On 7/3/2022 10:32 AM, WM wrote:
> zelos...@gmail.com schrieb am Sonntag, 3. Juli 2022 um 16:55:34 UTC+2:
>> söndag 3 juli 2022 kl. 15:08:01 UTC+1 skrev WM:
>>> William schrieb am Samstag, 2. Juli 2022 um 20:14:21 UTC+2:
>>>> On Saturday, July 2, 2022 at 3:02:01 PM UTC-3, WM wrote:
>>>>> William schrieb am Samstag, 2. Juli 2022 um 19:47:26 UTC+2:
>>>>>> . You do not say that something is true for an infinite set.
>>>>> I say that it is true for [each] natural number, none of which is infinite.
>>>>
>>>> Correct. You say that something is true for the elements of |N.
>>>
>>> For all elements of ℕ!
>>>
>>>> You do not say that something is true for |N
>>>
>>> Of course not. ℕ does not appear in
>>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
>> this is still false
>
> If it was, then you could find a first definable number k with E(k) being finite. You cannot, can you?
>
> Regards, WM

"definable" is meaningless, so your question is meaningless.

On Sunday, July 3, 2022 at 12:37:50 PM UTC-3, WM wrote:
> William schrieb am Sonntag, 3. Juli 2022 um 16:55:45 UTC+2:
> > On Sunday, July 3, 2022 at 11:32:04 AM UTC-3, WM wrote:
>
> > > > > For [each] element of ℕ!
> > > >
> > > > Correct, but not for the set |N.
> > > ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k} (#)
> > An infinite number of statements, (each statement about an element of |N)
> Yes, one statement for every natnumber. There are infinitely many natnumbers. That implies infinitely many statements.
> >
> > There is no contradiction with the statement
> > {∩E(k) : k ∈ ℕ} = { }
> > which is one statement about the set |N.
> No, it is about the elements of ℕ.

It is one statement about one collection.

--
William Hughes

söndag 3 juli 2022 kl. 16:29:10 UTC+1 skrev WM:
> William schrieb am Sonntag, 3. Juli 2022 um 16:32:09 UTC+2:
> > On Sunday, July 3, 2022 at 11:23:43 AM UTC-3, WM wrote:
> > > Jim Burns schrieb am Sonntag, 3. Juli 2022 um 00:10:53 UTC+2:
> >
> >
> > > > each element of ℕ can be reached by
> > > > a terminating process of '+1' steps.
> > >
> > > Then try to reach elements with less than ℵ₀ successors.
> > This cannot be done. So what? Each *element* of |N has ℵ₀ successors each of which can be reached by a terminating process of '+1' steps. The *set* |N cannot be reached by a terminating process of '+1' steps.
> The reason is that ℕ consists mainly of dark numbers. Note: ℕ is infinite but every reached number belongs to a comparatively tiny finite set.
>
> It is silly to claim that all natnumbers can be defined whereas almost all natnumbers are undefined successors of defined natnumbers.
>
> Regards, WM
THERE ARE NO DARK NUMBERS!

you cannot prove they exist because they don't!

söndag 3 juli 2022 kl. 16:32:05 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Sonntag, 3. Juli 2022 um 16:55:34 UTC+2:
> > söndag 3 juli 2022 kl. 15:08:01 UTC+1 skrev WM:
> > > William schrieb am Samstag, 2. Juli 2022 um 20:14:21 UTC+2:
> > > > On Saturday, July 2, 2022 at 3:02:01 PM UTC-3, WM wrote:
> > > > > William schrieb am Samstag, 2. Juli 2022 um 19:47:26 UTC+2:
> > > > >>. You do not say that something is true for an infinite set.
> > > > > I say that it is true for [each] natural number, none of which is infinite.
> > > >
> > > > Correct. You say that something is true for the elements of |N.
> > >
> > > For all elements of ℕ!
> > >
> > > > You do not say that something is true for |N
> > >
> > > Of course not. ℕ does not appear in
> > > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> > this is still false
> If it was, then you could find a first definable number k with E(k) being finite. You cannot, can you?
>
> Regards, WM

False again, your statement is false because N_def=N and the intersection is empty.

