Gus Gassmann schrieb am Sonntag, 15. August 2021 um 18:01:54 UTC+2:

> Every integer is definable.

Wrong.
If you subtract all individually definable integers from |N then aleph_0 natural numbers remain.
If you subtract all natuarl numbers from |N then none remains.
This demonstrates a difference.

> You have been shown that |N_def = |N.

Wrong.
If |N_def = |N then the definable endsegments are all endsegments. The definable endsegments have an infinite intersection. All endsegments have an empty intersection. Contradiction

Regards, WM

William schrieb am Sonntag, 15. August 2021 um 18:19:34 UTC+2:
> 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 you subtract all individually definable integers from |N then aleph_0 natural numbers remain.
If you subtract all natuarl numbers from |N then none remains.
This demonstrates a difference.

> You have been shown that |N_def = |N.

If |N_def = |N then the definable endsegments are all endsegments. The definable endsegments have an infinite intersection. All endsegments have an empty intersection. Contradiction

Regards, WM

On 8/15/2021 1:01 PM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 15. August 2021 um 18:01:54 UTC+2:
>
>> Every integer is definable.
>
> Wrong.
>1. If you subtract all individually definable integers from |N then aleph_0 natural numbers remain.
>2. If you subtract all natuarl numbers from |N then none remains.
> 3 This demonstrates a difference.

no, your conclusion is incorrect. Your mistake is in line 1.
You need to show or prove every integer is not definable, first.

>
>> You have been shown that |N_def = |N.
>
> Wrong.
> If |N_def = |N then the definable endsegments are all endsegments.

wrong.

The definable endsegments have an infinite intersection. All endsegments
have an empty intersection. Contradiction

still wrong.

>
> Regards, WM
>

Gus Gassmann schrieb am Sonntag, 15. August 2021 um 17:59:46 UTC+2:
> 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.)

Every endsegment reduces the contents of the intersection

E(1) = E(1)
E(1) ∩ E(2) = E(2)
E(1) ∩ E(2) ∩ E(3) = E(3)
E(1) ∩ E(2) ∩ E(3) ∩ E(4) = E(4)
E(1) ∩ E(2) ∩ E(3) ∩ E(4) ∩ E(5) = E(5)
....

An empty intersection requires an empty endsegment. Of course it is dark.

Regards, WM

Endsegments are inclusion monotonic.
Every endsegment reduces the intersection to its own set. No empty set without an empty endsegment.

Regards, WM

Python schrieb am Sonntag, 15. August 2021 um 03:01:46 UTC+2:
> 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
Endsegments are inclusion monotonic.
Every endsegment reduces the intersection to its own set. No empty intersection without an empty endsegment.

Regards, WM

On Sunday, 15 August 2021 at 15:07:35 UTC-3, WM wrote:
[...]
> An empty intersection requires an empty endsegment. Of course it is dark.

Makes no difference! An end segment E(n) is by definition a set of the form {k: k >= n}. Since n >= n, whether n is dark or not, the end segment E(n) is not empty. How stupid can you be to not understand that? To me it can be either dementia or malice, but tertium non datur. You pick.

On Sunday, August 15, 2021 at 2:03:17 PM UTC-4, WM wrote:
> William schrieb am Sonntag, 15. August 2021 um 18:19:34 UTC+2:
> > 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 you subtract all individually definable integers from |N then aleph_0 natural numbers remain.
Yes, the set of numbers you can write down is finite So what?
If you subtract the set of every |N_F none remain that need to be followed by aleph_0 others.

--
William Hughes

On Sunday, 15 August 2021 at 15:01:20 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 15. August 2021 um 18:01:54 UTC+2:
>
> > Every integer is definable.
>
> Wrong.
> If you subtract all individually definable integers from |N then aleph_0 natural numbers remain.
> If you subtract all natuarl numbers from |N then none remains.
> This demonstrates a difference.
> > You have been shown that |N_def = |N.
> Wrong.
> If |N_def = |N then the definable endsegments are all endsegments.
Indeed they are. But you knew last week that |N_def = |N, so the "If" should be redundant.
> The definable endsegments have an infinite intersection.

Nope. Elements n in |N_def (or |N, which is the same thing) have the property that intersection{E(k): k <= n} has cardinality aleph_0. Nothing said about intersection{E(k): k in |N_def (or |N, which is the same thing)}, which is provably empty.
> All endsegments have an empty intersection.
The intersection of end segments over |N_def (or |N, which is the same thing) is empty. Congratulations on this revolutionary insight.

