On 4/6/2022 1:37 PM, WM wrote:
> William schrieb am Mittwoch, 6. April 2022 um 17:54:29 UTC+2:
>> On Wednesday, April 6, 2022 at 10:15:42 AM UTC-3, FromTheRafters wrote:
>
>>>> Every FISON has been unioned together with all others
>> Nope the union of any FISON with "all others" is infinite.
>
> Show at least one FISON that is in the union of all FISONs but not in any finite union of the FISONs in the sequence.
>
> Regards, WM

Proof you are wrong again;

All FISONs are finite sets of ordered natural numbers from 1 to some k. The FISON(k) is {1,2,3,...,k}

Since there is an infinite number of natural numbers, there are an infinite number of FISONs.

there is no FISON (a finite set) that is the union of all FISONs (an infinite set)

On 4/6/2022 1:39 PM, WM wrote:
> William schrieb am Mittwoch, 6. April 2022 um 17:44:16 UTC+2:
>> On Wednesday, April 6, 2022 at 10:00:20 AM UTC-3, WM wrote:
>>
>>> Every FISON has been unioned together with all others in the sequence already:
>> Nope the set "Every FISON" is infinite. Each element of the sequence is finite. "already" is nonsense.
>
> "Already" means unioned in the sequence. Find a FISON that is not unioned already in the sequence. Fail.
>
> Regards, WM

baloney; *use MATH equations to describe it*, or go post in alt.fake.math

On 4/6/2022 1:43 PM, WM wrote:
> FromTheRafters schrieb am Mittwoch, 6. April 2022 um 15:52:34 UTC+2:
>> WM wrote on 4/6/2022 :
>>> FromTheRafters schrieb am Mittwoch, 6. April 2022 um 15:20:47 UTC+2:
>>>> WM wrote :
>>>>> Jim Burns schrieb am Dienstag, 5. April 2022 um 22:58:24 UTC+2:
>>>>>> On 4/5/2022 2:06 PM, WM wrote:
>>>>>>
>>>>>>> It is only possible to express numbers finitely.
>>>>>> It is possible to make a claim which is true of
>>>>>> each one of infinitely-many.
>>>>>
>>>>> For instance this one: Every FISON that exists has been unioned together
>>>>> with all others in the sequence already:
>>>>>
>>>>> {1} = {1}
>>>>> {1} U {1, 2} = {1, 2}
>>>>> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
>>>>> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4,}
>>>>> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
>>>>> ...
>>>>>
>>>>> That holds for all the infinitely many FISONs of the sequence as the ...
>>>>> indicate.
>>>> So, FISONs are finite then?
>>>
>>> That is dictated by their name: Finite initial segment.
>> We're finally getting somewhere*. Why do you think things which apply
>> to finite objects must necessarily also apply to infinite ones like the
>> set of *all* such finite objects?
>
> All such objects satisfy this specification: They are finite. They are unions of all predecessors. Find a FISON that is in the union of all FISONs but not in the sequence. Fail.
>
> Regards, WM

the union of all FISONs is not a FISON. Don't you know that ?

On 4/6/2022 1:46 PM, WM wrote:
> Jim Burns schrieb am Mittwoch, 6. April 2022 um 16:36:03 UTC+2:
>> On 4/6/2022 9:05 AM, WM wrote:
>
>> We know that
>> a FISON equals the union of itself and all earlier
>> FISONs, but not any later FISON
>
> But we know that this holds also for all later FISONs. None is left out.
>>
>>> Another union cannot increase the result.
>> We know that
>> a FISON does not equals the union of all FISONs
>
> Find a FISON that violates my claim. Find a FISON in the union of all which is not unioned in the sequence.
>
> Regards WM

The union of all FISONs is not a FISON. google for it.

On Wednesday, April 6, 2022 at 3:37:18 PM UTC-3, WM wrote:

> Show at least one FISON that is in the union of all FISONs but not in any finite union of the FISONs in the sequence.

Piffle. As I have pointed out many times it is not a singleton that is not in a finite union, but an infininte subset.

--
William Hughes

On Wednesday, April 6, 2022 at 3:39:20 PM UTC-3, WM wrote:

> "Already" means unioned in the sequence. Find a FISON that is not unioned already in the sequence. Fail.

