On 8/1/2021 8:21 AM, WM wrote:
> Greg Cunt schrieb
> am Samstag, 31. Juli 2021 um 23:32:07 UTC+2:
>> On Friday, July 30, 2021 at 6:16:33 PM UTC+2,
>> WM wrote:

>>> ∩{E(k) | k ∈ ℕ} = { }
>>> together with
>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
>>> proves finite endsegments.
>>
>> No, it doesn't.
>
> Stop your delusions! Try mathematics!
> One and only one k per step is removed.

Each k is removed by a step.
∀k ∈ ℕ: E(k+1) = E(k) \ {k}

What is left when each k is removed?

> The empty intersection cannot come into being
> in any other way.

The empty intersection cannot come into being
in your (WM's) way.

Your objection is that it is impossible to step one-by-one
from N to {}, removing one k with each step.
∀k ∈ ℕ: E(k+1) = E(k) \ {k}

Your "proof" that dark numbers exist is that adding
dark numbers to N make it possible-in-principle to
step one-by-one from N to {}, removing one k with each step.

One really big hole in that argument is that adding
dark numbers to N *DOESN'T* make it possible-in-principle to
step one-by-one from N to {}, removing one k with each step.

Describe what it means for it to be possible-in-principle
to step one-by-one from E(j) to E(k).
(See "steppable".)

Consider
E(1), E(2), ... , {}
Check it against this description.
Fail.

Consider
E(1), E(2), ... , {grossone-1,grossone}, {grossone}, {}
Check it against this description.
ALSO fail.

How E(1), E(2), ... , {} fails is that it can be split
into a first part and a last part that does not have
a crossing-pair -- a pair j,k such that,
crossing from j to k crosses from first part to last part.

Split it so:
E(1), E(2), ... (the definable end segments only)
and
{}

( I think requiring crossing-pairs captures the essence of
( being able to step one-by-one.
( Consider {1,2,3,4} which does not fail.
( ( Split as {1}, {2,3,4}, the pair is 1,2
( Split as {1,2}, {3,4}, the pair is 2,3
( Split as {1,2,3}, {4}, the pair is 3,4
( ( Missing one pair would be enough to conclude we cannot step
( one-by-one from 1 to 4.
( With one pair (hypothetically) missing, we would be
( locked away from one last part, unable to cross and complete
( our journey.

E(1), E(2), ... , {grossone-1,grossone}, {grossone}, {}
fails for the same reason E(1), E(2), ... , {} fails.
Split is so:
E(1), E(2), ... (definable end segments only)
and
... , {grossone-1,grossone}, {grossone}, {} (everything else)

*AS IN THE OTHER CASE*
every definable end segment has another definable
end segment as an immediate successor. There is no
crossing-pair from the definable end segments to
everything else, whether "everything else" includes
grossone and its ilk, or whether it's only {}.

Whatever you conceive your dark end segments to be,
their inclusion doesn't make stepping one-by-one from N to {}
any more possible-in-principle.

----
Your (WM's) whole shtick is born from confusion about
what it is to be able to step one-by-one form one end of
something to its other end.

_It is not enough to have over-all start and stop steps

The correct description is that,
whenever the steps are split into a first part and a last part,
there is a unique pair that steps from the first to the last.

----
> The empty intersection cannot come into being
> in any other way.

The empty intersection does not come into being as the
final step of a sequence of one-element deletions.

The empty set is the answer to the question
| What is the collection of elements in all end segments?

Because no finite number is in all end segments,
and no post-finite number is in N.

Greg Cunt schrieb am Sonntag, 1. August 2021 um 16:48:44 UTC+2:
> On Sunday, August 1, 2021 at 4:29:12 PM UTC+2, WM wrote:

> > You cannot have each one as the last one but nevertheless combine all as an infinite set.
> Sure we "can". :-)

A gang of fools may adhere to this delusion.
It is not mathematics since mathematics proves the contrary. Each one which appears as a last one is definable and provably insufficient for an infinite set (it could be removed without changing the cardinal number).

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

Unless you confess that you have uttered a foolish opinion, EOD.