I do not need to find a number that is such for it to be true. Stop being retarded

söndag 3 juli 2022 kl. 16:40:50 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Sonntag, 3. Juli 2022 um 16:57:00 UTC+2:
> > söndag 3 juli 2022 kl. 15:23:43 UTC+1 skrev WM:
> > > Jim Burns schrieb am Sonntag, 3. Juli 2022 um 00:10:53 UTC+2:
> > > > On 7/2/2022 12:42 PM, WM wrote:
> > >
> > > > > For all definable numbers we obtain
> > > > > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> > > > No, you don't obtain that.
> > > >
> > > > That definition is ill-formed.
> > >
> > > |∩{E(k) : k <10}| = ℵ₀
> > > |∩{E(k) : k <100}| = ℵ₀
> > > |∩{E(k) : k <10^10}| = ℵ₀
> > > |∩{E(k) : k <10^10^10}| = ℵ₀
> > > and so on for all definable numbers k.
> > no, this goes on for ALL natural numbers k, it does not matter which one.
> Then this does not go on for all natnumbers?
> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}

Yes, but it does not go where you want it to.

> > >
> > > Can you find an exception?
> > nope because it applies to ALL natural numbers. "definable" has no definition and is not needed as it is ALL NATURAL NUMBERS!
> Then explain how this fails:
> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}

It doesn't, the issue is that this property is not powerful enough to give what you want.

> > >
> > > I use only such endsegments which satisfy
> > > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀.
>
> > Nope, you juse ALL endsegments.
> why should I? I use only such endsegments which satisfy

There is no "Why should I", there is only that you DO!

> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀.

FALSE!

|∩{E(k) : k ∈ ℕ_def}| = 0

From your definition of N_def because N_def=N from your definition!

You are such a fucking retarded crank

> Regards, WM

On Sunday, July 3, 2022 at 12:44:45 PM UTC-3, WM wrote:
> William schrieb am Sonntag, 3. Juli 2022 um 17:02:07 UTC+2:
> > On Sunday, July 3, 2022 at 11:42:46 AM UTC-3, WM wrote:
> > > not reached
> >
> > "reached" is another one of your many many many ways of saying "can be written down".
> No.

Yes. That is what you use it to mean

> > So what if the ℵ₀ successors cannot be written down. Each of the successors has the property that it is the outcome of a terminating process of '+1' steps.
> No, then all would be used up, none would remain.

Nope. I said that each of them [without exception] not "all of them". For *each* of them [without exception] ℵ₀ successors remain.

> But what is wrong here?
>
> 1, ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
> 2, {∩E(k) : k ∈ ℕ} = { }
Nothing. Each of the infinite number of statements in 1. is true. The statement 2. is true, Each of the statements of 1. is consistent with 2.

--
William Hughes

On 7/3/2022 10:23 AM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 3. Juli 2022 um 00:10:53 UTC+2:
>> On 7/2/2022 12:42 PM, WM wrote:

>>> For all definable numbers we obtain
>>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
>>
>> No, you don't obtain that.
>>
>> That definition is ill-formed.
>
> |∩{E(k) : k <10}| = ℵ₀
> |∩{E(k) : k <100}| = ℵ₀
> |∩{E(k) : k <10^10}| = ℵ₀
> |∩{E(k) : k <10^10^10}| = ℵ₀
> and so on for all definable numbers k.
>
> Can you find an exception?

So, you want to say that ℕ_def is
the collection of all m such that
|∩{E(k) : k < m }| = ℵ₀

|∩{E(k) : k ∈ ℕ_def}| = ℵ₀
is not how one says that.
How one says that (in one way) is
ℕ_def =
{ m ∈ ℕ : |∩{E(k) : k < m }| = ℵ₀ }

>> If we correct your definition to what you say,
>> then you obtain |∩{E(k) : k ∈ ℕ_def}| = 0
>
> That is no correction but a misleading change,
> because I use only such endsegments which satisfy
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀.

The intersection of all the end segments
_which you intend to use_ is empty.

{E(k) : k <10} does not have all the end segments.
{E(k) : k <100} does not have all the end segments.
{E(k) : k <10^10} does not have all the end segments.
{E(k) : k <10^10^10} does not have all the end segments.

The intersection of _all_ the end segments is empty.
|⋂{ E(j) : | j ∈ {m∈ℕ:|∩{E(k):k<m}|=ℵ₀} )| = 0

>> As a consequence,
>> each element of ℕ can be reached by
>> a terminating process of '+1' steps.
>
> Then try to reach elements with less than
> ℵ₀ successors.