On Sunday, 15 August 2021 at 15:13:29 UTC-3, WM wrote:
> Endsegments are inclusion monotonic.
> Every endsegment reduces the intersection to its own set. No empty set without an empty endsegment.
Wrong, as every nontrivial mathematical statement you ever posted. You continually neglect the difference between finite and infinite collections. For someone pretending to know *ANYTHING* at all about infinity, that should be a showstopper. (Copy and pasting your answer in two different threads does not make a difference in that regard.)

On 8/15/2021 1:15 PM, WM wrote:
> Python schrieb am Sonntag, 15. August 2021 um 03:01:46 UTC+2:
>> 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
>
> Endsegments are inclusion monotonic.
> Every endsegment reduces the intersection to its own set. No empty intersection without an empty endsegment.
>
> Regards, WM
>

wrong. How many time do I have to tell you troll ?

you cannot prove there is any natural number in the intersection of all
endsegments.

proof the intersection is empty;

pick any natural number, k
k is not in E(k+1)
therefore k is not in the intersection of all endsegments.
therefore the intersection of all endsegments is empty.

On 8/15/2021 1:13 PM, WM wrote:
> Endsegments are inclusion monotonic.
> Every endsegment reduces the intersection to its own set. No empty set without an empty endsegment.
>
> Regards, WM
>

you cannot show there is an element in the intersection of all
endsegments. Fail.

Gus Gassmann schrieb am Sonntag, 15. August 2021 um 20:19:17 UTC+2:
> On Sunday, 15 August 2021 at 15:07:35 UTC-3, WM wrote:
> [...]
> > An empty intersection requires an empty endsegment. Of course it is dark.
> Makes no difference! An end segment E(n) is by definition a set of the form {k: k >= n}. Since n >= n, whether n is dark or not, the end segment E(n) is not empty.

And the intersection is infinite, for every definable n. No definable n is available to get an infinite set or an empty intersection. Endsegments are inclusion monotonic.
Every endsegment reduces the intersection to its own set. No empty set without an empty endsegment.

Regards, WM

Gus Gassmann schrieb am Sonntag, 15. August 2021 um 20:42:49 UTC+2:
> On Sunday, 15 August 2021 at 15:13:29 UTC-3, WM wrote:
> > Endsegments are inclusion monotonic.
> > Every endsegment reduces the intersection to its own set. No empty set without an empty endsegment.
> Wrong

> > Every endsegment reduces the intersection to its own set. No empty set without an empty endsegment.

E(1) = E(1)
E(1) ∩ E(2) = E(2)
E(1) ∩ E(2) ∩ E(3) = E(3)
E(1) ∩ E(2) ∩ E(3) ∩ E(4) = E(4)
E(1) ∩ E(2) ∩ E(3) ∩ E(4) ∩ E(5) = E(5)
....

> You continually neglect the difference between finite and infinite collections.

Inclusion monotony is ruling over all.

Regards, WM

William schrieb am Sonntag, 15. August 2021 um 20:30:06 UTC+2:
> On Sunday, August 15, 2021 at 2:03:17 PM UTC-4, WM wrote:
> > William schrieb am Sonntag, 15. August 2021 um 18:19:34 UTC+2:
> > > 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 you subtract all individually definable integers from |N then aleph_0 natural numbers remain.
> Yes, the set of numbers you can write down is finite So what?

It is a revolutionary new insight: the set of numbers you can define by thinking them or writing them down, independent of time and ressources and eagerness, is finite. Ask some matheologians after this topic.

> If you subtract the set of every |N_F none remain that need to be followed by aleph_0 others.

Therefore also the aleph_0 unwritable followers of every definable number can be removed.

Regards, WM

On Sunday, 15 August 2021 at 17:29:27 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 15. August 2021 um 20:42:49 UTC+2:
[...]
> > You continually neglect the difference between finite and infinite collections.
> Inclusion monotony is ruling over all.
This would be even more convincing if you actually had any clue how inclusion monotony works with infinite sets and infinite collections.

