Crank Wolfgang Mueckenheim, 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.

What *you* (and you only) mean by "inclusion monotic" is WRONG. (E_k)
is a counter-example to your silly claim.

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

All endsegments (any infinite subset of them by the way) are enough to
remove all natural numbers.

On 8/15/2021 5:58 PM, Python wrote:
> Crank Wolfgang Mueckenheim, 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.
>
> What *you* (and you only) mean by "inclusion monotic" is WRONG. (E_k)
> is a counter-example to your silly claim.
>
>> Every endsegment reduces the intersection to its own set. No empty
>> intersection without an empty endsegment.
>
> All endsegments (any infinite subset of them by the way) are enough to
> remove all natural numbers.

WM seems to be ALWAYS wrong, not just once and a while, but with every
post.

Post Ants

On Monday, August 16, 2021 at 1:04:44 AM UTC+2, Sergio wrote:

> WM seems to be ALWAYS wrong, not just once and a while, but with every
> post.

Actually, with each and every non-trivial claim. (I'm serious!)

Greg Cunt wrote:
> On Monday, August 16, 2021 at 1:04:44 AM UTC+2, Sergio wrote:
>
>> WM seems to be ALWAYS wrong, not just once and a while, but with every
>> post.
>
> Actually, with each and every non-trivial claim. (I'm serious!)

Definitely. Trivially true, but pointless, almost only claim these
days is "(whatever) belongs to a finite set".

On Sunday, 15 August 2021 at 20:30:53 UTC-3, Greg Cunt wrote:
> On Monday, August 16, 2021 at 1:04:44 AM UTC+2, Sergio wrote:
>
> > WM seems to be ALWAYS wrong, not just once and a while, but with every
> > post.
> Actually, with each and every non-trivial claim. (I'm serious!)

Second that.

WM - Consistently wrong since at least 2005. (I have *NEVER* seen a nontrivial fact treated correctly by that clown.)

Gus Gassmann wrote:
> On Sunday, 15 August 2021 at 20:30:53 UTC-3, Greg Cunt wrote:
>> On Monday, August 16, 2021 at 1:04:44 AM UTC+2, Sergio wrote:
>>
>>> WM seems to be ALWAYS wrong, not just once and a while, but with every
>>> post.
>> Actually, with each and every non-trivial claim. (I'm serious!)
>
> Second that.
>
> WM - Consistently wrong since at least 2005. (I have *NEVER* seen a nontrivial fact treated correctly by that clown.)

Which wouldn't be a problem if only Usenet was involved. Unfortunately
Hochschule Augsburg is involved too, clown Wolfgang Mueckenheim teaches
there.

On 8/15/2021 4:53 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.

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

Infinite end segment E is NOT a subset of
infinite end segment EE.

> Every set of infinite endsegments has an infinite intersection.

Assume that IAIES, the Intersection of All Infinite End
Segments, is an infinite end segment.

On the one hand, there is another infinite end segment EE
such that EE is a proper subset of IAIES, and
infinite end segment IAIES *IS NOT* a subset of
infinite end segment EE.

On the other hand, because IAIES is the Intersection of
All Infinite End Segments, IAIES is a subset of each
infinite end segment and IAIES *IS* a subset of
infinite end segment EE,

Contradiction.

Therefore,
IAIES, the Intersection of All Infinite End Segments,
is not an infinite end segment.

> 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.

It is like the sequence of circles of radius 1/n.
Each circle in the sequence contains infinitely-many points.
The limit of the sequence is a single point.

But there's no need to argue about what it's like.
We know what it is. It is the empty set.

WM presented the following explanation :
> 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.

You're talking about proper subsets, correct?

William schrieb am Sonntag, 15. August 2021 um 23:59:06 UTC+2:
> 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 element of |N_F.

Yes, every single element that can be distiguished from all others, aka can be written, described, or defined, is followed by aleph_0 unwritable elements. This means that the collection of writable elements has not aleph_0 elements. Otherwise we had two consecutive sets of cardinality aleph_0 in |N. Contradiction.