Each element of ℕ has |ℕ|-many successors in ℕ.

For each end segment,
there are |ℕ|-many element in common with
each _on the way_ to that end segment.

Each element of ℕ has fewer than |ℕ|-many
predecessors in ℕ.

Each element is only in end segments _on the way_
to its end segment,
which are fewer than all the end segments,
so not-in some of them,
so not-in the intersection of all.

Infinity is not
a reallyreallyreallyreallyreallyreally
large number.

sergi o used his keyboard to write :
> On 7/3/2022 10:32 AM, WM wrote:
>> zelos...@gmail.com schrieb am Sonntag, 3. Juli 2022 um 16:55:34 UTC+2:
>>> söndag 3 juli 2022 kl. 15:08:01 UTC+1 skrev WM:
>>>> William schrieb am Samstag, 2. Juli 2022 um 20:14:21 UTC+2:
>>>>> On Saturday, July 2, 2022 at 3:02:01 PM UTC-3, WM wrote:
>>>>>> William schrieb am Samstag, 2. Juli 2022 um 19:47:26 UTC+2:
>>>>>>> . You do not say that something is true for an infinite set.
>>>>>> I say that it is true for [each] natural number, none of which is
>>>>>> infinite.
>>>>>
>>>>> Correct. You say that something is true for the elements of |N.
>>>>
>>>> For all elements of ℕ!
>>>>
>>>>> You do not say that something is true for |N
>>>>
>>>> Of course not. ℕ does not appear in
>>>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
>>> this is still false
>>
>> If it was, then you could find a first definable number k with E(k) being
>> finite. You cannot, can you?
>>
>> Regards, WM
>
> "definable" is meaningless, so your question is meaningless.

That and the endsegments are defined as not finite.

Jim Burns schrieb am Sonntag, 3. Juli 2022 um 19:43:14 UTC+2:
> On 7/3/2022 10:23 AM, WM wrote:
> > Jim Burns schrieb
> > am Sonntag, 3. Juli 2022 um 00:10:53 UTC+2:
> >> On 7/2/2022 12:42 PM, WM wrote:
>
> >>> For all definable numbers we obtain
> >>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> >>
> >> No, you don't obtain that.
> >>
> >> That definition is ill-formed.
> >
> > |∩{E(k) : k <10}| = ℵ₀
> > |∩{E(k) : k <100}| = ℵ₀
> > |∩{E(k) : k <10^10}| = ℵ₀
> > |∩{E(k) : k <10^10^10}| = ℵ₀
> > and so on for all definable numbers k.
> >
> > Can you find an exception?
> So, you want to say that ℕ_def is
> the collection of all m such that
> |∩{E(k) : k < m }| = ℵ₀
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> is not how one says that.

It is how one says that.

> How one says that (in one way) is
> ℕ_def =
> { m ∈ ℕ : |∩{E(k) : k < m }| = ℵ₀ }

Say it that way if you like.

> >> If we correct your definition to what you say,
> >> then you obtain |∩{E(k) : k ∈ ℕ_def}| = 0
> >
> > That is no correction but a misleading change,
> > because I use only such endsegments which satisfy
> > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀.
> The intersection of all the end segments
> _which you intend to use_ is empty.

You are a liar. The intersection |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ is infinite by definition.
>
> {E(k) : k <10} does not have all the end segments.
> {E(k) : k <100} does not have all the end segments.
> {E(k) : k <10^10} does not have all the end segments.
> {E(k) : k <10^10^10} does not have all the end segments.
>
> The intersection of _all_ the end segments is empty.
> |⋂{ E(j) : | j ∈ {m∈ℕ:|∩{E(k):k<m}|=ℵ₀} )| = 0

Then ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k} would be violated.
>
> >> As a consequence,
> >> each element of ℕ can be reached by
> >> a terminating process of '+1' steps.
> >
> > Then try to reach elements with less than
> > ℵ₀ successors.
> Each element of ℕ has |ℕ|-many successors in ℕ.

Each definable element. That are those which I collect in ℕ_def.
But if ℕ_def = ℕ, then you do not cover each element. The successors are missing.
>
> For each end segment,
> there are |ℕ|-many element in common with
> each _on the way_ to that end segment.

On the way to ∩{E(k) : k ∈ ℕ} = { } there is a hurdle:

∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}

It prevents the empty intersection without finite intersections.

