Two alternatives?

[WM]

Set theorists confess that every natural number n has an infinite distance to ω
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
but that the set ℕ has no distance to ω. So either the distance is removed by some magic action when forming the set, or undefinable numbers are remaining and are simply included as individually unspecified elements in "the infinite set ℕ". These are the alternatives known to me. Are there more?

Regards, WM

[Dan Christensen]

Maybe you didn't know, but, there is no greatest natural number. And that, for any natural number n, there are infinitely many natural numbers are greater than n. There is no end to them. Really!

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[WM]

Dan Christensen schrieb am Donnerstag, 11. November 2021 um 16:33:51 UTC+1:
> On Thursday, November 11, 2021 at 5:16:19 AM UTC-5, WM wrote:
> > Set theorists confess that every natural number n has an infinite distance to ω

> > ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo (*)

> > but that the set ℕ has no distance to ω. So either the distance is removed by some magic action when forming the set, or undefinable numbers are remaining and are simply included as individually unspecified elements in "the infinite set ℕ". These are the alternatives known to me. Are there more?
> >

> for any natural number n, there are infinitely many natural numbers are greater than n.

That distance to ω is expressed in (*).

> There is no end to them. Really!

Taking the set ℕ ends them. The distance to ω vanishes completely although for every single n appearing in (*) the distance remains the same. What is the reason?

Regards, WM

[Dan Christensen]

On Thursday, November 11, 2021 at 12:54:35 PM UTC-5, WM wrote:
> Dan Christensen schrieb am Donnerstag, 11. November 2021 um 16:33:51 UTC+1:
> > On Thursday, November 11, 2021 at 5:16:19 AM UTC-5, WM wrote:
> > > Set theorists confess that every natural number n has an infinite distance to ω
> > > ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo (*)
> > > but that the set ℕ has no distance to ω. So either the distance is removed by some magic action when forming the set, or undefinable numbers are remaining and are simply included as individually unspecified elements in "the infinite set ℕ". These are the alternatives known to me. Are there more?
> > >

> > Maybe you didn't know, but, there is no greatest natural number. And that, for any natural number n, there are infinitely many natural numbers are greater than n. There is no end to them. Really!

> That distance to ω is expressed in (*).

> > There is no end to them. Really!

> Taking the set ℕ ends them.

The set N contains them. That doesn't mean there exists a greatest element of N.

> The distance to ω vanishes completely although for every single n appearing in (*) the distance remains the same.
>

I think you mean that, for ANY n, m in N, we have: |{n, n+1, n+2, ... }| = |{m, m+1, m+2, ... }| = ℵo

Makes sense to me. What is your problem?

[Jim Burns]

On 11/11/2021 5:16 AM, WM wrote:

> Set theorists confess that every natural number n
> has an infinite

It's not finite.
What does "finite" mean?

_My answer_
|
| A set B is _finite_ iff
| a total order '<<' of B exists such that,
| for each split of B, there is a crossing-pair,
| and B contains a lower end and an upper end.
| or B is empty.

Define
{j,...,k} such that
a total order '<' of {j,...,k} exists
for each split of B, there is a crossing-pair,
and B contains a lower end and an upper end
and also
'<' is such that i < i+1
for each crossing-pair j',k', k' = j' + 1
and the lower and upper ends are j and k

Lemma.
{j,...,k} is finite.


_Your answer_
|
| <crickets>

> Set theorists confess that every natural number n
> has an infinite distance to ω
> ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo

Define
ℕ such that
∀n ( n ∈ ℕ <-> {1,...,n} exists )

Lemma.
∀n ∈ ℕ ( ℕ\{1,...,n} is not finite )

( The standard order '<', i < i+1, does not have
( an upper end in ℕ\{1,...,n}, because any candidate
( for its upper end m is disqualified by m+1, also
( in ℕ\{1,...,n}.
(
( What is more, any other candidate-order for an order '<<'
( according to which
( the splits of ℕ\{1,...,n} have their own crossing-pairs
( and ℕ\{1,...,n} has both a lower and an upper end
( is a disqualified order.
( A qualified '<<' would imply that there _is_ some m in
( ℕ\{1,...,n} without m+1 in ℕ\{1,...,n}. Which is false.
(
( _According to the definition above_ ℕ\{1,...,n} is
( not finite because there is no qualified order, with
( crossing-pairs and ends, and it is not empty.
(
( Generalizing,
( ∀n ∈ ℕ ( ℕ\{1,...,n} is not finite )

> but that the set ℕ has no distance to ω.

ω is defined to be the first infinite ordinal,
the first after all finite ordinals.
ℕ is defined as the set of all finite ordinals.

It would take changing definitions to have "distance"
between ℕ and ω
Changing definitions doesn't change any facts about
what has been defined. It only changes how we refer
to them.

Define
ω such that
∀n ∈ ℕ ( n < ω ) ∧
∀α ( ∀n ∈ ℕ ( n < α ) -> ω =< α )

Lemma
~∃α, ∀n ∈ ℕ ( n < α < ω )

( ∀α ( ∀n ∈ ℕ ( n < α ) -> ω =< α )
(
( ~∃α ~( ∀n ∈ ℕ ( n < α ) -> ω =< α )
(
( ~∃α ( ∀n ∈ ℕ ( n < α ) ∧ ~(ω =< α) )
(
( ~∃α ( ∀n ∈ ℕ ( n < α ) ∧ (α < ω) )
(
( ~∃α ( ∀n ∈ ℕ ( n < α < ω ) )

> So either the distance is removed by some magic action
> when forming the set,

It's a definition.
∀n ( n < ω <-> n ∈ ℕ )

∀n ( n < ω <-> {1,...,n} exists )

where
for each split of {1,...,n},
there is a crossing-pair i,j, j = i+1,
and the lower and upper ends are 1 and n

> or undefinable numbers are remaining and
> are simply included as individually unspecified
> elements in "the infinite set ℕ".

What does "finite" mean?
Note that ℕ is _not that_

> These are the alternatives known to me.
> Are there more?

Worry less about the sequence of letters i n f i n i t e
Worry more about what that letter sequence means.

Use n to refer to _any possible_ end of {1,...,n},
which has crossing-pairs and two ends.

Reason truth-preserving-ly about n from the
incomplete description of it as ending {1,...,n},
which has crossing-pairs and two ends.

[WM]

Dan Christensen schrieb am Donnerstag, 11. November 2021 um 19:23:37 UTC+1:
> On Thursday, November 11, 2021 at 12:54:35 PM UTC-5, WM wrote:
> > Dan Christensen schrieb am Donnerstag, 11. November 2021 um 16:33:51 UTC+1:
> > > On Thursday, November 11, 2021 at 5:16:19 AM UTC-5, WM wrote:
> > > > Set theorists confess that every natural number n has an infinite distance to ω
> > > > ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo (*)
> > > > but that the set ℕ has no distance to ω. So either the distance is removed by some magic action when forming the set, or undefinable numbers are remaining and are simply included as individually unspecified elements in "the infinite set ℕ". These are the alternatives known to me. Are there more?
> > > >
> > > Maybe you didn't know, but, there is no greatest natural number. And that, for any natural number n, there are infinitely many natural numbers are greater than n. There is no end to them. Really!
> > That distance to ω is expressed in (*).
>
> > > There is no end to them. Really!
>
> > Taking the set ℕ ends them.
> The set N contains them. That doesn't mean there exists a greatest element of N.

But it means that the distance has vanished. The set contains what cannot be observed individually.