Regards, WM

William schrieb am Sonntag, 1. August 2021 um 17:13:45 UTC+2:
> On Sunday, August 1, 2021 at 9:54:37 AM UTC-4, WM wrote:
>
> > The intersection of all endsegments is empty. But the intersection of
> all endsegments which can be the last in a finite sequence is infinite.
> Nope, The sequence of all endsegments which can be the last in a finite sequence is indexed by|N_F.

Wrong. All that can be last is irrelevant for actual infinity. Proof:

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

Regards, WM

Jim Burns schrieb am Sonntag, 1. August 2021 um 20:52:20 UTC+2:
> On 8/1/2021 8:21 AM, WM wrote:

> > Stop your delusions! Try mathematics!
> > One and only one k per step is removed.
> Each k is removed by a step.
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> What is left when each k is removed?

Nothing. But before each k is removed, finite endsegments appear, or better do not appear because they are dark, but are passed.

> > The empty intersection cannot come into being
> > in any other way.
>
> The empty intersection cannot come into being
> in your (WM's) way.

It is not *my way* but the *general definition*:
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
If it cannot exist in set theory then set theory is wrong.
>
> Your objection is that it is impossible to step one-by-one
> from N to {}, removing one k with each step.
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> Your "proof" that dark numbers exist is that adding
> dark numbers to N make it possible-in-principle to
> step one-by-one from N to {}, removing one k with each step.

So it is.
>
> One really big hole in that argument is that adding
> dark numbers to N *DOESN'T* make it possible-in-principle to
> step one-by-one from N to {}, removing one k with each step.

If this were true, then no aleph_0 could exist at all.
>
> Describe what it means for it to be possible-in-principle
> to step one-by-one from E(j) to E(k).
> (See "steppable".)
>
> Consider
> E(1), E(2), ... , {}
> Check it against this description.
> Fail.
>
Therefore I use dark numbers. So the step one-by-one cannot be excluded.

> Consider
> E(1), E(2), ... , {grossone-1,grossone}, {grossone}, {}
> Check it against this description.
> ALSO fail.

Therefore I use dark numbers. So the step one-by-one cannot be excluded.

> How E(1), E(2), ... , {} fails is that it can be split
> into a first part and a last part that does not have
> a crossing-pair -- a pair j,k such that,
> crossing from j to k crosses from first part to last part.

Right. The first part is potentially infinite.
>
> Split it so:
> E(1), E(2), ... (the definable end segments only)
> and
> {}
>
> ( I think requiring crossing-pairs captures the essence of
> ( being able to step one-by-one.
> ( Consider {1,2,3,4} which does not fail.
> (
> ( Split as {1}, {2,3,4}, the pair is 1,2
> ( Split as {1,2}, {3,4}, the pair is 2,3
> ( Split as {1,2,3}, {4}, the pair is 3,4
> (
> ( Missing one pair would be enough to conclude we cannot step
> ( one-by-one from 1 to 4.
> ( With one pair (hypothetically) missing, we would be
> ( locked away from one last part, unable to cross and complete
> ( our journey.
>
> E(1), E(2), ... , {grossone-1,grossone}, {grossone}, {}
> fails for the same reason E(1), E(2), ... , {} fails.
> Split is so:
> E(1), E(2), ... (definable end segments only)
> and
> ... , {grossone-1,grossone}, {grossone}, {} (everything else)

Nice that you go into detail. But for potential infinity this is not required because it is not possible. You cannot express 20-digit numbers on a usual pocket calculator. You cannot express a crossing pair by any means.
>
> *AS IN THE OTHER CASE*
> every definable end segment has another definable
> end segment as an immediate successor.

Nevertheless you cannot express all natural numbers. It should be easy, given eneough time and ressources, but it is impossible.

> There is no
> crossing-pair from the definable end segments to
> everything else, whether "everything else" includes
> grossone and its ilk, or whether it's only {}.

There is no crossing pair from what you can define and what you cannot define with and without dark numbers existing. The question can only be: Is there something what I call dark or is there nothing?
>
> Whatever you conceive your dark end segments to be,
> their inclusion doesn't make stepping one-by-one from N to {}
> any more possible-in-principle.