Regards, WM

On 7/3/2022 4:35 PM, WM wrote:
> Jim Burns schrieb am Sonntag, 3. Juli 2022 um 19:43:14 UTC+2:
>> On 7/3/2022 10:23 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Sonntag, 3. Juli 2022 um 00:10:53 UTC+2:
>>>> On 7/2/2022 12:42 PM, WM wrote:
>>
>>>>> For all definable numbers we obtain
>>>>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
>>>>
>>>> No, you don't obtain that.
>>>>
>>>> That definition is ill-formed.
>>>
>>> |∩{E(k) : k <10}| = ℵ₀
>>> |∩{E(k) : k <100}| = ℵ₀
>>> |∩{E(k) : k <10^10}| = ℵ₀
>>> |∩{E(k) : k <10^10^10}| = ℵ₀
>>> and so on for all definable numbers k.
>>>
>>> Can you find an exception?
>> So, you want to say that ℕ_def is
>> the collection of all m such that
>> |∩{E(k) : k < m }| = ℵ₀
>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
>> is not how one says that.
>
> It is how one says that.
>
>> How one says that (in one way) is
>> ℕ_def =
>> { m ∈ ℕ : |∩{E(k) : k < m }| = ℵ₀ }
>
> Say it that way if you like.
>
>>>> If we correct your definition to what you say,
>>>> then you obtain |∩{E(k) : k ∈ ℕ_def}| = 0
>>>
>>> That is no correction but a misleading change,
>>> because I use only such endsegments which satisfy
>>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀.
>> The intersection of all the end segments
>> _which you intend to use_ is empty.
>
> You are a liar. The intersection |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ is infinite by definition.

no, not by definition, by Math. If k is always finite, then ℵ₀ .

>>
>> {E(k) : k <10} does not have all the end segments.
>> {E(k) : k <100} does not have all the end segments.
>> {E(k) : k <10^10} does not have all the end segments.
>> {E(k) : k <10^10^10} does not have all the end segments.
>>
>> The intersection of _all_ the end segments is empty.
>> |⋂{ E(j) : | j ∈ {m∈ℕ:|∩{E(k):k<m}|=ℵ₀} )| = 0
>
> Then ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k} would be violated.

that simply reduces to E(k+1) = E(k) \ {k}, or the relationship between adjacent endsegments.

Nothing gets violated.

>>
>>>> As a consequence,
>>>> each element of ℕ can be reached by
>>>> a terminating process of '+1' steps.
>>>
>>> Then try to reach elements with less than
>>> ℵ₀ successors.
>> Each element of ℕ has |ℕ|-many successors in ℕ.
>
> Each definable element. That are those which I collect in ℕ_def.

which shows ℕ_def is undefined, and unusable

> But if ℕ_def = ℕ, then you do not cover each element. The successors are missing.

nope.

>>
>> For each end segment,
>> there are |ℕ|-many element in common with
>> each _on the way_ to that end segment.
>
> On the way to ∩{E(k) : k ∈ ℕ} = { } there is a hurdle:

1. there is no "On way to" it is simply an equation, there is no hurdle, Miss Director

2. It is not an endsegment either.

>
> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}

which is;

3. E(k+1) = E(k) \ {k}

>
> It prevents the empty intersection without finite intersections.

Wrong. above 1,2,3 are all just fine with each other.

your counting rocks math, step by step, has led you astray, or you are playing us for suckers.

>
> Regards, WM
>
>

Wm

Polite honest serious brief question.

Is there any chance that you might be able to come up with the smallest
consistent list of axioms that forms the foundation and distinguishes
your math from what other people are using?
And get everyone to agree to preface all future posts with
"Assuming this list of axioms:..."?

I'm wondering whether those axioms might make some of your claims
much simpler to present and support, might avoid some errors, might
avoid some claims of errors, might make some look at that preface and
think they don't accept those axioms and thus have nothing to dispute,

Like any other foundation, that would have to be really really really well
thought out checked to see that it doesn't include any errors or contradictions.
And do all that substantial work before the foundation is presented.

If the answer is "no" then I would understand completely.
I was only hoping to find some way quit the endless arguments about
things that do not seem to really address the deeper questions.