Find a subset of FISONs, K, such that K is not contained in a single element of the sequence, Succeed.

--
William Hughes

WM was thinking very hard :
> FromTheRafters schrieb am Mittwoch, 6. April 2022 um 15:52:34 UTC+2:
>> WM wrote on 4/6/2022 :
>>> FromTheRafters schrieb am Mittwoch, 6. April 2022 um 15:20:47 UTC+2:
>>>> WM wrote :
>>>>> Jim Burns schrieb am Dienstag, 5. April 2022 um 22:58:24 UTC+2:
>>>>>> On 4/5/2022 2:06 PM, WM wrote:
>>>>>>
>>>>>>> It is only possible to express numbers finitely.
>>>>>> It is possible to make a claim which is true of
>>>>>> each one of infinitely-many.
>>>>>
>>>>> For instance this one: Every FISON that exists has been unioned together
>>>>> with all others in the sequence already:
>>>>>
>>>>> {1} = {1}
>>>>> {1} U {1, 2} = {1, 2}
>>>>> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
>>>>> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4,}
>>>>> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4,
>>>>> 5} ...
>>>>>
>>>>> That holds for all the infinitely many FISONs of the sequence as the ...
>>>>> indicate.
>>>> So, FISONs are finite then?
>>>
>>> That is dictated by their name: Finite initial segment.
>> We're finally getting somewhere*. Why do you think things which apply
>> to finite objects must necessarily also apply to infinite ones like the
>> set of *all* such finite objects?
>
> All such objects satisfy this specification: They are finite. They are unions
> of all predecessors. Find a FISON that is in the union of all FISONs but not
> in the sequence. Fail.

I asked the why of it, not the fact of it.

onsdag 6 april 2022 kl. 14:58:23 UTC+2 skrev WM:
> William schrieb am Dienstag, 5. April 2022 um 21:33:11 UTC+2:
> > On Tuesday, April 5, 2022 at 3:32:59 PM UTC-3, WM wrote:
> > > potentially infinite set
> >
> > Piffle.
> >
> Every element that exists has been unioned together with all others in the sequence already:
> {1} = {1}
> {1} U {1, 2} = {1, 2}
> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4,}
> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
> ...
> That holds for all the infinitely many FISONs of the sequence as the ... indicate. Another union cannot increase the result. Learn: cannot cannot cannot!
>
> Regards, WM
The union of all those is N however, which is not in your list

onsdag 6 april 2022 kl. 15:10:44 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 6. April 2022 um 06:20:40 UTC+2:
> > tisdag 5 april 2022 kl. 19:59:44 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Dienstag, 5. April 2022 um 06:33:56 UTC+2:
> > > > måndag 4 april 2022 kl. 12:48:38 UTC+2 skrev WM:
> > > > > zelos...@gmail.com schrieb am Montag, 4. April 2022 um 09:48:17 UTC+2:
> > >
> > > > > > You assume only finite number of FISONs here, which is a false assumption
> > > > > No, less than ℵo FISONs is dictated by mathematics.
> > > >
> > > > And the union of all is N
> > > Fact is that
> > the union of ALL FISONs, which is INFINITELY MANY, is N
> If ℕ is potentially infinite.

N is infinite, there is no "potentially infinite" in mathematics.

>If however |ℕ| = ℵo

Which it is

>, then all FISONs are not enough

Empty assertion by a crank.

>because never a union of all FISONs is larger than all unions of FISONs contained in the sequence of FISONs:

False, you meant to say a FINITE union of FISONs. The union of ALL FISONs, which is an INFINITE union, is N

> > > {1} = {1}
> > > {1} U {1, 2} = {1, 2}
> > > {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> > > {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4,}
> > > {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
> > > ...
> > Repeating wrong stuff doesn't make it more right.
> Why then do you?

I don't repeat wrong stuff, YOU are doing it.

>How should ℵo result if all unios of existing FISONs are in the sequence of FISONs and are therefore finite?

All FINITE unions you crank, I say union of ALL INFINITELY MANY FISONs is N

> > Union of all FISONs is N
> Repeating wrong stuff doesn't make it right.