It makes it possible in principle that endsegments exist and does not prevent reaching the empty set.

> The correct description is that,
> whenever the steps are split into a first part and a last part,
> there is a unique pair that steps from the first to the last.

But without steps in the adkness you have to claim that all endsegments which appear at last position leave infinite intersections while otherwise they leave empty intersections. Since the position of their appearance is irrelevant, this claim is nonsense.

The only alternative is that actual infinity and endsegments do not exist.
>
> ----
> > The empty intersection cannot come into being
> > in any other way.
> The empty intersection does not come into being as the
> final step of a sequence of one-element deletions.

Then the defining formula is wrong. Then no endsegment exists.
>
> The empty set is the answer to the question
> | What is the collection of elements in all end segments?
>
> Because no finite number is in all end segments,
> and no post-finite number is in N.

If no finite number is in all endsegments, then not infinitely many numbers can remain in all endsegments. But this is claimed. What numbers should that be which remain in all endsegments?

Regards, WM

On Sunday, August 1, 2021 at 8:58:52 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Sonntag, 1. August 2021 um 16:48:44 UTC+2:
> > On Sunday, August 1, 2021 at 4:29:12 PM UTC+2, WM wrote:
> > >
> > > You cannot have each one as the last one but nevertheless combine all as an infinite set.
> > >
> > Sure we "can". :-)

Look, I just did:

U{{n} : n e IN} = IN .

:-)

> It is <bla bla>
>
> Each one which appears as a last one is definable

Great, this implies that each and every natural number is "definable" (WM). Something we claimed all along.

Of course, each and every natural number

> [can] be "removed" without changing the cardinal number [i.e. such that the cardinal number of the "resulting" set is idential with the cardinal number of IN].

Right. After all, aleph_0 - 1 = aleph_0 .

> ∀n ∈ IN: |IN \ {1, 2, 3, ..., n}| = ℵo.

Right. After all, An e IN: aleph_0 - n = aleph_0 .

> Unless you confess that you have uttered a foolish opinion, EOD.

lol. There never WAS a "discussion", dumbo.

All I do, is correcting your errors (adding some comments from time to time).

Greg Cunt schrieb am Sonntag, 1. August 2021 um 21:34:26 UTC+2:
> On Sunday, August 1, 2021 at 8:58:52 PM UTC+2, WM wrote:
> > Greg Cunt schrieb am Sonntag, 1. August 2021 um 16:48:44 UTC+2:
> > > On Sunday, August 1, 2021 at 4:29:12 PM UTC+2, WM wrote:
> > > >
> > > > You cannot have each one as the last one but nevertheless combine all as an infinite set.
> > > >
> > > Sure we "can". :-)

If all endsegments which as last ones leave an infinite intersection
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
could be applied together with empty intersection
∩{E(k) | k ∈ ℕ_def} = { }
then there is a contradiction, because for the intersection the position does not matter.

Only if there are more with indices in ℕ than in ℕ_def, then this is possible
∩{E(k) | k ∈ ℕ} = { } .

Regards, WM

On 8/1/2021 1:58 PM, WM wrote:
> Greg Cunt schrieb am Sonntag, 1. August 2021 um 16:48:44 UTC+2:
>> On Sunday, August 1, 2021 at 4:29:12 PM UTC+2, WM wrote:
>
>>> You cannot have each one as the last one but nevertheless combine all as an infinite set.
>> Sure we "can". :-)
>
> A gang of fools may adhere to this delusion.
> It is not mathematics since mathematics proves the contrary. Each one which appears as a last one is definable and provably insufficient for an infinite set (it could be removed without changing the cardinal number).
>
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
>
> Unless you confess that you have uttered a foolish opinion, EOD.
>
> Regards, WM
>

there is no ℕ_def, because "defined, etc" is fake math.