Thank you

söndag 3 juli 2022 kl. 22:35:54 UTC+1 skrev WM:
> Jim Burns schrieb am Sonntag, 3. Juli 2022 um 19:43:14 UTC+2:
> > On 7/3/2022 10:23 AM, WM wrote:
> > > Jim Burns schrieb
> > > am Sonntag, 3. Juli 2022 um 00:10:53 UTC+2:
> > >> On 7/2/2022 12:42 PM, WM wrote:
> >
> > >>> For all definable numbers we obtain
> > >>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> > >>
> > >> No, you don't obtain that.
> > >>
> > >> That definition is ill-formed.
> > >
> > > |∩{E(k) : k <10}| = ℵ₀
> > > |∩{E(k) : k <100}| = ℵ₀
> > > |∩{E(k) : k <10^10}| = ℵ₀
> > > |∩{E(k) : k <10^10^10}| = ℵ₀
> > > and so on for all definable numbers k.
> > >
> > > Can you find an exception?
> > So, you want to say that ℕ_def is
> > the collection of all m such that
> > |∩{E(k) : k < m }| = ℵ₀
> > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> > is not how one says that.
> It is how one says that.

No it isn't you fucking retard! This demonstrates that you do not understand mathematical notation!

> > How one says that (in one way) is
> > ℕ_def =
> > { m ∈ ℕ : |∩{E(k) : k < m }| = ℵ₀ }
> Say it that way if you like.

Say it as IT SHOULD BE SAID!

> > >> If we correct your definition to what you say,
> > >> then you obtain |∩{E(k) : k ∈ ℕ_def}| = 0
> > >
> > > That is no correction but a misleading change,
> > > because I use only such endsegments which satisfy
> > > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀.
> > The intersection of all the end segments
> > _which you intend to use_ is empty.
> You are a liar. The intersection |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ is infinite by definition.

Nope, you do not define it as such at all! You are the liar here.

> >
> > {E(k) : k <10} does not have all the end segments.
> > {E(k) : k <100} does not have all the end segments.
> > {E(k) : k <10^10} does not have all the end segments.
> > {E(k) : k <10^10^10} does not have all the end segments.
> >
> > The intersection of _all_ the end segments is empty.
> > |⋂{ E(j) : | j ∈ {m∈ℕ:|∩{E(k):k<m}|=ℵ₀} )| = 0
> Then ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k} would be violated.

Nope, you are ONCE AGAIN demonstrating you do not understand mathematics.

> >
> > >> As a consequence,
> > >> each element of ℕ can be reached by
> > >> a terminating process of '+1' steps.
> > >
> > > Then try to reach elements with less than
> > > ℵ₀ successors.
> > Each element of ℕ has |ℕ|-many successors in ℕ.
> Each definable element. That are those which I collect in ℕ_def.

"Definable" has no meaning!

> But if ℕ_def = ℕ, then you do not cover each element. The successors are missing.

By all your definitions, N_def=N, you cannot get around this!

> >
> > For each end segment,
> > there are |ℕ|-many element in common with
> > each _on the way_ to that end segment.
> On the way to ∩{E(k) : k ∈ ℕ} = { } there is a hurdle:
> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
> It prevents the empty intersection without finite intersections.

It does not you fucking retard! This proves yet again that you do not understand mathematical definitions!

>
> Regards, WM

William schrieb am Sonntag, 3. Juli 2022 um 18:15:46 UTC+2:
> On Sunday, July 3, 2022 at 12:37:50 PM UTC-3, WM wrote:

> > > There is no contradiction with the statement
> > > {∩E(k) : k ∈ ℕ} = { }
> > > which is one statement about the set |N.
> > No, it is about the elements of ℕ.
> It is one statement about one collection.

Above you can read: k ∈ ℕ. That means if all endsegments E(k) defined by an element k of ℕ are intersected, then the result is the empty set. There is no endsegment E(ℕ).

∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵ₀ is also a statement for all elements of ℕ. Accordingf to you it is about the set ℕ.

But let us start from the fact that you accept
|{∩E(k) : k =< mn ∈ ℕ}| = ℵ₀ .
This says that until *every* natural number n the intersection of endsegments is ℵ₀-infinite. {∩E(k) : k ∈ ℕ} = { }
says that the intersection over *all* endsegments is empty:

But no-one can say how the ℵ₀ elements are lost between every and all. It happens in an uncontrollable way, whether or not we accept the basic equation
∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
I call this state of ignorance dark. You may call it different, but you accept it.