> > The distance to ω vanishes completely although for every single n appearing in (*) the distance remains the same.
> >
> I think you mean that, for ANY n, m in N, we have: |{n, n+1, n+2, ... }| = |{m, m+1, m+2, ... }| = ℵo
>
> Makes sense to me. What is your problem?

∀ n∈ℕ ∃ m∈ℕ: m > n .
That is correct. By exchanging the quantifiers
∃ m∈ℕ ∀ n∈ℕ : m > n
we get a wrong statement. Why? In the first line first n∈ℕ has to be chosen, in the second line the whole set ℕ is available to surpass m.

Now consider the endsegments E(n) = {n, n+1, n+2, ...}:
∀ n∈ℕ ∃ M⊂ℕ: M ⊂ E(n) ∧ |M| = ℵo .
That is correct. By exchanging the quantifiers
∃ M⊂ℕ ∀ n∈ℕ : M ⊂ E(n) ∧ |M| = ℵo
we get a wrong statement.

Therefore the M must vanish when the whole set ℕ is considered (that means ∀ n∈ℕ appears at the second position). How can that happen if M exists for every chosen natural number? The answer is clear: It is not possible to choose all natural numbers. The numbers in M cannot be chosen. They are dark.

Regards, WM

[WM]

Jim Burns schrieb am Donnerstag, 11. November 2021 um 20:17:35 UTC+1:

> ℕ is defined as the set of all finite ordinals.
>
> It would take changing definitions to have "distance"
> between ℕ and ω

There is no ordinal number between ℕ and ω.

> Worry less about the sequence of letters i n f i n i t e
> Worry more about what that letter sequence means.

No, the basic terms are finite and infinite.
Do you know why quantifier exchange is forbidden?

∀ n∈ℕ ∃ m∈ℕ: m > n .
That is correct. By exchanging the quantifiers
∃ m∈ℕ ∀ n∈ℕ : m > n

we get a wrong statement. Why? In the upper line, first n∈ℕ has to be chosen, it can easily be surpassed. In the second line the whole set ℕ is available to surpass m.

Now consider the endsegments E(n) = {n, n+1, n+2, ...}:
∀ n∈ℕ ∃ M⊂ℕ: M ⊂ E(n) ∧ |M| = ℵo .
That is correct. By exchanging the quantifiers
∃ M⊂ℕ ∀ n∈ℕ : M ⊂ E(n) ∧ |M| = ℵo
we get a wrong statement.

Therefore the set M must vanish (at least cannot be infinite) when the whole set ℕ is considered (that means ∀ n∈ℕ appears at the second position). How can that happen if M exists for every chosen natural number? The answer is clear: It is not possible to choose all natural numbers. The numbers in M cannot be chosen. They are dark.

Regards, WM

[Serg io]

On 11/11/2021 4:16 AM, WM wrote:
> Set theorists confess that every natural number n has an infinite distance to ω
> ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo

that is bogus, simplify dude;

|ℕ| = ℵo

> but that the set ℕ has no distance to ω.

how does a "set" have a "distance" ?

> So either the distance is removed by some magic action when forming the set, or undefinable numbers are remaining and are simply included as individually unspecified elements in "the infinite set ℕ".

no in both cases. Try again.

> These are the alternatives known to me. Are there more?

Isn't your premise wrong?

>
> Regards, WM
>

[Python]

Crank Wolfgang Mückenheim, aka WM wrote:
...

> ∀ n∈ℕ ∃ m∈ℕ: m > n .
> That is correct. By exchanging the quantifiers
> ∃ m∈ℕ ∀ n∈ℕ : m > n
> we get a wrong statement. Why?

Because it usually so, crank Wolfgang Mückenheim, from Hochschule
Augsburg. Nice to see you confess one of your usual sophistry by
the way...

[FredJeffries]

On Thursday, November 11, 2021 at 1:00:36 PM UTC-8, WM wrote:

> Do you know why quantifier exchange is forbidden?
> ∀ n∈ℕ ∃ m∈ℕ: m > n .
> That is correct. By exchanging the quantifiers
> ∃ m∈ℕ ∀ n∈ℕ : m > n
> we get a wrong statement. Why?

A more interesting question is why you (or anyone) would think that it should be a CORRECT statement? Why should 'quantifier exchange' NOT be forbidden?

[WM]

It is a scientific attitude to ask for reasons of results. Here we have the result that quantifier exchange is wrong. Why? The answer is this:

Instead of the above, we could also write
∀ n∈ℕ_def ∃ m∈ℕ_def: m > n
∃ m∈ℕ_def ∀ n∈ℕ_def : m > n
in order to emphasize that only natural numbers must appear which can be used as individuals. The set ℕ_def is a potentially infinite collection. Therefore fpr every n we can find a larger one, and if we take m out of this collection, we can find a larger one. In case of potential infinity, there is no quantifier exchange allowed, bceuas there is no largest element.

The scene changes in cse of fiished infinity. If there are all natnumbers, then we can talk about endsegments. Of course only definable natnumbers n∈ℕ_def can be chosen. Therefore the first parts remains as above
∀ n∈ℕ_def ∃ M⊂ℕ: M ⊂ E(n) ∧ |M| = ℵo .
Each one has an infinite endsegment E(n), an infinite part of which is dark. We call it M. Please note that now we drop the restriction but consider all natural numbers, because at the second position we need not choose any individual but assume the whole set, knowing well that, by definition, there is no subset M of natural numbers larger than every natural number
∃ M⊂ℕ ∀ n∈ℕ : M ⊂ E(n) ∧ |M| = ℵo .
Hence the set M is present between every definable natnumber and omega, but not between every natnumber and omega. There is only one conclusion possible. ℕ and ℕ _def are different sets.

Regards, WM

[FredJeffries]

On Friday, November 12, 2021 at 4:32:38 AM UTC-8, WM wrote:
> FredJeffries schrieb am Freitag, 12. November 2021 um 00:38:36 UTC+1:
> > On Thursday, November 11, 2021 at 1:00:36 PM UTC-8, WM wrote:
> >
> > > Do you know why quantifier exchange is forbidden?
> > > ∀ n∈ℕ ∃ m∈ℕ: m > n .
> > > That is correct. By exchanging the quantifiers
> > > ∃ m∈ℕ ∀ n∈ℕ : m > n
> > > we get a wrong statement. Why?
> > A more interesting question is why you (or anyone) would think that it should be a CORRECT statement? Why should 'quantifier exchange' NOT be forbidden?
> It is a scientific attitude to ask for reasons of results. Here we have the result that quantifier exchange is wrong. Why?

We have a kindergarten class with 12 students.

Every child in the class is eating an ice cream cone. But reversing the quantifiers we get: There is an ice cream cone that every child is eating. Is that a correct statement? Why? According to YOUR 'reasoning' it must be correct since there are only finitely many students.

[FredJeffries]

On Friday, November 12, 2021 at 4:32:38 AM UTC-8, WM wrote:

> ℕ and ℕ _def are different sets.

Your ℕ _def is not a set.

[FredJeffries]

Any sufficiently advanced technology is indistinguishable from magic.
-- Arthur C Clarke

https://en.wikipedia.org/wiki/Clarke%27s_three_laws

[WM]

You misunderstood. I did not say that quantifier exchange is correct. It is not correct in my examples let alone in your finitary example. But in your example, or the more famous dancers or lovers, the reason is very simple to understand. In my second example however, the reason, dark numbers, is very difficult to see.

We should investigate it in particular because the reversion is not always false. There is a mother of all men, at least according to the bible and, if we define a certain standard for homo sapiens sapiens, also in reality.

Regards, WM

[WM]

It is not a set according to the definition of set theory. But it is tedious to use always "or collection". Further set theory has not patented the name. I use set also in my book about classical mathematics without alephs: "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015).