On 8/15/2021 3:29 PM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 15. August 2021 um 20:42:49 UTC+2:
>> On Sunday, 15 August 2021 at 15:13:29 UTC-3, WM wrote:
>>> Endsegments are inclusion monotonic.
>>> Every endsegment reduces the intersection to its own set. No empty set without an empty endsegment.
>> Wrong
>
>>> Every endsegment reduces the intersection to its own set. No empty set without an empty endsegment.
>
> E(1) = E(1)
> E(1) ∩ E(2) = E(2)
> E(1) ∩ E(2) ∩ E(3) = E(3)
> E(1) ∩ E(2) ∩ E(3) ∩ E(4) = E(4)
> E(1) ∩ E(2) ∩ E(3) ∩ E(4) ∩ E(5) = E(5)
> ...
>
>> You continually neglect the difference between finite and infinite collections.
>
> Inclusion monotony is ruling over all.
>
> Regards, WM
>

inclusion monotony is a red herring, diversion.

It is proven the intersection of all endsegments has no elements in it.

On 8/15/2021 1:47 PM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 15. August 2021 um 20:19:17 UTC+2:
>> On Sunday, 15 August 2021 at 15:07:35 UTC-3, WM wrote:
>> [...]
>>> An empty intersection requires an empty endsegment. Of course it is dark.
>> Makes no difference! An end segment E(n) is by definition a set of the form {k: k >= n}. Since n >= n, whether n is dark or not, the end segment E(n) is not empty.
>
> And the intersection is infinite, for every definable n. No definable n is available to get an infinite set or an empty intersection. Endsegments are inclusion monotonic.
> Every endsegment reduces the intersection to its own set. No empty set without an empty endsegment.
>
> Regards, WM
>

wrong. repeat it, and it is till wrong, and you know it.

Gus Gassmann schrieb am Sonntag, 15. August 2021 um 22:38:59 UTC+2:
> On Sunday, 15 August 2021 at 17:29:27 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Sonntag, 15. August 2021 um 20:42:49 UTC+2:
> [...]
> > > You continually neglect the difference between finite and infinite collections.
> > Inclusion monotony is ruling over all.
> This would be even more convincing if you actually had any clue how inclusion monotony works with infinite sets and infinite collections.

It is based upon the fact that every endsegment is a subset of its direct predecessor and a superset of its direct successor. Every set of infinite endsegments has an infinite intersection. It is like the sequence f(n) = 1 +1/n. It is never zero and cannot have the limit zero. You believe that the terms of the sequence are never less than 1 but that its limit is zero.

Regards, WM

On Sunday, August 15, 2021 at 4:35:02 PM UTC-4, WM wrote:
> aleph_0 unwritable follow [ers of] every definable number
Indeed, as aleph_0 unwritable follow every single 3element of |N_F. And when you have used all the element of |N_F there is no problem because there are no elements that need aleph_9 elements to follow them

--
William Hughes

On Sunday, August 15, 2021 at 1:41:54 PM UTC-7, Sergio wrote:
> On 8/15/2021 1:47 PM, WM wrote:
> > Gus Gassmann schrieb am Sonntag, 15. August 2021 um 20:19:17 UTC+2:
> >> On Sunday, 15 August 2021 at 15:07:35 UTC-3, WM wrote:
> >> [...]
> >>> An empty intersection requires an empty endsegment. Of course it is dark.
> >> Makes no difference! An end segment E(n) is by definition a set of the form {k: k >= n}. Since n >= n, whether n is dark or not, the end segment E(n) is not empty.
> >
> > And the intersection is infinite, for every definable n. No definable n is available to get an infinite set or an empty intersection. Endsegments are inclusion monotonic.
> > Every endsegment reduces the intersection to its own set. No empty set without an empty endsegment.
> >
> > Regards, WM
> >
> wrong. repeat it, and it is till wrong, and you know it.

Counterexample to "not an at least countably infinite continuous domain":
sweep principle.

(Courtesy bridges.)

An "end segment" is for symmetry as about infinity, and, zero.

It's almost fair to say that counting both ways they meet.