Regards, WM

William schrieb am Sonntag, 3. Juli 2022 um 18:29:34 UTC+2:
> On Sunday, July 3, 2022 at 12:44:45 PM UTC-3, WM wrote:

> > But what is wrong here?
> >
> > 1, ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
> > 2, {∩E(k) : k ∈ ℕ} = { }
>
> Nothing. Each of the infinite number of statements in 1. is true. The statement 2. is true, Each of the statements of 1. is consistent with 2.

According to (1) the empty set in (2) cannot be realized without a finite predecessor.

Regards, WM

William schrieb am Sonntag, 3. Juli 2022 um 18:29:34 UTC+2:

> I said that each of them [without exception] not "all of them". For *each* of them [without exception] ℵ₀ successors remain.

For all of them they are gone. This change cannot be investigated.
The change from E(n) to E(n+1) can be investigated for definable n.

Regards, WM

On Monday, July 4, 2022 at 2:55:40 PM UTC+2, WM wrote:
> William schrieb am Sonntag, 3. Juli 2022 um 18:15:46 UTC+2:
> > On Sunday, July 3, 2022 at 12:37:50 PM UTC-3, WM wrote:
> > > >
> > > > There is no contradiction with the statement
> > > > {∩E(k) : k ∈ ℕ} = { }

But there's a typo in it. It should read:
∩{E(k) : k ∈ ℕ} = { }

> > It is one statement about one collection.

Well, it states that the set ∩{E(k) : k ∈ ℕ} equals the set { }. In other words, that the set ∩{E(k) : k ∈ ℕ} is identical with the set { }.

Hint: Frege considered such statements in detail in his seminal article: "Über Sinn und Bedeutung".
See: https://en.wikipedia.org/wiki/Sense_and_reference

> Above you can read: k ∈ ℕ. That means ...

No, THAT doesn't mean anything "in isolation".

> if all endsegments E(k) [with] k of ℕ are intersected, then the result is the empty set.

Yeah, "∩{E(k) : k ∈ ℕ} = { }" can be INTERPRETED that way, right.

> <nonsense deleted>

On Monday, July 4, 2022 at 10:00:56 AM UTC-3, WM wrote:
> William schrieb am Sonntag, 3. Juli 2022 um 18:29:34 UTC+2:
>
> > I said that each of them [without exception] not "all of them". For *each* of them [without exception] ℵ₀ successors remain.
> For all of them they are gone.

The statement "For 'all of them', the colleciton, ℵ₀ successors remain." is not made. The statement that is made is: For *each* of them [without exception] ℵ₀ successors remain.

--
William Hughes

On Monday, July 4, 2022 at 2:58:04 PM UTC+2, WM wrote:
> William schrieb am Sonntag, 3. Juli 2022 um 18:29:34 UTC+2:
> > On Sunday, July 3, 2022 at 12:44:45 PM UTC-3, WM wrote:
> > >
> > > But what is wrong here?
> > >
> > > 1, ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}

1. can be expressed the following way too - i. e. MUCH SIMPLER:

1', ∀k ∈ ℕ : E(k+1) = E(k) \ {k} .

A trivial fact. Now

> > > 2, ∩{E(k) : k ∈ ℕ} = { }

can easily be derived from 1'.

Since 1' implies:

∀k ∈ ℕ: En ∈ ℕ: k !e E(n). (For each and every natural number there's an ensegement not containing it.)

Hence

∀k ∈ ℕ: k !e ∩{E(k) : k ∈ IN}

by definition of ∩.

Hence

~Ek ∈ ℕ: k e ∩{E(k) : k ∈ IN}

Hence

∩{E(k) : k ∈ IN} = { }

(since ∩{E(k) : k ∈ IN} c ℕ).

qed

> > Nothing.

Except a typo in you original statement: "2, {∩E(k) : k ∈ ℕ} = { }".

> According to (1) the empty set in (2) cannot be realized without a finite predecessor.

Nonsense. (You see: You once more uttered a "statement" which is _not even wrong_.)

__________________________________________________________________

Why are you so bad in math, Mückenheim?

bwr fml schrieb am Montag, 4. Juli 2022 um 04:21:06 UTC+2:

> Polite honest serious brief question.
>
> Is there any chance that you might be able to come up with the smallest
> consistent list of axioms that forms the foundation and distinguishes
> your math from what other people are using?