On 8/1/2021 2:47 PM, WM wrote:
> Greg Cunt schrieb am Sonntag, 1. August 2021 um 21:34:26 UTC+2:
>> On Sunday, August 1, 2021 at 8:58:52 PM UTC+2, WM wrote:
>>> Greg Cunt schrieb am Sonntag, 1. August 2021 um 16:48:44 UTC+2:
>>>> On Sunday, August 1, 2021 at 4:29:12 PM UTC+2, WM wrote:
>>>>>
>>>>> You cannot have each one as the last one but nevertheless combine all as an infinite set.
>>>>>
>>>> Sure we "can". :-)
>
> If all endsegments which as last ones leave an infinite intersection
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> could be applied together with empty intersection
> ∩{E(k) | k ∈ ℕ_def} = { }
> then there is a contradiction, because for the intersection the position does not matter.
>
> Only if there are more with indices in ℕ than in ℕ_def, then this is possible
> ∩{E(k) | k ∈ ℕ} = { } .
>
> Regards, WM
>

repeating it, does not fix your broken math.

On Sunday, August 1, 2021 at 3:01:39 PM UTC-4, WM wrote:
> William schrieb am Sonntag, 1. August 2021 um 17:13:45 UTC+2:
> > On Sunday, August 1, 2021 at 9:54:37 AM UTC-4, WM wrote:
> >
> > > The intersection of all endsegments is empty. But the intersection of
> > all endsegments which can be the last in a finite sequence is infinite.
> > Nope, The sequence of all endsegments which can be the last in a finite sequence is indexed by|N_F.
> Wrong.

Nope. Every element of |N_F the last element of a FISON. So if an endsegment is indexed by am element of |N_F it is the last of a finite sequence. No element of U_F is contained in every endsegment indexed by an element of }N_F

--
William Hughes

On 8/1/2021 3:24 PM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 1. August 2021 um 20:52:20 UTC+2:

>> Describe what it means for it to be possible-in-principle
>> to step one-by-one from E(j) to E(k).
>> (See "steppable".)
>>
>> Consider
>> E(1), E(2), ... , {}
>> Check it against this description.
>> Fail.
>>
> Therefore I use dark numbers.
> So the step one-by-one cannot be excluded.

There are dark numbers in the next case,
but not in this one.

Here, no definable E(k) has its E(k+1) = {}.
So, each E(k+1) is definable.

>> Consider
>> E(1), E(2), ... , {grossone-1,grossone}, {grossone}, {}
>> Check it against this description.
>> ALSO fail.
>
> Therefore I use dark numbers.
> So the step one-by-one cannot be excluded.

What is different here from the previous case?
There are dark numbers between {} and
the definable end segments.

What is NOT different?
Each definable E(k) has has the same E(k+1) as before.
Before, each E(k+1) was definable.
They are all still definable.

Insert dark end segments between the defined and {}.
AS BEFORE, each adjacent pair which starts with a definable
end segment ends with a definable end segment.
There is no crossing-pair from the definables to
after-the-definables. (Inserting dark end segments didn't
change the definables.)

What else is NOT different?
It is still not possible-in-principle to step
one-by-one from N to {}.
Inserting dark end segments didn't change that.

>> How E(1), E(2), ... , {} fails is that it can be split
>> into a first part and a last part that does not have
>> a crossing-pair -- a pair j,k such that,
>> crossing from j to k crosses from first part to last part.
>
> Right. The first part is potentially infinite.
>
>> Split it so:
>> E(1), E(2), ... (the definable end segments only)
>> and
>> {}
>>
>> ( I think requiring crossing-pairs captures the
>> ( essence of being able to step one-by-one.
>> ( Consider {1,2,3,4} which does not fail.
>> (
>> ( Split as {1}, {2,3,4}, the pair is 1,2
>> ( Split as {1,2}, {3,4}, the pair is 2,3
>> ( Split as {1,2,3}, {4}, the pair is 3,4
>> (
>> ( Missing one pair would be enough to conclude we cannot step
>> ( one-by-one from 1 to 4.
>> ( With one pair (hypothetically) missing, we would be
>> ( locked away from one last part, unable to cross and complete
>> ( our journey.
>>
>> E(1), E(2), ... , {grossone-1,grossone}, {grossone}, {}
>> fails for the same reason E(1), E(2), ... , {} fails.
>> Split is so:
>> E(1), E(2), ... (definable end segments only)
>> and
>> ... , {grossone-1,grossone}, {grossone}, {} (everything else)
>
> Nice that you go into detail.
>
> But for potential infinity this is not required because
> it is not possible.
> You cannot express 20-digit numbers on a usual pocket calculator.
> You cannot express a crossing pair by any means.