Nope, so shut up already.

Union of all FISONs is N and it is trivial to prove.

>
> Regards, WM

On 4/6/2022 2:46 PM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 6. April 2022 um 16:36:03 UTC+2:
>> On 4/6/2022 9:05 AM, WM wrote:

>> We know that
>> a FISON equals the union of itself and all earlier
>> FISONs, but not any later FISON
>
> But we know that this holds also for all later FISONs.
> None is left out.

A FISON which is equal to the union of
itself and earlier FISONs *but not later FISONs*
is not the union of all FISONs

We know that
a FISON does not equals the union of all FISONs
....a claim true of one of infinitely-many FISONs
without needing to choose which one.

>>>> If
>>>> there is always some finite number k for which
>>>> k replications of distance y passes distance x, k*y > x
>>>> then
>>>
>>> this is given by a finite formula:
>>> "there is always some finite number k for which
>>> k replications of distance y passes distance x, k*y > x"
>>> with k and y to be chosen.
>>
>> Choosing is not needed.
>>
>> We know that,
>> for distances x and y,
>> a finite number k exists such that k*y > x
>>
>> ...a claim true of two distances
>> without needing to choose which two.

>>> Another union cannot increase the result.
>>
>> We know that
>> a FISON does not equals the union of all FISONs
>
> Find a FISON that violates my claim.

You're referring to your claim that
a FISON exists equal to the union of all FISONs?

That's an existential claim.
Existential claims are not disproved by counter-examples.

There are two opposing claims.

| A FISON is equal to the union of all FISONs.

| A FISON is not equal to the union of all FISONs
| and it's not needed to choose which FISON.

Any FISON equal to the union of all FISONs
is enough to prove the first claim and
disprove the second claim.

It's not symmetric, though.

A FISON _not_ equal to the union of all FISONs
is _not_ enough to prove the second claim and
disprove the first claim.

Providing that FISON typically involves choosing.
So, it wouldn't show that choosing isn't needed.

A typical strategy is to start with a _complete_
claim, one for which choosing isn't needed
(what is usually called _valid_ ),
and then preserve that completeness through a
sequence of claims ending at the second claim.

A good choice for a complete claim to start with is
| A FISON is a collection with a counting-order
| which begins at 1 and ends somewhere
| and it's not needed to choose which FISON.

We know that claim, but not by seeing an example.
Not that there aren't plenty of examples.
It says what we mean by FISON, and we know that.

It will only be true of FISONS,
but here we are only concerned with FISONs

William schrieb am Mittwoch, 6. April 2022 um 21:11:56 UTC+2:
> On Wednesday, April 6, 2022 at 3:37:18 PM UTC-3, WM wrote:
>
> > Show at least one FISON that is in the union of all FISONs but not in any finite union of the FISONs in the sequence.
> Piffle. As I have pointed out many times it is not a singleton that is not in a finite union, but an infininte subset.

If there is not even any FISON not in a finite union, then you are wrong with an infinite subset.

Better you can see dark numbers with the complementary endsegments E(n) = {n, n+1, n+2, ...}. The sequence of endsegments is inclusion-monotonic. All endsegments are infinite. Hence their intersection is infinite. But you cannot find any natnumber remaining. So you say the intersection is empty. From the mathematical point of view this would be nonense.

Regards, WM

FromTheRafters schrieb am Mittwoch, 6. April 2022 um 21:31:52 UTC+2:
> WM was thinking very hard :

> > All such objects satisfy this specification: They are finite. They are unions
> > of all predecessors. Find a FISON that is in the union of all FISONs but not
> > in the sequence. Fail.
> I asked the why of it, not the fact of it.

The reason is this: All FISONs are finite by definition. All unions of FISONs are unions of finitely many FISONs, because by the pigeonhole principle, only finitely many can be distinct by finitely many symbols.

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 7. April 2022 um 06:46:30 UTC+2:
> onsdag 6 april 2022 kl. 15:10:44 UTC+2 skrev WM:

> >How should ℵo result if all unios of existing FISONs are in the sequence of FISONs and are therefore finite?
>
> All FINITE unions, I say union of ALL INFINITELY MANY FISONs is N

There are not ℵo FISONs without an infinite FISON. Therefore there is no ℵo-union.