On Sunday, August 15, 2021 at 3:19:20 PM UTC-7, Ross A. Finlayson wrote:
> On Sunday, August 15, 2021 at 1:41:54 PM UTC-7, Sergio wrote:
> > On 8/15/2021 1:47 PM, WM wrote:
> > > Gus Gassmann schrieb am Sonntag, 15. August 2021 um 20:19:17 UTC+2:
> > >> On Sunday, 15 August 2021 at 15:07:35 UTC-3, WM wrote:
> > >> [...]
> > >>> An empty intersection requires an empty endsegment. Of course it is dark.
> > >> Makes no difference! An end segment E(n) is by definition a set of the form {k: k >= n}. Since n >= n, whether n is dark or not, the end segment E(n) is not empty.
> > >
> > > And the intersection is infinite, for every definable n. No definable n is available to get an infinite set or an empty intersection. Endsegments are inclusion monotonic.
> > > Every endsegment reduces the intersection to its own set. No empty set without an empty endsegment.
> > >
> > > Regards, WM
> > >
> > wrong. repeat it, and it is till wrong, and you know it.
> Counterexample to "not an at least countably infinite continuous domain":
> sweep principle.
>
> (Courtesy bridges.)
>
> An "end segment" is for symmetry as about infinity, and, zero.
>
> It's almost fair to say that counting both ways they meet.

Including that it has to be zero and infinity for them not to.

On 8/15/2021 3:34 PM, WM wrote:
> William schrieb am Sonntag, 15. August 2021 um 20:30:06 UTC+2:
>> On Sunday, August 15, 2021 at 2:03:17 PM UTC-4, WM wrote:
>>> William schrieb am Sonntag, 15. August 2021 um 18:19:34 UTC+2:
>>>> 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 you subtract all individually definable integers from |N then aleph_0 natural numbers remain.
>> Yes, the set of numbers you can write down is finite So what?
>
> It is a revolutionary new insight: the set of numbers you can define by thinking them or writing them down, independent of time and ressources and eagerness, is finite.

new ?? it is silly, it is *Observer Dependent Math*

In grade school Children are forced to write down all the numbers they
will ever use in life, using small crayons and scraps of toilet paper...

>
>> If you subtract the set of every |N_F none remain that need to be followed by aleph_0 others.
>
> Therefore also the aleph_0 unwritable followers of every definable number can be removed.

all this writable, defining, and removal is simply diversion.

there is no math that supports it, you have proven that.

It is a dead end.

>
> Regards, WM
>

Diverted Ants
The Dead End Ants
No Math Ants

On 8/15/2021 3:53 PM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 15. August 2021 um 22:38:59 UTC+2:
>> On Sunday, 15 August 2021 at 17:29:27 UTC-3, WM wrote:
>>> Gus Gassmann schrieb am Sonntag, 15. August 2021 um 20:42:49 UTC+2:
>> [...]
>>>> You continually neglect the difference between finite and infinite collections.
>>> Inclusion monotony is ruling over all.
>> This would be even more convincing if you actually had any clue how inclusion monotony works with infinite sets and infinite collections.
>
> It is based upon the fact that every endsegment is a subset of its direct predecessor and a superset of its direct successor. Every set of infinite endsegments has an infinite intersection. It is like the sequence f(n) = 1 +1/n. It is never zero and cannot have the limit zero. You believe that the terms of the sequence are never less than 1 but that its limit is zero.
>
> Regards, WM
>

....and you cannot prove there is a number in the intersection of all
every endsegment, so it is empty.

you are just not knowledgeable of taking the limit in Math. You dont
have the math skills.

On Sunday, 15 August 2021 at 17:53:34 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 15. August 2021 um 22:38:59 UTC+2:
> > On Sunday, 15 August 2021 at 17:29:27 UTC-3, WM wrote:
> > > Gus Gassmann schrieb am Sonntag, 15. August 2021 um 20:42:49 UTC+2:
> > [...]
> > > > You continually neglect the difference between finite and infinite collections.
> > > Inclusion monotony is ruling over all.
> > This would be even more convincing if you actually had any clue how inclusion monotony works with infinite sets and infinite collections.
> It is based upon the fact that every endsegment is a subset of its direct predecessor and a superset of its direct successor. Every set of infinite endsegments has an infinite intersection.
Bullshit. You even used to know how one takes intersections of an infinite number of sets. Sad to see that falling victim to your dementia.

Once again: Every *finite* set of end segments has an intersection with cardinality aleph_0.

> It is like the sequence f(n) = 1 +1/n. It is never zero and cannot have the limit zero.
Another thing that fell victim to your dementia is the ability to compute set limits. (For inclusion monotone sequences the limit equals the intersection, and for the set of all end segments, both are the empty set.)