I don't need to express 20-digit numbers to make some true claims
about a crossing-pair, some things true no matter which numbers
j,k refer to or which end segments E(j),E(k) refer to.

Not every claim is true-no-matter-which.
But SOME claims ARE true-no-matter-which.
We can work with THOSE without knowing which
of infinitely-many we are referring to.
Because it doesn't matter which.

----
This is where it matters whether we include dark numbers
among the things we are talking about. A claim
true-no-matter-which _about definables_ might not be
true-no-matter-which about _dark or definables_
What we choose to talk about which claims we can and
cannot make.

So, the wheels go 'round and 'round, and we end up back
at the same place. We _choose_ to talk about those
things we count with because, well, we count with them.
The things we count with are not dark numbers.
No thing we count with is post-finite
among the things-we-count-with.
No thing we count with is last
among the things-we-count-with.

That is why your dark numbers are not natural numbers.

----

söndag 1 augusti 2021 kl. 21:47:31 UTC+2 skrev WM:
> Greg Cunt schrieb am Sonntag, 1. August 2021 um 21:34:26 UTC+2:
> > On Sunday, August 1, 2021 at 8:58:52 PM UTC+2, WM wrote:
> > > Greg Cunt schrieb am Sonntag, 1. August 2021 um 16:48:44 UTC+2:
> > > > On Sunday, August 1, 2021 at 4:29:12 PM UTC+2, WM wrote:
> > > > >
> > > > > You cannot have each one as the last one but nevertheless combine all as an infinite set.
> > > > >
> > > > Sure we "can". :-)
> If all endsegments which as last ones leave an infinite intersection
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> could be applied together with empty intersection
> ∩{E(k) | k ∈ ℕ_def} = { }
> then there is a contradiction, because for the intersection the position does not matter.
>
> Only if there are more with indices in ℕ than in ℕ_def, then this is possible
> ∩{E(k) | k ∈ ℕ} = { } .
>
> Regards, WM

There is no contradiction because the first is always a finite intersection but the latter is not a finite intersection given that your N_def=N

On Sunday, August 1, 2021 at 9:47:31 PM UTC+2, WM wrote:

> If all endsegments which [...] leave an infinite intersection
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo

Man, you are quite confused.

Look, if S is a FINITE set of endsegements, then

|∩S| = ℵo ,

since ∩S is just the smallest endsegment in S (and for all endsegements e: |e| = ℵo).

If on the other hand S is an INFINITE set of endsegements, then

|∩S| = 0 ,

since in this case ∩S = { }.

William schrieb am Sonntag, 1. August 2021 um 23:12:29 UTC+2:
> On Sunday, August 1, 2021 at 3:01:39 PM UTC-4, WM wrote:
> > William schrieb am Sonntag, 1. August 2021 um 17:13:45 UTC+2:
> > > On Sunday, August 1, 2021 at 9:54:37 AM UTC-4, WM wrote:
> > >
> > > > The intersection of all endsegments is empty. But the intersection of
> > > all endsegments which can be the last in a finite sequence is infinite.
> > > Nope, The sequence of all endsegments which can be the last in a finite sequence is indexed by|N_F.
> > Wrong.
> Nope. Every element of |N_F the last element of a FISON.

Well, then |N_F is what I call |N_def.

> So if an endsegment is indexed by am element of |N_F it is the last of a finite sequence. No element of U_F

What was U_F?

> is contained in every endsegment indexed by an element of }N_F

Regards, WM

Jim Burns schrieb am Montag, 2. August 2021 um 05:20:51 UTC+2:
> On 8/1/2021 3:24 PM, WM wrote:
> > Jim Burns schrieb

> > Therefore I use dark numbers.
> > So the step one-by-one cannot be excluded.
> What is different here from the previous case?
> There are dark numbers between {} and
> the definable end segments.
>
> What is NOT different?