> Union of all FISONs is N and it is trivial to prove.
>
It is trivial to prove that ZF is self-contradictory.

Regards, WM

Jim Burns schrieb am Donnerstag, 7. April 2022 um 16:27:27 UTC+2:
> On 4/6/2022 2:46 PM, WM wrote:
> > Jim Burns schrieb
> > am Mittwoch, 6. April 2022 um 16:36:03 UTC+2:
> >> On 4/6/2022 9:05 AM, WM wrote:
>
> >> We know that
> >> a FISON equals the union of itself and all earlier
> >> FISONs, but not any later FISON
> >
> > But we know that this holds also for all later FISONs.
> > None is left out.
> A FISON which is equal to the union of
> itself and earlier FISONs *but not later FISONs*
> is not the union of all FISONs

But since the sequence contains all unions of FISONs, it contains the union of all FISONs.
> We know that
> a FISON does not equals the union of all FISONs

But we know that the union of all FISONs does not contain more than is unioned already in the sequence.

> ...a claim true of one of infinitely-many FISONs
> without needing to choose which one.

So it is.

> > Find a FISON that violates my claim.
> You're referring to your claim that
> a FISON exists equal to the union of all FISONs?

No, there is no such FISON because of potential infinity. There are all unions in the sequence already. That is my claim.
>
> That's an existential claim.

All FISONs are unioned in the sequence would be disproved by a FISON not in any union.

> A typical strategy is to start with a _complete_
> claim, one for which choosing isn't needed

See above. My claim is proved by the impossibility to find such a FISON and by the inclusion-monotony of FISONs.

But even simpler is this: If the union of FISONs is ℕ, then the intersection of endsegments is empty. But all endsegments are infinite and inclusion-monotonic. Contradiction.

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 7. April 2022 um 06:43:30 UTC+2:
> onsdag 6 april 2022 kl. 14:58:23 UTC+2 skrev WM:

> > Every element that exists has been unioned together with all others in the sequence already:
> > {1} = {1}
> > {1} U {1, 2} = {1, 2}
> > {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> > {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4,}
> > {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
> > ...
> > That holds for all the infinitely many FISONs of the sequence as the .... indicate. Another union cannot increase the result. Learn: cannot cannot cannot!

> The union of all those is N however, which is not in your list

If the union of FISONs is ℕ, then the intersection of endsegments is empty. But all endsegments are infinite and inclusion-monotonic. Contradiction.

Regards, WM

On 4/7/2022 11:37 AM, WM wrote:
> William schrieb am Mittwoch, 6. April 2022 um 21:11:56 UTC+2:
>> On Wednesday, April 6, 2022 at 3:37:18 PM UTC-3, WM wrote:
>>
>>> Show at least one FISON that is in the union of all FISONs but not in any finite union of the FISONs in the sequence.
>> Piffle. As I have pointed out many times it is not a singleton that is not in a finite union, but an infininte subset.
>
> If there is not even any FISON not in a finite union, then you are wrong with an infinite subset.
>
> Better you can see dark numbers with the complementary endsegments E(n) = {n, n+1, n+2, ...}. The sequence of endsegments is inclusion-monotonic. All endsegments are infinite. Hence their intersection is infinite. But you cannot find any natnumber remaining. So you say the intersection is empty. From the mathematical point of view this would be nonense.
>
> Regards, WM

Deal with it.

On Thursday, April 7, 2022 at 1:37:51 PM UTC-3, WM wrote:
> William schrieb am Mittwoch, 6. April 2022 um 21:11:56 UTC+2:
> > On Wednesday, April 6, 2022 at 3:37:18 PM UTC-3, WM wrote:
> >
> > > Show at least one FISON that is in the union of all FISONs but not in any finite union of the FISONs in the sequence.
> > Piffle. As I have pointed out many times it is not a singleton that is not in a finite union, but an infininte subset.
> If there is not even any FISON not in a finite union, then you are wrong with an infinite subset.

Piffle. It is trivial to note the every FISON is a finite union. It is trivial to see that an infinite set of FISONs is not in a finite union.