Therefore the intersection of all endsegments indexed by such elements cannot be empty:
|∩{E(k) : k ∈ ℕ_def}| = ℵo .
Do you agree?

Regards, WM

Gus Gassmann schrieb am Montag, 16. August 2021 um 00:47:26 UTC+2:
> On Sunday, 15 August 2021 at 17:53:34 UTC-3, WM wrote:
> > 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.
>
> Once again: Every *finite* set of end segments has an intersection with cardinality aleph_0.

Now collect all endsegments of such finite sets for the intersection. It cannot be empty.

> (For inclusion monotone sequences the limit equals the intersection, and for the set of all end segments, both are the empty set.)

Of course. But the definable endsegments are not all endsegments, because all definable endsegments have aleph_0 successors.

Regards, WM

Python schrieb am Montag, 16. August 2021 um 00:57:49 UTC+2:

> > Endsegments are inclusion monotonic.
> What *you* (and you only) mean by "inclusion monotic" is WRONG. (E_k)
> is a counter-example

No.
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)
....
The intersection is analyzed step by step as long as possible. At no definable step it is empty.

> > Every endsegment reduces the intersection to its own set. No empty intersection without an empty endsegment.
> All endsegments (any infinite subset of them by the way) are enough to
> remove all natural numbers.

Of course. But the definable endsegments are only those which are followed by aleph_0 endsegments. Therefore there cannot be aleph_0 definable edsegments.

Regards, WM

FromTheRafters schrieb am Montag, 16. August 2021 um 11:34:33 UTC+2:
> WM presented the following explanation :

> > It is based upon the fact that every endsegment is a subset of its direct
> > predecessor and a superset of its direct successor.
> You're talking about proper subsets, correct?

Yes.

Regards, WM

On Monday, 16 August 2021 at 08:02:37 UTC-3, WM wrote:
[...]
> Of course. But the definable endsegments are not all endsegments, because all definable endsegments have aleph_0 successors.

Boohoo! In ZFC every natural number, and hence every end segment, is definable. And every end segment contains aleph_0 elements. Try again. (Or better, don''t bother; you'll never get it, anyway.)

Jim Burns schrieb am Montag, 16. August 2021 um 05:32:18 UTC+2:
> On 8/15/2021 4:53 PM, WM wrote:

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

Yes. But not every infinite endsegment must be definable. Further EE can be the last of a set of infinite endsegments. Then it is the intersection of this set.
>
> Infinite end segment E is NOT a subset of
> infinite end segment EE.

Unless EE is the intersection of a set of endsegments. Then it is the last one.

> > Every set of infinite endsegments has an infinite intersection.
> Assume that IAIES, the Intersection of All Infinite End
> Segments, is an infinite end segment.

It is with no doubt the last of the intersected endsegments.
>
> On the one hand, there is another infinite end segment EE
> such that EE is a proper subset of IAIES, and
> infinite end segment IAIES *IS NOT* a subset of
> infinite end segment EE.

If all infinite endsegments had been intersected, then EE is the last infinite endsegment of the set.
>
> On the other hand, because IAIES is the Intersection of
> All Infinite End Segments, IAIES is a subset of each
> infinite end segment and IAIES *IS* a subset of
> infinite end segment EE,

Not necessarily a proper subset.
>
> Contradiction.

Why? EE is simply the last endsegment of the set.
>
> Therefore,
> IAIES, the Intersection of All Infinite End Segments,
> is not an infinite end segment.

Non sequitur.

Regards, WM

Gus Gassmann schrieb am Montag, 16. August 2021 um 13:52:05 UTC+2:
> On Monday, 16 August 2021 at 08:02:37 UTC-3, WM wrote:
> [...]
> > Of course. But the definable endsegments are not all endsegments, because all definable endsegments have aleph_0 successors.
> Boohoo! In ZFC every natural number, and hence every end segment, is definable.

That is the reason why ZFC is inconsistent. It is an obviously wrong claim.
Proof: Every definable number has aleph_0 successors.
Not every number has aleph_0 successors because |N contains only numbers and if they are collectively subtracted from |N by |N \ |N, then nothing remains.