Regards, WM

[Jim Burns]

On 11/11/2021 4:00 PM, WM wrote:

> Do you know why quantifier exchange is forbidden?
> ∀ n∈ℕ ∃ m∈ℕ: m > n .
> That is correct.
> By exchanging the quantifiers
> ∃ m∈ℕ ∀ n∈ℕ : m > n
> we get a wrong statement. Why?

Because '>' is totally ordered.

How to see that's the reason...
|
|| ∀n ∈ ℕ, ∃m ∈ ℕ ( m > n )
| iff
|| ∀m ∈ ℕ, ∃n ∈ ℕ ( n > m )
|
| and
|| ∀m ∈ ℕ, [∃n ∈ ℕ ( n > m )]
| iff
|| ~∃m ∈ ℕ, ~ [∃n ∈ ℕ ( n > m )]
|
| and
|| ~∃m ∈ ℕ, [ ~∃n ∈ ℕ [( n > m )] ]
| iff
|| ~∃m ∈ ℕ, [ ∀n ∈ ℕ ~ [( n > m )] ]
|
| and
|| ~∃m ∈ ℕ, ∀n ∈ ℕ [ ~( n > m ) ]
| iff
|| ~∃m ∈ ℕ, ∀n ∈ ℕ [ ( m >= n ) ]

Thus,
| ∀n ∈ ℕ, ∃m ∈ ℕ ( m > n )
and
| ∃m ∈ ℕ, ∀n ∈ ℕ ( m >= n )

is a contradiction.

> Why?
> In the upper line, first n∈ℕ has to be chosen,
> it can easily be surpassed.
> In the second line the whole set ℕ is available
> to surpass m.

No, that's not the reason.
The reasons are

(i)
'<' is a total order

Thus,
x > y xor y >= x

(ii)
1 ≠ 0

Either
P(x) is true for
at least _one_ possible referent of x,
xor
P(x) is true for
_no_ possible referents of x
( and P(x) is false for
( all possible referents of x

Thus,
∃x P(x) xor ∀x ~P(x)

> Now consider the endsegments E(n) = {n, n+1, n+2, ...}:
> ∀ n∈ℕ ∃ M⊂ℕ: M ⊂ E(n) ∧ |M| = ℵo .
> That is correct. By exchanging the quantifiers

You might have noticed me mention "truth-preserving",
"preserving truth", "truth-preserving-ly", and so on
a few bazillion times.

I have a reason for all those mentions.

By partially describing an indefinite *one* of
infinitely-many and reasoning *truth-preserving-ly*
from its partial description, we can reach conclusions
which we will know preserves the truth of the description,
even when we don't know which one of the *infinitely-many*
we are reasoning about.

There is a progression here:
... one ... truth-preserving-ly ... infinitely-many ...

Limiting ourselves to truth-preserving operations
is an essential aspect of how we finite beings
are learning about the infinite by reasoning
about the one.

That quantifier exchange does not preserve truth.

Because of
... one ... truth-preserving-ly ... infinitely-many ...
there appears to be no point to making that exchange.

Why are you making it?

[FredJeffries]

On Friday, November 12, 2021 at 6:38:36 AM UTC-8, WM wrote:
> FredJeffries schrieb am Freitag, 12. November 2021 um 14:01:31 UTC+1:
> > On Friday, November 12, 2021 at 4:32:38 AM UTC-8, WM wrote:
> > > FredJeffries schrieb am Freitag, 12. November 2021 um 00:38:36 UTC+1:
> > > > On Thursday, November 11, 2021 at 1:00:36 PM UTC-8, WM wrote:
> > > >
> > > > > Do you know why quantifier exchange is forbidden?
> > > > > ∀ n∈ℕ ∃ m∈ℕ: m > n .
> > > > > That is correct. By exchanging the quantifiers
> > > > > ∃ m∈ℕ ∀ n∈ℕ : m > n
> > > > > we get a wrong statement. Why?
> > > > A more interesting question is why you (or anyone) would think that it should be a CORRECT statement? Why should 'quantifier exchange' NOT be forbidden?
> > > It is a scientific attitude to ask for reasons of results. Here we have the result that quantifier exchange is wrong. Why?
> > We have a kindergarten class with 12 students.
> >
> > Every child in the class is eating an ice cream cone. But reversing the quantifiers we get: There is an ice cream cone that every child is eating. Is that a correct statement? Why? According to YOUR 'reasoning' it must be correct since there are only finitely many students.
> You misunderstood. I did not say that quantifier exchange is correct.

You misunderstand my (deliberate) misunderstanding. You ask 'Why?'

You then proceed to use your gibberish about 'infinite' 'sets' to try to provide an answer.

But the answer has NOTHING to do with 'infinite' 'sets'. It is much more fundamental than that. And once you understand THAT fundamental, the surprising happening is when quantifier switch DOES result in a correct statement.

You are noticing a red apple falling to the ground and ask 'Why?' And you search for an answer by examining the color red or the nutritional value of apples.

EOD

[FredJeffries]

On Friday, November 12, 2021 at 6:43:35 AM UTC-8, WM wrote:
> FredJeffries schrieb am Freitag, 12. November 2021 um 14:03:35 UTC+1:
> > On Friday, November 12, 2021 at 4:32:38 AM UTC-8, WM wrote:
> > > ℕ and ℕ _def are different sets.
> > Your ℕ _def is not a set.
> It is not a set according to the definition of set theory.

So, once again, you ADMIT that your twenty years of gibberish postings has NOTHING to do with your stated aim of demonstrating the contradictions of (what is implicitly accepted on this group (notice the name of the group) as) set theory. Thank you for once again affirming that you are a charlatan foisting a shell game.

> But it is tedious to use always "or collection". Further set theory has not patented the name. I use set also in my book about classical mathematics without alephs: "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015).

And you, once again, clearly (sarcasm intended) state that you have absolutely no interest in clarity and precision and actually-being-understood. Again, thank you for once again affirming that you are a charlatan foisting a shell game.

EOD

[Python]

Being a charlatan here is harmless as most of the small audience is made
either of other cranks or educated people who can spot the sophistries
at first read.

Doing the same as a "teacher" at Hochschule Augsburg is a CRIME.

[Jim Burns]

On 11/12/2021 9:38 AM, WM wrote:
> FredJeffries schrieb
> am Freitag, 12. November 2021 um 14:01:31 UTC+1:
>> On Friday, November 12, 2021 at 4:32:38 AM UTC-8,
>> WM wrote:
>>> FredJeffries schrieb
>>> am Freitag, 12. November 2021 um 00:38:36 UTC+1:
>>>> On Thursday, November 11, 2021 at 1:00:36 PM UTC-8,
>>>> WM wrote:

>>>>> Do you know why quantifier exchange is forbidden?
>>>>> ∀ n∈ℕ ∃ m∈ℕ: m > n .
>>>>> That is correct. By exchanging the quantifiers
>>>>> ∃ m∈ℕ ∀ n∈ℕ : m > n
>>>>> we get a wrong statement. Why?

I am assuming you intended to say

| ∀n ∈ ℕ, ∃ m∈ℕ ( m > n )
and
| ∃m ∈ ℕ, ∀n ∈ ℕ ( m >= n )

----
Note that
| ∃m ∈ ℕ, ∀n ∈ ℕ ( m > n )
implies
| ∃m ∈ ℕ ( m > m )

If I'm wrong about your intentions,
a better answer to _why_
| ∃m ∈ ℕ, ∀n ∈ ℕ ( m > n )

is wrong is that ~( m > m )

> You misunderstood.
> I did not say that quantifier exchange is correct.