If all ℵo endsegments are definable and all have ℵo elements, then ℵo endsegments have deleted ℵo natnumbers already and nevertheless contain ℵo natnumbers. That is a contradiction. With dark numbers ℵo endsegments do not contain ℵo natnumbers each.

> That is why your dark numbers are not natural numbers.
>
But if all were visible, then the statements that ℵo endsegments contain ℵo natnumbers woulde yield a contradiction.

Regards, WM

zelos...@gmail.com schrieb am Montag, 2. August 2021 um 06:53:50 UTC+2:
> söndag 1 augusti 2021 kl. 21:47:31 UTC+2 skrev WM:
> > Greg Cunt schrieb am Sonntag, 1. August 2021 um 21:34:26 UTC+2:
> > > On Sunday, August 1, 2021 at 8:58:52 PM UTC+2, WM wrote:
> > > > Greg Cunt schrieb am Sonntag, 1. August 2021 um 16:48:44 UTC+2:
> > > > > On Sunday, August 1, 2021 at 4:29:12 PM UTC+2, WM wrote:
> > > > > >
> > > > > > You cannot have each one as the last one but nevertheless combine all as an infinite set.
> > > > > >
> > > > > Sure we "can". :-)
> > If all endsegments which as last ones leave an infinite intersection
> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> > could be applied together with empty intersection
> > ∩{E(k) | k ∈ ℕ_def} = { }
> > then there is a contradiction, because for the intersection the position does not matter.
> >
> > Only if there are more with indices in ℕ than in ℕ_def, then this is possible
> > ∩{E(k) | k ∈ ℕ} = { } .

> There is no contradiction because the first is always a finite intersection but the latter is not a finite intersection given that your N_def=N

That is not given. N_def is finite, N is infinite.

Regards, WM

On Monday, August 2, 2021 at 9:26:15 AM UTC-4, WM wrote:
> William schrieb am Sonntag, 1. August 2021 um 23:12:29 UTC+2:
> > On Sunday, August 1, 2021 at 3:01:39 PM UTC-4, WM wrote:
> > > William schrieb am Sonntag, 1. August 2021 um 17:13:45 UTC+2:
> > > > On Sunday, August 1, 2021 at 9:54:37 AM UTC-4, WM wrote:
> > > >
> > > > > The intersection of all endsegments is empty. But the intersection of
> > > > all endsegments which can be the last in a finite sequence is infinite.
> > > > Nope, The sequence of all endsegments which can be the last in a finite sequence is indexed by|N_F.
> > > Wrong.
> > Nope. Every element of |N_F the last element of a FISON.
> Well, then |N_F is what I call |N_def.

Nope. |N_def is finite. |N_F is a Peano set and has cardinality aleph_0

--
William Hughes

On 8/2/2021 8:29 AM, WM wrote:
> Jim Burns schrieb am Montag, 2. August 2021 um 05:20:51 UTC+2:
>> On 8/1/2021 3:24 PM, WM wrote:
>>> Jim Burns schrieb
>
>>> Therefore I use dark numbers.
>>> So the step one-by-one cannot be excluded.
>> What is different here from the previous case?
>> There are dark numbers between {} and
>> the definable end segments.
>>
>> What is NOT different?
>
> If all ℵo endsegments are definable and all have ℵo elements, then ℵo endsegments have deleted ℵo natnumbers already and nevertheless contain ℵo natnumbers. That is a contradiction. With dark numbers ℵo endsegments do not contain ℵo natnumbers each.

you cannot count dark numbers, your conclusion is false

>
>> That is why your dark numbers are not natural numbers.
>>
> But if all were visible, then the statements that ℵo endsegments contain ℵo natnumbers woulde yield a contradiction.

there is no contradiction

>
> Regards, WM
>

On Monday, 2 August 2021 at 10:30:03 UTC-3, WM wrote:
> Jim Burns schrieb am Montag, 2. August 2021 um 05:20:51 UTC+2:
> > On 8/1/2021 3:24 PM, WM wrote:
> > > Jim Burns schrieb
> > > Therefore I use dark numbers.
> > > So the step one-by-one cannot be excluded.
> > What is different here from the previous case?
> > There are dark numbers between {} and
> > the definable end segments.
> >
> > What is NOT different?
> If all ℵo endsegments are definable and all have ℵo elements, then ℵo endsegments have deleted ℵo natnumbers already and nevertheless contain ℵo natnumbers.
I hope you don't do your own taxes! Not only are you hopelessly inept at mathematics, you can't even count correctly!
[...]