Regards, WM

On Monday, 16 August 2021 at 09:07:16 UTC-3, WM wrote:
> Gus Gassmann schrieb am Montag, 16. August 2021 um 13:52:05 UTC+2:
> > On Monday, 16 August 2021 at 08:02:37 UTC-3, WM wrote:
> > [...]
> > > Of course. But the definable endsegments are not all endsegments, because all definable endsegments have aleph_0 successors.
> > Boohoo! In ZFC every natural number, and hence every end segment, is definable.
> That is the reason why ZFC is inconsistent. It is an obviously wrong claim.
> Proof: Every definable number has aleph_0 successors.
> Not every number has aleph_0 successors because |N contains only numbers and if they are collectively subtracted from |N by |N \ |N, then nothing remains.

This is your usual bullshit, again. For *EVERY* S you have S \ S = {}. For *EVERY* infinite set T you can't get there by subtracting elements one at a time. Again, how stupid do you have to be not to get that? (Sixteen years of malfeasance --- at least --- really ought to tell you that you are in the wrong profession.)

måndag 16 augusti 2021 kl. 14:07:16 UTC+2 skrev WM:
> Gus Gassmann schrieb am Montag, 16. August 2021 um 13:52:05 UTC+2:
> > On Monday, 16 August 2021 at 08:02:37 UTC-3, WM wrote:
> > [...]
> > > Of course. But the definable endsegments are not all endsegments, because all definable endsegments have aleph_0 successors.
> > Boohoo! In ZFC every natural number, and hence every end segment, is definable.
> That is the reason why ZFC is inconsistent. It is an obviously wrong claim.
> Proof: Every definable number has aleph_0 successors.
> Not every number has aleph_0 successors because |N contains only numbers and if they are collectively subtracted from |N by |N \ |N, then nothing remains.
>
> Regards, WM

>That is the reason why ZFC is inconsistent. It is an obviously wrong claim..

WRONG!

You have not once shown ANY contradiction in it!

>Not every number has aleph_0 successors because |N contains only numbers and if they are collectively subtracted from |N by |N \ |N, then nothing remains.

WRONG AGAIN!

This is empty assertions by you.

WM laid this down on his screen :
> FromTheRafters schrieb am Montag, 16. August 2021 um 11:34:33 UTC+2:
>> WM presented the following explanation :
>
>>> It is based upon the fact that every endsegment is a subset of its direct
>>> predecessor and a superset of its direct successor.
>> You're talking about proper subsets, correct?
>
> Yes.

What if they are not proper subsets?

FromTheRafters schrieb am Montag, 16. August 2021 um 15:39:07 UTC+2:
> WM laid this down on his screen :
> > FromTheRafters schrieb am Montag, 16. August 2021 um 11:34:33 UTC+2:
> >> WM presented the following explanation :
> >
> >>> It is based upon the fact that every endsegment is a subset of its direct
> >>> predecessor and a superset of its direct successor.
> >> You're talking about proper subsets, correct?
> >
> > Yes.
> What if they are not proper subsets?

Then they are not endsegments:
∀k ∈ ℕ: E(k+1) = E(k) \ {k}

Regards, WM

On Monday, August 16, 2021 at 6:57:37 AM UTC-4, WM wrote:

> |∩{E(k) : k ∈ ℕ_def}| = ℵo .
> Do you agree?

You have defined |N_def to be a finite set. Yes this is true. Trivial, but true. So what?

If we define "writable" in the usual manner (no resource bound) there are no "unwritable" elements of |N_F. You may wish to insist upon such a bound but then the interesting set becomes the nr-writable numbers (the elements of |N_F which have the property that they can be written down except possibly for resource reasons). There are no elements of |N_F that are not nr-writable.