> We should investigate it in particular because the
> reversion

"reversion"? "quantifier exchange" was clear enough.

Too clear, perhaps.
I think that you are trying to vague up the discussion.

> We should investigate it in particular because the
> reversion is not always false.

It needs to be always true in order to argue
truth-preserving-ly. "Not always false" is useless.

This is always true:

| ∃x ∈ B, ∀k ∈ C, P(x,k)
implies
| ∀k ∈ C, ∃x ∈ B, P(x,k)

This is NOT always true:

| ∀k ∈ C, ∃x ∈ B, P(x,k)
implies
| ∃x ∈ B, ∀k ∈ C, P(x,k)

> We should investigate it in particular because the
> reversion is not always false.

In the case of '>', it is always false.
(Not always true is reason enough to not use it.)

> There is a mother of all men, at least according to
> the bible and, if we define a certain standard for
> homo sapiens sapiens, also in reality.

You have changed what "mother" means.
This is _equivocation_ which is a fallacy.
Just to be clear:
*DON'T* use fallacies in your reasoning.

There is no "mother of all men" in the first sense.
My mother's mother is not my mother.
That's proof enough that there isn't.

In the second sense, my mother's mother is
"a mother of me".
She isn't my mother in the first sense..

[WM]

Jim Burns schrieb am Freitag, 12. November 2021 um 16:29:32 UTC+1:
> On 11/11/2021 4:00 PM, WM wrote:
>
> > Do you know why quantifier exchange is forbidden?
> > ∀ n∈ℕ ∃ m∈ℕ: m > n .
> > That is correct.
> > By exchanging the quantifiers
> > ∃ m∈ℕ ∀ n∈ℕ : m > n
> > we get a wrong statement. Why?

> > In the upper line, first n∈ℕ has to be chosen,
> > it can easily be surpassed.
> > In the second line the whole set ℕ is available
> > to surpass m.
> No, that's not the reason.
> The reasons are
>
> (i)
> '<' is a total order

That is a preconditon but not a reason. In the set {1, 2, 3} there is also total order.

> > Now consider the endsegments E(n) = {n, n+1, n+2, ...}:
> > ∀ n∈ℕ ∃ M⊂ℕ: M ⊂ E(n) ∧ |M| = ℵo .
> > That is correct. By exchanging the quantifiers
> You might have noticed me mention "truth-preserving",
> "preserving truth", "truth-preserving-ly", and so on
> a few bazillion times.
>
> I have a reason for all those mentions.

Quantifier exchange is not truth preserving. Please stay with the topic.

>
> That quantifier exchange does not preserve truth.

Yes, that is clear without all your evasions.

>
> Because of
> ... one ... truth-preserving-ly ... infinitely-many ...
> there appears to be no point to making that exchange.
>
> Why are you making it?

That is the point! I make it in order to show that for all definable natnumbers
∀ n∈ℕ_def ∃ M⊂ℕ: M ⊂ E(n) ∧ |M| = ℵo
there is an actually infinite endsegment. But there is no infinite endsegment for all natnumbers. This statement
∃ M⊂ℕ ∀ n∈ℕ : M ⊂ E(n) ∧ |M| = ℵo
is wrong. Hence infinitely many elements of M disappear. What can cause this loss? It is the well-known mechanism
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
over the set ℕ \ ℕ_def. We have the ultimate equation

|ℕ| = oo + ℵo .

Regards, WM

[WM]

FredJeffries schrieb am Freitag, 12. November 2021 um 16:54:51 UTC+1:
> On Friday, November 12, 2021 at 6:38:36 AM UTC-8, WM wrote:

> > You misunderstood. I did not say that quantifier exchange is correct.

> You are noticing a red apple falling to the ground and ask 'Why?' And you search for an answer by examining the color red or the nutritional value of apples.

No, I have discovered the reason. But when Newton talked about his apple experience, most of the audience may have reacted in the same spirit as you here.

Regards, WM

[WM]

FredJeffries schrieb am Freitag, 12. November 2021 um 17:00:13 UTC+1:
> On Friday, November 12, 2021 at 6:43:35 AM UTC-8, WM wrote:

> > But it is tedious to use always "or collection". Further set theory has not patented the name. I use set also in my book about classical mathematics without alephs: "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015).
> And you,

So EOD lasts 6 minutes in your case?

> once again, clearly (sarcasm intended) state that you have absolutely no interest in clarity and precision

Yes that must be sarcasm. In mathematics, as you should know, many words are used with different meanings. To use set for a collection is not the worst crime.

Regards, WM

[Dan Christensen]

On Thursday, November 11, 2021 at 3:54:59 PM UTC-5, WM wrote:
> Dan Christensen schrieb am Donnerstag, 11. November 2021 um 19:23:37 UTC+1:
> > On Thursday, November 11, 2021 at 12:54:35 PM UTC-5, WM wrote:
> > > Dan Christensen schrieb am Donnerstag, 11. November 2021 um 16:33:51 UTC+1:
> > > > On Thursday, November 11, 2021 at 5:16:19 AM UTC-5, WM wrote:
> > > > > Set theorists confess that every natural number n has an infinite distance to ω
> > > > > ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo (*)
> > > > > but that the set ℕ has no distance to ω. So either the distance is removed by some magic action when forming the set, or undefinable numbers are remaining and are simply included as individually unspecified elements in "the infinite set ℕ". These are the alternatives known to me. Are there more?
> > > > >
> > > > Maybe you didn't know, but, there is no greatest natural number. And that, for any natural number n, there are infinitely many natural numbers are greater than n. There is no end to them. Really!
> > > That distance to ω is expressed in (*).
> >
> > > > There is no end to them. Really!
> >
> > > Taking the set ℕ ends them.

> > The set N contains them. That doesn't mean there exists a greatest element of N.

> But it means that the distance has vanished. The set contains what cannot be observed individually.

Pure gibberish!

> > > The distance to ω vanishes completely although for every single n appearing in (*) the distance remains the same.
> > >
> > I think you mean that, for ANY n, m in N, we have: |{n, n+1, n+2, ... }| = |{m, m+1, m+2, ... }| = ℵo
> >
> > Makes sense to me. What is your problem?

> ∀ n∈ℕ ∃ m∈ℕ: m > n .

Yes.

> That is correct. By exchanging the quantifiers
> ∃ m∈ℕ ∀ n∈ℕ : m > n
> we get a wrong statement. Why?

Because you made an elementary mistake in logic, Mucke. Every person has a mother. That does not mean that every person has the same mother. Think about it. Draw a picture.

Time to pack it in, Mucke. Math just isn't your thing. Try knitting instead.

Regards,
Dan

[WM]

Python schrieb am Freitag, 12. November 2021 um 17:30:30 UTC+1:

> Being a charlatan here is harmless as most of the small audience is made
> either of other cranks or educated people who can spot the sophistries
> at first read.

And you think to be educated?

>
> Doing the same as a "teacher" at Hochschule Augsburg is a CRIME.

And having sold thousands of books with the corrected mathematical contents is certainly also a crime in the eyes of matheologians who mutilate the despised mathematics by violating inclusion monotony, pigeonhole principle and accepting two consecutive actually infinite sets in the natural order of |N.