On 8/2/2021 9:29 AM, WM wrote:
> Jim Burns schrieb
> am Montag, 2. August 2021 um 05:20:51 UTC+2:
>> On 8/1/2021 3:24 PM, WM wrote:

>>> Therefore I use dark numbers.
>>> So the step one-by-one cannot be excluded.
>>
>> What is different here from the previous case?
>> There are dark numbers between {} and
>> the definable end segments.
>>
>> What is NOT different?

What else is NOT different?
It is still not possible-in-principle to step
one-by-one from N to {}.
Inserting dark end segments didn't change that.

> If all ℵo endsegments are definable and
> all have ℵo elements,
> then ℵo endsegments have deleted ℵo natnumbers already and

For each element k, the sequence 0,1,2,...,k is finite.

For each end segment N\{0,...,k}, the sequence
N\{},
N\{0},
N\{0,1},
N\{0,1,2},
....,,
N\{0,...,k},
is finite.

No more than finitely-many are deleted from any end segment.

> nevertheless contain ℵo natnumbers.
> That is a contradiction.

That is a mistake.
No more than finitely-many are deleted from any end segment.

----
There is no post-finite number
because, for each k, {0,...,k} is finite.
There is no last number
because, for each {0,...,k}, k+1 is not in it.

Define N = Union{ all {0,...,k} }.
There is no last number in N.

For each k, there is no last number in N\{0,...,k}.

Intrsct{ all N\{0,...,k} } =
N\Union{ all {0,...,k} } =
N\N =
{}

> With dark numbers ℵo endsegments do not contain
> ℵo natnumbers each.

You can have aleph_0 end segments which do not contain
aleph_0 elements each.
But that won't be the natural numbers.
And it still won't be possible-in-principle to step
one-by-one to {}.

Take the integers { ...,-3,-2,-1,0,1,2,3,... }
and define a different order << on them such that
all the positives come before all the negatives and 0.
1,2,3,...,-3,-2,-1,0

End segments:
{ 1,2,3,...,-3,-2,-1,0 }
{ 2,3,4,...,-3,-2,-1,0 }
{ 3,4,5,...,-3,-2,-1,0 }
....
{ -3,-2,-1,0 }
{ -2,-1,0 }
{ -1,0 }
{ 0 }
{}

The successor of each positive is still positive,
and the predecessor of each negative is still negative.
Split them so
{ 1,2,3,... }
and
{ ...,-3,-2,-1,0 }
and there is no crossing-pair from positive to negative.

It is not possible-in-principle to step one-by-one
beyond the positives, the natural numbers.

That's not a flaw.
We match natural numbers to unmatched things-being-counted.
If we ever ran out of unmatched natural numbers before
we ran out of unmatched things-being-counted, that would
be a flaw. So, by design, it is not possible to step beyond
the natural numbers.

>> That is why your dark numbers are not natural numbers.
>
> But if all were visible, then the statements that
> ℵo endsegments contain ℵo natnumbers woulde yield
> a contradiction.

Treat 'ℵo' like a clause that says "no last one of these".

What we have is
There is no last end segment which contains no last element.

Do you need a proof of this?
Consider
Ak in N: E(k+1) = E(k)\{k}

If E(k+1) contains a last element,
then E(k) contains that same last element.

....is equivalent to...

If E(k) does not contain a last element,
then E(k+1) does not contain a last element.

But E(k+1) is after E(k). E(k) is not last.

No end segment which does not contain a last element is
the last end segment which does not contain a last element.

This is not a contradiction.