--
William Hughes

On Thursday, April 7, 2022 at 2:23:08 PM UTC-3, WM wrote:

> If the union of FISONs is ℕ, then the intersection of endsegments is empty.

Correct, Indeed both statements are trivially true
(The sequence of endsegments has no last element).

--
William Hughes

On Thursday, April 7, 2022 at 6:41:33 PM UTC+2, WM wrote:
>
> All unions of FISONs are unions of finitely many FISONs

No, you psychotic asshole full of shit, the union off *all* FISONs is an union of infinitely many FISONs, because there are infinitely many FISONs.

You are dumb like shit.

On Thursday, April 7, 2022 at 6:41:33 PM UTC+2, WM wrote:
>
> All unions of FISONs are unions of finitely many FISONs

No, you psychotic asshole full of shit, the union of *all* FISONs is an union of infinitely many FISONs, because there are infinitely many FISONs.

You are dumb like shit.

On Thursday, April 7, 2022 at 7:22:54 PM UTC+2, WM wrote:
>
> But since the sequence contains all unions of FISONs

Nope. It does not contain the union of *all* FISONs.

> If the union of FISONs is ℕ, then the intersection of endsegments is empty.

Well, the union of all FISONs is ℕ and the intersection of all endsegments is empty.

So your implication is indeed true. (Since T -> T == T.)

WM wrote :
> William schrieb am Mittwoch, 6. April 2022 um 21:11:56 UTC+2:
>> On Wednesday, April 6, 2022 at 3:37:18 PM UTC-3, WM wrote:
>>
>>> Show at least one FISON that is in the union of all FISONs but not in any
>>> finite union of the FISONs in the sequence.
>> Piffle. As I have pointed out many times it is not a singleton that is not
>> in a finite union, but an infininte subset.
>
> If there is not even any FISON not in a finite union, then you are wrong with
> an infinite subset.
>
> Better you can see dark numbers with the complementary endsegments E(n) = {n,
> n+1, n+2, ...}. The sequence of endsegments is inclusion-monotonic.

Each successive endsegmment in your sequence excludes the element "n-1"
which the previous endsegment had. How does this make the sequence
*inclusion* monotonic?

torsdag 7 april 2022 kl. 18:41:33 UTC+2 skrev WM:
> FromTheRafters schrieb am Mittwoch, 6. April 2022 um 21:31:52 UTC+2:
> > WM was thinking very hard :
> > > All such objects satisfy this specification: They are finite. They are unions
> > > of all predecessors. Find a FISON that is in the union of all FISONs but not
> > > in the sequence. Fail.
> > I asked the why of it, not the fact of it.
> The reason is this: All FISONs are finite by definition. All unions of FISONs are unions of finitely many FISONs,

False, You can take the union of INFINITELY many FISONs.

>because by the pigeonhole principle, only finitely many can be distinct by finitely many symbols.
>
> Regards, WM

torsdag 7 april 2022 kl. 19:12:02 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 7. April 2022 um 06:46:30 UTC+2:
> > onsdag 6 april 2022 kl. 15:10:44 UTC+2 skrev WM:
>
> > >How should ℵo result if all unios of existing FISONs are in the sequence of FISONs and are therefore finite?
> >
> > All FINITE unions, I say union of ALL INFINITELY MANY FISONs is N
>
> There are not ℵo FISONs without an infinite FISON.

False, that is a pure non-sequitor.

>Therefore there is no ℵo-union.
> > Union of all FISONs is N and it is trivial to prove.
> >
> It is trivial to prove that ZF is self-contradictory.

Yet you have never succeeded.

>
> Regards, WM

It happens that WM formulated :
> zelos...@gmail.com schrieb am Donnerstag, 7. April 2022 um 06:46:30 UTC+2:
>> onsdag 6 april 2022 kl. 15:10:44 UTC+2 skrev WM:
>
>>> How should ℵo result if all unios of existing FISONs are in the sequence of
>>> FISONs and are therefore finite?
>>
>> All FINITE unions, I say union of ALL INFINITELY MANY FISONs is N
>
> There are not ℵo FISONs without an infinite FISON.

There you go with your infinite FISONs again. Did you forget the
definition of FISON?