Regards, WM

[Jim Burns]

On 11/12/2021 1:26 PM, WM wrote:
> Jim Burns schrieb
> am Freitag, 12. November 2021 um 16:29:32 UTC+1:
>> On 11/11/2021 4:00 PM, WM wrote:

>>> Do you know why quantifier exchange is forbidden?
>>> ∀ n∈ℕ ∃ m∈ℕ: m > n .
>>> That is correct.
>>> By exchanging the quantifiers
>>> ∃ m∈ℕ ∀ n∈ℕ : m > n
>>> we get a wrong statement. Why?
>>> In the upper line, first n∈ℕ has to be chosen,
>>> it can easily be surpassed.
>>> In the second line the whole set ℕ is available
>>> to surpass m.
>>
>> No, that's not the reason.
>> The reasons are
>> (i)
>> '<' is a total order
>
> That is a preconditon but not a reason.
> In the set {1, 2, 3} there is also total order.

In {1,2,3},
| ∀n ∈ {1,2,3}, ∃m ∈ {1,2,3}, m > n

is wrong.

| ∀n ∈ {1,2,3}, ∃m ∈ {1,2,3}, m > n

is still equivalent to
| ~∃m ∈ {1,2,3}, ∀n ∈ {1,2,3}, m >= n

But now they're both false.
With ℕ, they're both true.

----
What you want *which you don't know that you want*
is a finite sequence of statements
H[1], H[2], ... , S[j-1], S[j], S[j+1}, ... , C[k]

in which each statement is
either
assumed to be true for the purpose of deriving C[k]
( the H[1], H[2], ... )
or
follows by _only truth-preserving inferences_ from
statements earlier in the finite sequence
( C[k] follows from statements in H[1],...,S[k-1]
( ...
( S[j+1] follows from statements in H[1],...,S[j]
( S[j] followsfrom statents in H[1],...,S[j-1]
( ...
( H[2] need not follow from anything in the sequence
( H[1] need not follow from anything in the sequence

If we have such a statement-sequence, we say that
conclusion C[k] _follows from_ assumptions H[1],H[2],...

----
| ∀n ∈ {1,2,3}, ∃m ∈ {1,2,3}, m > n

is not true.

However amusing it might be to use only _truth-preserving_
inferences from that assumption, if there is
no truth to preserve in the H[1],H[2],... then we have
no justification from H[1],...,C[k] for the truth of C[k].

[WM]

Jim Burns schrieb am Freitag, 12. November 2021 um 19:25:27 UTC+1:
> On 11/12/2021 9:38 AM, WM wrote:

> >>>>> Do you know why quantifier exchange is forbidden?
> >>>>> ∀ n∈ℕ ∃ m∈ℕ: m > n .
> >>>>> That is correct. By exchanging the quantifiers
> >>>>> ∃ m∈ℕ ∀ n∈ℕ : m > n
> >>>>> we get a wrong statement. Why?
> I am assuming you intended to say
> | ∀n ∈ ℕ, ∃ m∈ℕ ( m > n )
> and
> | ∃m ∈ ℕ, ∀n ∈ ℕ ( m >= n )
> ----
> Note that
> | ∃m ∈ ℕ, ∀n ∈ ℕ ( m > n )
> implies
> | ∃m ∈ ℕ ( m > m )

I know, but since it is wrong in every case, I have not bothered.

>
> > I did not say that quantifier exchange is correct.
> > We should investigate it in particular because the
> > reversion
> "reversion"? "quantifier exchange" was clear enough.

Thought it was synonymous.

>
> I think that you are trying to vague up the discussion.
> > We should investigate it in particular because the
> > reversion is not always false.
> It needs to be always true in order to argue
> truth-preserving-ly. "Not always false" is useless.

It make the question interesting why it is false in some cases.

> There is no "mother of all men" in the first sense.

But you can grasp the sense I use. And certainly you can find examples where quantifier exchange is not false.

Regards, WM

[Jim Burns]

On 11/12/2021 2:20 PM, WM wrote:
> Jim Burns schrieb
> am Freitag, 12. November 2021 um 19:25:27 UTC+1:
>> On 11/12/2021 9:38 AM, WM wrote:

>>> We should investigate it in particular because the
>>> reversion is not always false.
>>
>> It needs to be always true in order to argue
>> truth-preserving-ly. "Not always false" is useless.
>
> It make the question interesting why it is false
> in some cases.

"De gustibus non est disputandum"

In the case of a total order '<',
the reason it's false is that x > y xor y >= x

What we've been calling a _quantifier_ exchange
is described better as a _variable-name_ exchange.

| ∀n ∈ ℕ, ∃m ∈ ℕ ( m > n )

can be written as
| ~∃n ∈ ℕ, ~∃m ∈ ℕ ( m > n )


| ~∃m ∈ ℕ, ∀n ∈ ℕ ( m >= n )

can be written as
| ~∃m ∈ ℕ, ~∃n ∈ ℕ ~( m >= n )

which is

| ~∃m ∈ ℕ, ~∃n ∈ ℕ ( n > m )

_exchanging variable-names_
| ~∃n ∈ ℕ, ~∃m ∈ ℕ ( m > n )

> And certainly you can find examples
> where quantifier exchange is not false.

| ∃x ∈ B, ∃k ∈ C, P(x,k)

forces
| ∃k ∈ C, ∃x ∈ B, P(x,k)

| ∀x ∈ B, ∀k ∈ C, P(x,k)
forces

| ∀k ∈ C, ∀x ∈ B, P(x,k)

| ∃x ∈ B, ∀k ∈ C, P(x,k)

forces

| ∀k ∈ C, ∃x ∈ B, P(x,k)

| ∀x ∈ B, ∃k ∈ C, P(x,k)
*does not force*
| ∃k ∈ C, ∀x ∈ B, P(x,k)

There are instances in which this last variety of
quantifier exchange produces a true final statement
from a true initial statement.

( Any instance in which ∃k ∈ C, ∀x ∈ B, P(x,k)
( is true springs immediately to mind.

The goal is to start from whatever initial statements
we need and then *force* the conclusion, whatever that
conclusion may be. Saying "C[k] might or might not
be true" is singularly unimpressive. It's where we
were before we began.

( It's conceivable that ∃k ∈ C, ∀x ∈ B, P(x,k)
( is forced by _other_ information we have, or by some
( _other_ statement-sequence. There's nothing toobject
( to in that. But, if that's so, _this quantifier_
( _exchange_ isn't playing any role.

[WM]

Jim Burns schrieb am Freitag, 12. November 2021 um 21:19:11 UTC+1:
> On 11/12/2021 2:20 PM, WM wrote:

> What we've been calling a _quantifier_ exchange
> is described better as a _variable-name_ exchange.
> | ∀n ∈ ℕ, ∃m ∈ ℕ ( m > n )

> | ~∃m ∈ ℕ, ∀n ∈ ℕ ( m >= n )

Let us stay with the correct statements.
>
> The goal is

to accept for all n that can be envisaged
∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
and to explain what
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
means.

Regards, WM

[WM]

Dan Christensen schrieb am Freitag, 12. November 2021 um 20:00:50 UTC+1:
> On Thursday, November 11, 2021 at 3:54:59 PM UTC-5, WM wrote:

> > > The set N contains them. That doesn't mean there exists a greatest element of N.
>
> > But it means that the distance has vanished. The set contains what cannot be observed individually.
> Pure gibberish!

Try wherther your software can understand and help you to interpret

∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo .

~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo

> Every person has a mother. That does not mean that every person has the same mother.

Look: Mother, Grandmother, Grandgrandmother, Grandgrandgrandmother, ... Think about it. Draw a picture.

Regards, WM

[Jim Burns]

On 11/12/2021 3:59 PM, WM wrote:
> Jim Burns schrieb
> am Freitag, 12. November 2021 um 21:19:11 UTC+1:

>> What we've been calling a _quantifier_ exchange
>> is described better as a _variable-name_ exchange.
>> | ∀n ∈ ℕ, ∃m ∈ ℕ ( m > n )
>> | ~∃m ∈ ℕ, ∀n ∈ ℕ ( m >= n )
>
> Let us stay with the correct statements.