--
William Hughes

zelos...@gmail.com schrieb am Montag, 16. August 2021 um 14:40:59 UTC+2:
> måndag 16 augusti 2021 kl. 14:07:16 UTC+2 skrev WM:
> > Gus Gassmann schrieb am Montag, 16. August 2021 um 13:52:05 UTC+2:
> > > On Monday, 16 August 2021 at 08:02:37 UTC-3, WM wrote:
> > > [...]
> > > > Of course. But the definable endsegments are not all endsegments, because all definable endsegments have aleph_0 successors.
> > > Boohoo! In ZFC every natural number, and hence every end segment, is definable.
> > That is the reason why ZFC is inconsistent. It is an obviously wrong claim.
> > Proof: Every definable number has aleph_0 successors.
> > Not every number has aleph_0 successors because |N contains only numbers and if they are collectively subtracted from |N by |N \ |N, then nothing remains.
>
> >That is the reason why ZFC is inconsistent. It is an obviously wrong claim.
> WRONG!
>
> You have not once shown ANY contradiction in it!
> >Not every number has aleph_0 successors because |N contains only numbers and if they are collectively subtracted from |N by |N \ |N, then nothing remains.
> WRONG AGAIN!

Fact.

You are a believer in nonsense. But they have modified your brain such that you cannot understand. You can only parrot: "not once shown ANY contradiction". Your judgement is therefore irrelevant.

Regards, WM

Gus Gassmann schrieb am Montag, 16. August 2021 um 14:37:34 UTC+2:
> On Monday, 16 August 2021 at 09:07:16 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Montag, 16. August 2021 um 13:52:05 UTC+2:
> > > On Monday, 16 August 2021 at 08:02:37 UTC-3, WM wrote:
> > > [...]
> > > > Of course. But the definable endsegments are not all endsegments, because all definable endsegments have aleph_0 successors.
> > > Boohoo! In ZFC every natural number, and hence every end segment, is definable.
> > That is the reason why ZFC is inconsistent. It is an obviously wrong claim.
> > Proof: Every definable number has aleph_0 successors.
> > Not every number has aleph_0 successors because |N contains only numbers and if they are collectively subtracted from |N by |N \ |N, then nothing remains.
> For *EVERY* S you have S \ S = {}. For *EVERY* infinite set T you can't get there by subtracting elements one at a time.

Very good. It is not even possible by subtractingone one at every half time although the process would last less than two units of time. Now try to find out the reason.

Regards, WM

William schrieb am Montag, 16. August 2021 um 16:14:49 UTC+2:
> On Monday, August 16, 2021 at 6:57:37 AM UTC-4, WM wrote:
>
> > |∩{E(k) : k ∈ ℕ_def}| = ℵo .
> > Do you agree?
> You have defined |N_def to be a finite set.

Of course. Every definable element is the last one of a finite initial segment. How could it be else?

> Yes this is true. Trivial, but true. So what?

Let the other s know your insight.
>
> If we define "writable" in the usual manner (no resource bound) there are no "unwritable" elements of |N_F.

Then the "usual" way is wrong. It may "prove" that every n is writable as it "proves" that every ne set can be well-ordered.

> You may wish to insist upon such a bound but then the interesting set becomes the nr-writable numbers (the elements of |N_F which have the property that they can be written down except possibly for resource reasons). There are no elements of |N_F that are not nr-writable.

You see that you are wrong but you don't wish to confess.

Every definable natnumber is the last one of a FISON. Therefore an infinite endsegment follows. Two consecutive aleph_0 sets in |N are impossible. Therefore an infinite set of definable numbers is impossible by definition.

Regards, WM

On Monday, August 16, 2021 at 10:29:42 AM UTC-4, WM wrote:

>
> Every definable natnumber is the last one of a FISON.

There is no element of |N_F that is not the "last one of a FISION". However, |N_F is a Peano set with cardinality aleph_0

--
William Hughes

William schrieb am Montag, 16. August 2021 um 16:54:23 UTC+2:
> On Monday, August 16, 2021 at 10:29:42 AM UTC-4, WM wrote:
>
> > Every definable natnumber is the last one of a FISON.
> There is no element of |N_F that is not the "last one of a FISION".

That means every definable element is followed by infinitely many undefinable elements. Otherwise you could remove the definable elements such that no element remained.. Like you can do with the undefinable elements. But you can't with definable elements.

> However, |N_F is a Peano set with cardinality aleph_0

A Peano set has no cardinality because it is not a set but only a collection.

Regards, WM