I do not use different axioms but only the axioms and theorems of ZFC. The only difference is that I take them seriously. For instance I take the bijection between ℕ and ℚ seriously in that I assume that every fraction has got an index. That is not a matter of "limit in the infinite".. I assume Cantor's function
k = (m + n - 1)(m + n - 2)/2 + m
with the resulting sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...

I represent the positive fractions by the well known matrix

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ...
....

and use the indices 1, 2, 3, ..., abbreviated by X, initially covering the first column

XOOO...
XOOO...
XOOO...
XOOO...
....

(in oder to prove by the bijection of the indices with the integer fractions n/1 that all natural numbers are applied). According to Cantos sequence they are distributed over the whole matrix step by step:

XXOO...
OOOO...
XOOO...
XOOO...
....

XXOO...
XOOO...
OOOO...
XOOO...
....

XXXO...
XOOO...
OOOO...
OOOO...
....

and so on.

The result is that the X will never cover the whole matrix although every definable fraction will get an index.

> And get everyone to agree to preface all future posts with
> "Assuming this list of axioms:..."?

As I said, I only abolish the imprecise application of "limits" in bijections.

> If the answer is "no" then I would understand completely.
> I was only hoping to find some way quit the endless arguments about
> things that do not seem to really address the deeper questions.

Also my most important argument does not apply axioms different from ZFC.
All mathematicians believe that for endegments E(k) = {k, k+1, k+2, ...}

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
and
∩{E(k) : k ∈ ℕ} = { }

I am only asking how the loss of ℵ₀ natural numbers between every finite intersection and the infinite intersection can be justified. It cannot be analyzed in detail Therefore I call it dark. And if

∀k ∈ ℕ: E(k+1) = E(k) \ {k}

is correct for all natural numbers which are lost in the final intersection, then there must be finite endsegments as predecessors of the empty set. Since they cannot be found, I call them dark. I know of no-one who had hitherto thought about that.

Regards, WM

Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 15:19:21 UTC+2:
> On Monday, July 4, 2022 at 2:55:40 PM UTC+2, WM wrote:
> > William schrieb am Sonntag, 3. Juli 2022 um 18:15:46 UTC+2:
> > > On Sunday, July 3, 2022 at 12:37:50 PM UTC-3, WM wrote:
> > > > >
> > > > > There is no contradiction with the statement
> > > > > {∩E(k) : k ∈ ℕ} = { }
> But there's a typo in it. It should read:
> ∩{E(k) : k ∈ ℕ} = { }

Yes.
> > > It is one statement about one collection.
> Well, it states that the set ∩{E(k) : k ∈ ℕ} equals the set { }. In other words, that the set ∩{E(k) : k ∈ ℕ} is identical with the set { }.
>
> Hint: Frege considered such statements in detail in his seminal article: "Über Sinn und Bedeutung".
> See: https://en.wikipedia.org/wiki/Sense_and_reference
>
> > Above you can read: k ∈ ℕ. That means ...
>
> No, THAT doesn't mean anything "in isolation".

It means that not the set ℕ but its elements k are applied.
>
> > if all endsegments E(k) [with] k of ℕ are intersected, then the result is the empty set.
>
> Yeah, "∩{E(k) : k ∈ ℕ} = { }" can be INTERPRETED that way, right.

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

I am only asking how the loss of ℵ₀ natural numbers between every finite intersection and the infinite intersection can be justified. It cannot be analyzed in detail.

Regards, WM

WM wrote on 7/4/2022 :
> bwr fml schrieb am Montag, 4. Juli 2022 um 04:21:06 UTC+2:
>
>> Polite honest serious brief question.
>>
>> Is there any chance that you might be able to come up with the smallest
>> consistent list of axioms that forms the foundation and distinguishes
>> your math from what other people are using?
>
> I do not use different axioms but only the axioms and theorems of ZFC.

ZFC does not have the "Axiom of Because I Said So" which is the one
that you use almost exclusively.

William schrieb am Montag, 4. Juli 2022 um 15:20:54 UTC+2:
> On Monday, July 4, 2022 at 10:00:56 AM UTC-3, WM wrote:
> > William schrieb am Sonntag, 3. Juli 2022 um 18:29:34 UTC+2:
> >
> > > I said that each of them [without exception] not "all of them". For *each* of them [without exception] ℵ₀ successors remain.
> > For all of them they are gone.
> The statement "For 'all of them', the colleciton, ℵ₀ successors remain." is not made.

