8/11/2021 2:46 PM, WM wrote:
>>>> William schrieb am Mittwoch, 11. August 2021 um 16:50:30 UTC+2:
>>>>> On Wednesday, August 11, 2021 at 9:57:42 AM UTC-4, WM wrote:
>>>>>> William schrieb am Dienstag, 10. August 2021 um 15:14:35 UTC+2:
>>>>>>> On Tuesday, August 10, 2021 at 5:35:20 AM UTC-4, WM wrote:
>>>>>>>> William schrieb am Montag, 9. August 2021 um 17:55:55 UTC+2:
>>>>>>>
>>>>>>>>> (Note, every element of |N_F is finite and followed by almost
>>>>>>>>> all elements of |N_F.)
>>>>>>>> Every definable index, yes
>>>>>>> Indeed, every index you can write down, and every index you cannot
>>>>>>> write down too.
>>>>>>>> But all indexes are not followed by indexes:
>>>>>>> Another use of the ambiguous "all".
>>>>>> It isn't ambiguous
>>>>> He said staring at two different meanings of the phrase " All
>>>>> indexes are not followed by indexes"
>>>>
>>>> There is only one meaning here. All indexes are used up in the
>>>> bijection of indexes and endsegments. No additional natural numbers
>>>> are available as contents of endsegments.
>>>
>>>
>>> you ran out of natural numbers ?  you can "re-use" them you know.
>>>
>>> In case you run out, here are some extras 1's  1 1 1 1  1 1  1 1  1 1 1
>>> 2's  2 2  2 2 2  2 2  2 2 2  2 2
>>> 3's  3 3 3  3 3  3 3 3 3
>>
>> LOL!
>>
>> How can one run out of an infinite source?
>
>
> I want to try running out, but I need to find that infinite source of
> money first ...
>

Just print it to your hearts desire. Every played Monopoly? ;^)

On Saturday, 14 August 2021 at 18:33:41 UTC-3, WM wrote:
[...]
> All elements are not followed by aleph_0 elements. If every element is used, then aleph_0 elements remain. If all elements are used, then none remains.

Another classic WM use of natural language to obfuscate and misdirect, in order to deceitfully set up a switch of quantifiers. The core truth of this word salad is that every "element" (i.e., natural number in the set |N imbued with the usual order) is followed by aleph_0 others. Everything else is smoke and mirrors.

On Saturday, 14 August 2021 at 18:40:39 UTC-3, WM wrote:
> Jim Burns schrieb am Samstag, 14. August 2021 um 20:58:59 UTC+2:
> > On 8/14/2021 10:20 AM, WM wrote:
>
> > > Therefore all endsegments satisfying
> > > ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
> > > can be combined to
> > > |∩{E(k) : k ∈ ℕ_def}| = ℵo .
> > The empty set is the intersection of all infinite end segments.
> Nionsense. Endsegments are inclusion monotonic. A set of infinite endsegments has an infinite intersection.

Another classic WM misuse of a concept. Monotone inclusion only means that the intersection of a set of nested sets is contained in every one of them. If the set of elements has infinite cardinality (as in the set { E(1), E(2), E(3), ... }) then the intersection may be empty, as the previous example shows: For every n, n ~in E(n+1).

[Remaining nonsense removed]

Gus Gassmann schrieb am Sonntag, 15. August 2021 um 12:59:23 UTC+2:
> On Saturday, 14 August 2021 at 18:40:39 UTC-3, WM wrote:
> > Endsegments are inclusion monotonic. A set of infinite endsegments has an infinite intersection.
> If the set of elements has infinite cardinality (as in the set { E(1), E(2), E(3), ... }) then the intersection may be empty, as the previous example shows: For every n, n ~in E(n+1).

This shows that empty endsegments must exist. All numbers lost! But not only in the intersection.

Regards, WM

Gus Gassmann schrieb am Sonntag, 15. August 2021 um 12:51:24 UTC+2:
> On Saturday, 14 August 2021 at 18:33:41 UTC-3, WM wrote:
> [...]
> > All elements are not followed by aleph_0 elements. If every element is used, then aleph_0 elements remain. If all elements are used, then none remains.
> every "element" (i.e., natural number in the set |N imbued with the usual order) is followed by aleph_0 others. Everything else is smoke and mirrors.

If all numbers are subtracted, then none remains. So not every number can be followed by aleph_0 others. But every definable.

Regards, WM

On Sunday, 15 August 2021 at 11:55:11 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 15. August 2021 um 12:59:23 UTC+2:
> > On Saturday, 14 August 2021 at 18:40:39 UTC-3, WM wrote:
> > > Endsegments are inclusion monotonic. A set of infinite endsegments has an infinite intersection.
> > If the set of elements has infinite cardinality (as in the set { E(1), E(2), E(3), ... }) then the intersection may be empty, as the previous example shows: For every n, n ~in E(n+1).
> This shows that empty endsegments must exist. All numbers lost! But not only in the intersection.

Nope! Every end segment E(n) contains at the very least the number n. (Otherwise it is not an end segment.) And once it contains n, it contains n+1, and n+2, and so on. Try again.

On Sunday, 15 August 2021 at 11:58:18 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 15. August 2021 um 12:51:24 UTC+2:
> > On Saturday, 14 August 2021 at 18:33:41 UTC-3, WM wrote:
> > [...]
> > > All elements are not followed by aleph_0 elements. If every element is used, then aleph_0 elements remain. If all elements are used, then none remains.
> > every "element" (i.e., natural number in the set |N imbued with the usual order) is followed by aleph_0 others. Everything else is smoke and mirrors.
> If all numbers are subtracted, then none remains. So not every number can be followed by aleph_0 others. But every definable.

Every integer is definable. You have been shown that |N_def = |N. Once again, your position has been vanquished. You are a blowhard, you use smoke and mirrors, and when that is not enough, misdirection, deceit, and lies.

On Sunday, August 15, 2021 at 10:58:18 AM UTC-4, WM wrote:

>> If all numbers are subtracted, then none remains.

So none remains that needs to be followed by aleph_0 otheres.
If not all numbers are subtracted then aleph_0 remains.

--
William Hughes

On 8/15/2021 9:58 AM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 15. August 2021 um 12:51:24 UTC+2:
>> On Saturday, 14 August 2021 at 18:33:41 UTC-3, WM wrote:
>> [...]
>>> All elements are not followed by aleph_0 elements. If every element is used, then aleph_0 elements remain. If all elements are used, then none remains.
>> every "element" (i.e., natural number in the set |N imbued with the usual order) is followed by aleph_0 others. Everything else is smoke and mirrors.
>
> If all numbers are subtracted, then none remains. So not every number can be followed by aleph_0 others. But every definable.
>
> Regards, WM
>

" If all numbers are subtracted, then none remains."

Do you mean from the set of natural numbers, you subtract, takeout,
remove all, or every, or each natural numbers, there is nothing left ?

"So not every number can be followed by aleph_0 others."

*wrong*. Each and every natural number "is followed by aleph_0 others"

On 8/15/2021 9:55 AM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 15. August 2021 um 12:59:23 UTC+2:
>> On Saturday, 14 August 2021 at 18:40:39 UTC-3, WM wrote:
>>> Endsegments are inclusion monotonic. A set of infinite endsegments has an infinite intersection.
>> If the set of elements has infinite cardinality (as in the set { E(1), E(2), E(3), ... }) then the intersection may be empty, as the previous example shows: For every n, n ~in E(n+1).
>
> This shows that empty endsegments must exist. All numbers lost!

you do not think in terms of math.

If you did you would post equations of you position.