<WM<JB>>

>>> I am assuming you intended to say

>>> | ∀n ∈ ℕ, ∃ m∈ℕ ( m > n )

>>> and

>>> | ∃m ∈ ℕ, ∀n ∈ ℕ ( m >= n )

>>> ----
>>> Note that
>>> | ∃m ∈ ℕ, ∀n ∈ ℕ ( m > n )
>>> implies
>>> | ∃m ∈ ℕ ( m > m )
>>
>> I know, but since it is wrong in every case,
>> I have not bothered.

It's not so much that you haven't bothered.
You've bothered to ask for the incorrect version,
after all.

It's that the correct

| ∀n ∈ ℕ, ∃ m∈ℕ ( m > n )

and

| ∃m ∈ ℕ, ∀n ∈ ℕ ( m >= n )

don't look like a quantifier exchange.

How did a quantifier exchange insert '='
inside there?

>> The goal is
>
> to accept for all n that can be envisaged
> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> and to explain what
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> means.

Here is my explanation.

| Set B ⊆ ℕ contains a maximum <->
| B ⊆ ℕ is non-empty finite

| Set B ⊆ ℕ does not contain a minimum <->
| B ⊆ ℕ is empty

----
Define
{j,...,k} to be the two-ended natural segment
with j and k as ends.

{j,...,k} contains a maximum and a minimum.

Define
{j,...} to be
the union over k of all {j,...,k} with minimum j.

j is the common minimum of all {j,...,k}
There is no common maximum of all {j,...,k}

{j,...} contains a minimum j but no maximum

Define
{...} to be the intersection over j of all {j,...}

There is no maximum to any {j,...}
There is no common minimum of all {j,...}

{...} contains no maximum and no minimum.

To sum up:
{j,...,k} {j,...} {...}
max Y N N
no min N N Y

fin≠{} Y N N
{} N N Y

infinite N Y N

Two-ended {j,...,k} is non-empty-finite.

One-ended {j,...} is not non-empty-finite and not empty.
{j,...}is infinite.

No-ended {...} is empty.

[Python]

Crank Wolfgang Mückenheim, aka WM wrote:

> Python schrieb am Freitag, 12. November 2021 um 17:30:30 UTC+1:
>
>> Being a charlatan here is harmless as most of the small audience is made
>> either of other cranks or educated people who can spot the sophistries
>> at first read.
>
> And you think to be educated?

FUCK YOU idiot Mückenheim. ONCE you'll pay for your crimes.

>> Doing the same as a "teacher" at Hochschule Augsburg is a CRIME.
>
> And having sold thousands of books

books full of mistakes and sophistries, you've earned money for that,
crank? And you're proud of it? You are a SHAME, a DISGRACE.

Sooner or later you'll pay for your crime Mückenheim.

[WM]

Jim Burns schrieb am Samstag, 13. November 2021 um 00:49:53 UTC+1:
> On 11/12/2021 3:59 PM, WM wrote:
> > Jim Burns schrieb

> It's that the correct
> | ∀n ∈ ℕ, ∃ m∈ℕ ( m > n )
> and
> | ∃m ∈ ℕ, ∀n ∈ ℕ ( m >= n )
> don't look like a quantifier exchange.
>
> How did a quantifier exchange insert '='
> inside there?

Insert it also in th first statement. Then you have recovered the quantifier exchange.

> >> The goal is
> >
> > to accept for all n that can be envisaged
> > ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> > and to explain what
> > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> > means.
> Here is my explanation.
>
> | Set B ⊆ ℕ contains a maximum <->
> | B ⊆ ℕ is non-empty finite
>
> | Set B ⊆ ℕ does not contain a minimum <->
> | B ⊆ ℕ is empty

Set M does not contain a maximum. Neither number n nor set M are the same in every statement above. But M is infinite in all statements. Further deviation deleted. Again the two statements
∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo

are contradictory for ℕ_def = ℕ. Infinitely many natural numbers cannot be chosen but only numbers of ℕ\M.

Regards, WM

[WM]

Python schrieb am Samstag, 13. November 2021 um 04:00:31 UTC+1:

> > And having sold thousands of books

> you've earned money for that,

of course.

> And you're proud of it?

of course. But mainly I am proud of having shown set theory wrong:

With the endsegments E(n) = {n, n+1, n+2, ...} we get

∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

Therefore it is impossible to apply most natural numbers (those of M) for counting or mapping.

Regards, WM

[Dan Christensen]

On Friday, November 12, 2021 at 4:04:36 PM UTC-5, WM wrote:
> Dan Christensen schrieb am Freitag, 12. November 2021 um 20:00:50 UTC+1:
> > On Thursday, November 11, 2021 at 3:54:59 PM UTC-5, WM wrote:
>
> > > > The set N contains them. That doesn't mean there exists a greatest element of N.
> >
> > > But it means that the distance has vanished. The set contains what cannot be observed individually.

> > Pure gibberish!

> Try wherther your software can understand and help you to interpret
> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo .
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo

Trying to change the subject, Mucke? Can't blame you. You are looking like a complete idiot here. Even the basic rules of logic seem to elude you.

[unsnipping]

> > > ∀ n∈ℕ ∃ m∈ℕ: m > n .

> > Yes.

> > > That is correct. By exchanging the quantifiers
> > > ∃ m∈ℕ ∀ n∈ℕ : m > n
> > > we get a wrong statement. Why?

> > Because you made an elementary mistake in logic, Mucke. Every person has a mother. That does not mean that every person has the same mother.

> Look: Mother, Grandmother, Grandgrandmother, Grandgrandgrandmother, ... Think about it. Draw a picture.
>

Confirming once again that you just don't understand even the basic rules of logic.

From ∀ x∈S ∃ y∈S : R(x,y), we cannot always infer that ∃ y∈S ∀ x∈S : R(x,y). Other information about R would be required to make such an inference.

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Jim Burns]

On 11/13/2021 3:51 AM, WM wrote:
> Jim Burns schrieb
> am Samstag, 13. November 2021 um 00:49:53 UTC+1:
>> On 11/12/2021 3:59 PM, WM wrote:
>>> Jim Burns schrieb

>> It's that the correct
>> | ∀n ∈ ℕ, ∃m ∈ ℕ ( m > n )
>> and
>> | ∃m ∈ ℕ, ∀n ∈ ℕ ( m >= n )
>> don't look like a quantifier exchange.
>>
>> How did a quantifier exchange insert '='
>> inside there?
>
> Insert it also in th first statement.
> Then you have recovered the quantifier exchange.

(i)
With '=', the first statement becomes trivially true
| ∀n ∈ ℕ, ∃m ∈ ℕ ( m >= n )

because ( n >= n )

It's true, but not because of ℕ
For example
| ∀n ∈ {1,2,3}, ∃m ∈ {1,2,3} ( m >= n )

Are there dark numbers in {1,2,3}? Which ones?

(ii)
Without '=', the second statement becomes trivially false
| ∃m ∈ ℕ, ∀n ∈ ℕ ( m > n )

because ~( m > m )

(iii)
However, the important point is that the WM-exchange
is not truth-preserving. That does not mean it always
gives a false result. It means it sometimes gives
a false result.

That's enough to disallow its use in an argument.
|
| The WM-exchange says BLAH
|
| But the WM-exchange is wrong sometimes.
|
| Not this time!
|
| Why isn't the WM-exchange wrong this time?
|
| ...