No, we know that for all of them they are gone.

> The statement that is made is: For *each* of them [without exception] ℵ₀ successors remain.

That begs the question: How are the gone if

∀k ∈ ℕ: E(k+1) = E(k) \ {k}

is valid for all gone natnumbers?

Regards, WM

On Monday, July 4, 2022 at 3:31:38 PM UTC+2, WM wrote:

> I do not use different axioms but only the axioms and theorems of ZFC.

[ Don't forget the usual definitions (in this context), dumbo. For, say, IN, omega, U, ∩, \, etc. ]

In this case you should be able to follow the proof below.

> I am only asking how the loss of ℵ₀ natural numbers between [any] finite intersection and [any] infinite intersection can be justified.

Well, in the first case the intersection only concerns FINITELY many endsegments, but in the other INFINITELY many endsegments.

You see: For all e IN: {1, 2, 3, ...,, n} is finite, while {1, 2, 3, ...} is infinite. In the same way: For all e IN: IN \ {1, 2, 3, ...,, n} is infinite, while IN \ {1, 2, 3, ...} is { }.

Hint: There's no "step" which leads from finite to infinite or from infinite to finite ({ }).

> if
> E(k+1) = E(k) \ {k}
> is correct for all natural numbers [k]

--- which is the case (by definition).

> <nonsense deleted>

Now,

∩{E(k) : k ∈ ℕ} = { }

can easily be derived from

Ak e IN: E(k+1) = E(k) \ {k} (*)

Since (*) implies:

∀k ∈ ℕ: En ∈ ℕ: k !e E(n). (For each and every natural number there's an ensegement not containing it.)

Hence

∀k ∈ ℕ: k !e ∩{E(k) : k ∈ IN}

by definition of ∩. [ Hint: x e ∩{E(k) : k ∈ IN} iff x e E(k) for ALL k in IN. In words: x is in ∩{E(k) : k ∈ IN} iff x is in each and every endsegment. ]

Hence

~Ek ∈ ℕ: k e ∩{E(k) : k ∈ IN}

Hence

∩{E(k) : k ∈ IN} = { }

(since ∩{E(k) : k ∈ IN} c ℕ).

qed

Why are you so bad in math, Mückenheim?

On Monday, July 4, 2022 at 3:37:25 PM UTC+2, WM wrote:

> I am only asking how the loss of ℵ₀ natural numbers between [any] finite intersection and [any] infinite intersection can be justified.

Well, in the first case the intersection only concerns FINITELY many endsegments, but in the other INFINITELY many endsegments.

You see: For all e IN: {1, 2, 3, ...,, n} is finite, while {1, 2, 3, ...} is infinite. In the same way: For all e IN: IN \ {1, 2, 3, ...,, n} is infinite, while IN \ {1, 2, 3, ...} is { }.

Hint: There's no "step" which leads from finite to infinite or from infinite to finite ({ }).

Now,

∩{E(k) : k ∈ ℕ} = { }

can easily be derived from

Ak e IN: E(k+1) = E(k) \ {k} (*)

Since (*) implies:

∀k ∈ ℕ: En ∈ ℕ: k !e E(n). (For each and every natural number there's an ensegement not containing it.)

Hence

∀k ∈ ℕ: k !e ∩{E(k) : k ∈ IN}

by definition of ∩. [ Hint: x e ∩{E(k) : k ∈ IN} iff x e E(k) for ALL k in IN. In words: x is in ∩{E(k) : k ∈ IN} iff x is in each and every endsegment. ]

Hence

~Ek ∈ ℕ: k e ∩{E(k) : k ∈ IN}

Hence

∩{E(k) : k ∈ IN} = { }

(since ∩{E(k) : k ∈ IN} c ℕ).

qed

_______________________________________

On the other hand, the proof for

∩{E(k) : k ∈ M} is infinite

for all finite M c IN, M =/= { } is trivial too.

[ Left as an exercise for the reader. ]

On Monday, July 4, 2022 at 3:42:48 PM UTC+2, WM wrote:

> That begs the question: How are they gone if
>
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
>
> is valid[.]

Well, see: https://www.youtube.com/watch?v=1y2SIIeqy34&ab_channel=Spadecaller