William schrieb am Montag, 2. August 2021 um 15:35:51 UTC+2:

> Nope. |N_def is finite. |N_F is a Peano set and has cardinality aleph_0

Then all endsegments are in bijection with N_F. Therefore all aleph_0 natural numbers have been deleted by them. But all are infinite. What numbers do they contain?

Regards, WM

Jim Burns schrieb am Montag, 2. August 2021 um 20:16:14 UTC+2:
> On 8/2/2021 9:29 AM, WM wrote:

> > If all ℵo endsegments are definable and
> > all have ℵo elements,
> > then ℵo endsegments have deleted ℵo natnumbers already and

> For each element k, the sequence 0,1,2,...,k is finite.

But the set of endsegments is infinite and has cardinality aleph_0. It has deleted all aleph_0 natural numbers.
>
> No more than finitely-many are deleted from any end segment.

But infinitely many are deleted by the set of endsegments.

> > nevertheless contain ℵo natnumbers.
> > That is a contradiction.
> That is a mistake.
> No more than finitely-many are deleted from any end segment.

That is irrelevant. We consider all endsegments which are claimed to be infinite.
>
> Define N = Union{ all {0,...,k} }.
> There is no last number in N.

But there are aleph_0 numbers in |N.
>
> For each k, there is no last number in N\{0,...,k}.
>
> Intrsct{ all N\{0,...,k} } =
> N\Union{ all {0,...,k} } =
> N\N =
> {}

That is nice but it is not the question. The question is this:

If all ℵo endsegments are definable and all have ℵo elements, then ℵo endsegments have deleted ℵo natnumbers already and nevertheless contain ℵo natnumbers.

> So, by design, it is not possible to step beyond
> the natural numbers.

But a bijection between natnumbers and endsegments is possible.
All natnumbers are used to enumerate all endsegments. But all of them contain aleph_0 natnumbers. What are these natnumbers?

>
> What we have is
> There is no last end segment which contains no last element.

All have deleted aleph_0 natnumbers. Yet all contain aleph_0 natnumbers. What are these?

Regards, WM

Gus Gassmann schrieb am Montag, 2. August 2021 um 17:43:58 UTC+2:
> On Monday, 2 August 2021 at 10:30:03 UTC-3, WM wrote:

> > If all ℵo endsegments are definable and all have ℵo elements, then ℵo endsegments have deleted ℵo natnumbers already and nevertheless contain ℵo natnumbers.
> I hope

Better try to understand.

Regards, WM

On Monday, 2 August 2021 at 17:49:50 UTC-3, WM wrote:
> Gus Gassmann schrieb am Montag, 2. August 2021 um 17:43:58 UTC+2:
> > On Monday, 2 August 2021 at 10:30:03 UTC-3, WM wrote:
>
> > > If all ℵo endsegments are definable and all have ℵo elements, then ℵo endsegments have deleted ℵo natnumbers already and nevertheless contain ℵo natnumbers.
> > I hope
> Better try to understand.

Unlike you, I do understand. I truly hope you do not do your own taxes, because your understanding of mathematics and numbers is rapidly tending to zero. You evidently do not know how to add any longer.

On 8/2/2021 3:37 PM, WM wrote:
> William schrieb am Montag, 2. August 2021 um 15:35:51 UTC+2:
>
>> Nope. |N_def is finite. |N_F is a Peano set and has cardinality aleph_0
>
> Then all endsegments are in bijection with N_F. Therefore all aleph_0 natural numbers have been deleted by them. But all are infinite. What numbers do they contain?
>
> Regards, WM
>

think it through, and you will find your mistake

On Monday, August 2, 2021 at 4:37:51 PM UTC-4, WM wrote:
> William schrieb am Montag, 2. August 2021 um 15:35:51 UTC+2:
>
> > Nope. |N_def is finite. |N_F is a Peano set and has cardinality aleph_0
> Then all endsegments are in bijection with N_F. Therefore all aleph_0 [elements of U_F] have been deleted by them. But all are infinite. What numbers do they contain?

Each endsegment "deletes". does not contain, a finite set of elements. So each endsegment is infinite. No single endsegment "deletes" every element of U_F. However, there is no element of U_F that is not "deleted" by some endsegment. So the GLB is the empty set.

--
William Hughes