That's what you need to tell us, if you can.
Give us what makes this time special,
and just don't mention the faulty WM-exchange.

>>>> The goal is
>>>
>>> to accept for all n that can be envisaged
>>> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
>>> and to explain what
>>> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
>>> means.
>>
>> Here is my explanation.

first rule:

>> | Set B ⊆ ℕ contains a maximum <->
>> | B ⊆ ℕ is non-empty finite
>>

second rule:

>> | Set B ⊆ ℕ does not contain a minimum <->
>> | B ⊆ ℕ is empty
>
> Set M does not contain a maximum.

That follows from |M| = ℵo and the first rule.

Note that {} also does not contain a maximum.

> Neither number n nor set M are the same in every
> statement above. But M is infinite in all statements.

And the second statement says
infinite M subset all end segments does not exist.

The second statement is true.
By the second rule,
non-empty M ⊂ ℕ contains a minimum min(M)
M is not subset E(min(M)+1)

Turning that around, we get,
if M subset all end segments,
then M ⊂ ℕ does not contain a minimum,
so M must be empty.

> Further deviation deleted.

Your loss.

> Again the two statements
> ∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> are contradictory for ℕ_def = ℕ.

| The WM-exchange is not truth-preserving.
| "Set theory" (first-order logic) doesn't claim
| that the WM-exchange is truth-preserving.
| Proving that the WM-exchange is not truth-preserving
| does not prove that "set theory" is wrong.

> Again the two statements
> ∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> are contradictory for ℕ_def = ℕ.
> Infinitely many natural numbers
> cannot be chosen but only numbers of ℕ\M.

Define
ℕ such that
∀n ( n ∈ ℕ <-> {1,...,n} exists )


Define
{j,...,k} such that
for each split, its crossing-pair i,i+1 exists,
{j,...,k} contains max k, min j

{j,...,k} ⊆ ℕ == max Y, min Y


Define {j,...} ⊆ ℕ as
the union over k ∈ ℕ of {j,...,k} ⊆ ℕ

{j,...} ⊆ ℕ == max N, min Y


Define {...} as
the intersection over j ∈ ℕ of {j,...} ⊆ ℕ

{...} ⊆ ℕ == max N, min N


max Y, min Y == finite, not empty

max N, min Y == not finite

max N, min N == empty

[WM]

Dan Christensen schrieb am Samstag, 13. November 2021 um 15:47:41 UTC+1:
> On Friday, November 12, 2021 at 4:04:36 PM UTC-5, WM wrote:
> > Dan Christensen schrieb am Freitag, 12. November 2021 um 20:00:50 UTC+1:
> > > On Thursday, November 11, 2021 at 3:54:59 PM UTC-5, WM wrote:
> >
> > > > > The set N contains them. That doesn't mean there exists a greatest element of N.
> > >
> > > > But it means that the distance has vanished. The set contains what cannot be observed individually.
>

> > Try wherther your software can understand and help you to interpret
> > ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo .
> > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> Trying to change the subject,

Chuckle. This is the subject.

Regards, WM

[WM]

Jim Burns schrieb am Samstag, 13. November 2021 um 17:49:34 UTC+1:
> On 11/13/2021 3:51 AM, WM wrote:
> > Jim Burns schrieb
> > am Samstag, 13. November 2021 um 00:49:53 UTC+1:
> >> On 11/12/2021 3:59 PM, WM wrote:
> >>> Jim Burns schrieb
>
> >> It's that the correct
> >> | ∀n ∈ ℕ, ∃m ∈ ℕ ( m > n )
> >> and
> >> | ∃m ∈ ℕ, ∀n ∈ ℕ ( m >= n )
> >> don't look like a quantifier exchange.
> >>
> >> How did a quantifier exchange insert '='
> >> inside there?
> >
> > Insert it also in th first statement.
> > Then you have recovered the quantifier exchange.
> (i)
> With '=', the first statement becomes trivially true

of course. But that doesn't matter.

> Are there dark numbers in {1,2,3}? Which ones?

None.

>
> (ii)
> Without '=', the second statement becomes trivially false

of course. But that doesn't matter.

>

> However, the important point is that the WM-exchange
> is not truth-preserving. That does not mean it always
> gives a false result. It means it sometimes gives
> a false result.

> That's what you need to tell us, if you can.
> Give us what makes this time special,

Here we have two correct statements, according to ZF

∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

> > Neither number n nor set M are the same in every
> > statement above. But M is infinite in all statements.
> And the second statement says
> infinite M subset all end segments does not exist.
>
> The second statement is true.

Of course. The first is also true. Between every chosen natnumber and ω there is an infinite set. Between ℕ or ∀ n∈ℕ there is nothing. Coclusion?

> Turning that around, we get,
> if M subset all end segments,
> then M ⊂ ℕ does not contain a minimum,
> so M must be empty.

M contains a minimum: It has ℵ₀ elements.
An example is the set of all fractions between 0 and (n+1)/n. It has ℵ₀ elements,

>
> > Again the two statements
> > ∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> > are contradictory for ℕ_def = ℕ.
> | The WM-exchange is not truth-preserving.
> | "Set theory" (first-order logic) doesn't claim
> | that the WM-exchange is truth-preserving.

"Set theory" claims that both statements are true however.

Regards, WM

[Dan Christensen]

Wrong again, Mucke. You incorrectly interchanged quantifiers and erroneously came to the conclusion that there must be a largest natural number. You are looking like a real idiot here, Mucke. Once again, you have failed to demonstrate any inconsistency in set theory.

[Jim Burns]

On 11/13/2021 12:33 PM, WM wrote:
> Jim Burns schrieb

> am Samstag, 13. November 2021 um 17:49:34 UTC+1:

>> However, the important point is that the WM-exchange
>> is not truth-preserving. That does not mean it always
>> gives a false result. It means it sometimes gives
>> a false result.
>
>> That's what you need to tell us, if you can.
>> Give us what makes this time special,
>
> Here we have two correct statements, according to ZF
> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

The rest of your argument for dark numbers is that,
contrary to ZF, *this* is right

| ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

| ∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

and therefore, you argue, ZF is wrong.

You use the WM-exchange to argue from the correct first
to the incorrect (so we say) second.

The WM-exchange does not preserve truth.

In order to make your case in this way, it is necessary
for the WM-exchange to preserve truth.

You cannot make your case this way.

>> Turning that around, we get,
>> if M subset all end segments,
>> then M ⊂ ℕ does not contain a minimum,
>> so M must be empty.
>
> M contains a minimum: It has ℵ₀ elements.

_What you want to show_ is that M has ℵ₀ elements.
You're trying to show it with the WM-exchange.

The WM-exchange is not truth-preserving.

The WM-exchange does not show that.

----
If M contains a minimum,
then M is not a subset of E(min(M)+1)

If M is a subset of all end segments,
then M does not contain a minimum,
so M is empty.

[Jim Burns]

On 11/13/2021 12:33 PM, WM wrote:

> Here we have two correct statements, according to ZF
> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

∀M ⊆ ℕ,
0 < |M| < ℵo <->
∃k ∈ M, ∀j ∈ M (j =< k)

∀M ⊆ ℕ,
0 < |M| <->
∃i ∈ M, ∀j ∈ M (i =< j)


∃k ∈ {m,...,n}, ∀j ∈ {m,...,n} (j =< k)
∃i ∈ {m,...,n}, ∀j ∈ {m,...,n} (i =< j)

0 < |{m,...,n}| < ℵo
0 < |{m,...,n}|
---------------------
0 < |{m,...,n}| < ℵo


~∃k ∈ {m,...}, ∀j ∈ {m,...} (j =< k)
∃i ∈ {m,...}, ∀j ∈ {m,...} (i =< j)

~( 0 < |{m,...}| < ℵo )
0 < |{m,...}|
------------------------
~( |{m,...}| < ℵo )


~∃k ∈ {...}, ∀j ∈ {...} (j =< k)
~∃i ∈ {...}, ∀j ∈ {...} (i =< j)

~( 0 < |{...}| < ℵo )
~( 0 < |{...}| )
------------------------
~( 0 < |{...}| )

[Serg io]

That Book ? Gad, give their money back. Shame on you!

only use numbers that are defined by beeps, flashes, honks, raps taps ? it is a book for non math idiots

[Serg io]

obviously wrong.

[Serg io]

more lies, you don't know any of that. *You dont know math at all*. Go ahead put some algebra out there...

[Mostowski Collapse]

Thats almost as hilarious as Dan Christensen thinking ZFC is
an alternative to Peano. Whereby every child knows that Peano
has omega induction and ZFC has transfinite induction.

Cantor, the father of set theory, was tralking about transfinite
all the time, and that was more than 100 years ago.

[WM]

Jim Burns schrieb am Samstag, 13. November 2021 um 20:51:56 UTC+1:
> On 11/13/2021 12:33 PM, WM wrote:
> > Jim Burns schrieb

> > Here we have two correct statements, according to ZF
> > ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
> The rest of your argument for dark numbers is that,
> contrary to ZF, *this* is right
> | ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> | ∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

No, you misunderstood. I never said that it would be right.
See my post of 11.11.2021, 22:00:36
****************************************
Now consider the endsegments E(n) = {n, n+1, n+2, ...}:
∀ n∈ℕ ∃ M⊂ℕ: M ⊂ E(n) ∧ |M| = ℵo .

That is correct. By exchanging the quantifiers

∃ M⊂ℕ ∀ n∈ℕ : M ⊂ E(n) ∧ |M| = ℵo

we get a wrong statement.

**********************************************

So your following arguing is in vain. I accept both statements of ZF.

> If M contains a minimum,
> then M is not a subset of E(min(M)+1)
>
> If M is a subset of all end segments,
> then M does not contain a minimum,
> so M is empty.

Wrong.
Theorem: The intersection of gapless actually infinite subsets of ℕ, like the infinite endsegments, is infinite. This is the infinite set M.
Proof trivial. There are not two consecutive actually infinite set in the natural order of ℕ. Hence the complement of every infinite set is finite. But ℕ is actually infinite.

An easily understandable example the set of positive fractions between 0 and 1 + 1/n. The sequence of sets has an infinite mininimum, namely the fractions in (0, 1].

Why ist your counter argument insufficient?
Every natnumber that you can apply in your argument (n not in E(n+1)) has infinitely many successors. Therefore your argument about the emptiness of the intersection of all infinite endsegments concerns only these numbers but it does not prove anything about the successors, namely the actually infinite set M.

Regards, WM

[WM]

Jim Burns schrieb am Samstag, 13. November 2021 um 22:47:38 UTC+1:
> On 11/13/2021 12:33 PM, WM wrote:
> > Here we have two correct statements, according to ZF
> > ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
> ∀M ⊆ ℕ,
> 0 < |M| < ℵo <->
> ∃k ∈ M, ∀j ∈ M (j =< k)

No. M is dark. All numbers that you can apply have ℵo successors. You cannot apply any other natnumber. Now, according to your arguing, the successors would also have a minimum M. But you cannot find it. Therefore, you say that in

∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

the set M cannot exist.
This is wrong by the simple and correct statement ∀ n∈ℕ: E(n) = ℵo.

So all E(n) and their subset M are actually infinite. But then not all natnumbers of ℕ can be applied, because ℕ is actually infinite too; all its elements form an infinite set of card ℵo. If they all could appear in

∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

then we had two infinite consecutive sets ℕ, M. Contradiction. Or could you define the point where ℕ ends and M begins?

Regards, WM

[Mostowski Collapse]

In ZFC the finite ordinal n has much more ordinals alpha
with n < alpha, they are at least uncountable many, actually

the ordinals alpha with n < alpha is a proper class, so
you cannot assign it a cardinal. This class here:

{ alpha | n < alpha }

Is not only dark. It is ultra dark, the darkest darks
of the darks, its more than dark than vantablack.

[Mostowski Collapse]

Well maybe you can assign it a cardinal with suitable
large cardinal axioms, and what ever. But since you
can view cardinals as selected ordinals, and ordinals

as selected sets, assigning a cardinal to it would say
putting { alpha | n < alpha } in bijection with a set.
Which is impossible for a proper class.

Its similar to Dans NOT PURPLE, i.e. V \ p, only
here the class difference is On \ n.

[Mostowski Collapse]

The purple paradox for On \ n is now, instead of a finite
ordinal n, you can also take a finite ot infinite ordinal
beta, this here will never have a cardinality:

On \ beta

Can DC Proof, prove that? A small correction On \ beta
gives { alpha | beta =< alpha }, for von Neuman ordinals
where < is the same as in, needs some thinking.

[Dan Christensen]

On Sunday, November 14, 2021 at 5:16:04 AM UTC-5, Mostowski Collapse wrote:
> Thats almost as hilarious as Dan Christensen thinking ZFC is
> an alternative to Peano.

Where did you get that idea, Jan Burse?

[Mostowski Collapse]

By your idiotic comments:
> Better that you should start with the basics. Again, you might start by constructing
> (i.e. proving the existence of) the set of natural numbers N.

Thats not the basics of ZFC. Check out the table of contents here:

Table of Contents - Basic Set Theory - Levy
https://books.google.ch/books?id=hwrmDAAAQBAJ&pg=PR11&hl=de&source=gbs_selected_pages&cad=3#v=onepage&q&f=false

Omega is just an ordinal. You need first a notion of ordinal.

[Mostowski Collapse]

Try proving proposition 2.20 (goes back to Cantor 1897):

If <A, <> is a well ordered class and B is subclass of A, then
<B, <> is isomorphic to <A,<> or to a section of <A, <>.

[Jim Burns]

On 11/14/2021 5:50 AM, WM wrote:
> Jim Burns schrieb
> am Samstag, 13. November 2021 um 22:47:38 UTC+1:
>> On 11/13/2021 12:33 PM, WM wrote:

>>> Here we have two correct statements, according to ZF
>>> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
>>> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
>>
>> ∀M ⊆ ℕ,
>> 0 < |M| < ℵo <->
>> ∃k ∈ M, ∀j ∈ M (j =< k)
>
> No. M is dark.

When you use M in the second statement,
|
| M is an infinite subset of ℕ which is a subset of
| each end segment of ℕ
|
is a reasonable rendering of what I take you
to mean by "M is dark".

Then that second statement is '~∃M, ( M is dark )'
"Dark M does not exist".

And you claim you agree with that second statement.
And you keep on arguing as though everyone else is
stupid or dishonest for saying dark M does not exist.
From my guy-on-the-internet point of view, that
looks unhealthy.

----
When I use M here, M is a subset of ℕ.
M either has a finite non-zero cardinality
( 0 < |M| < ℵo ) or it doesn't.
M either contains a maximum
( ∃k ∈ M, ∀j ∈ M (j =< k) ) or it doesn't.

My claim is that either property without the other
is impossible.
My claim is about numbers with FISONs and sets of
numbers with FISONs.
My claim applies to any "dark numbers" with FISONs.
My claim DOES NOT apply to any "dark numbers"
WITHOUT FISONs.

> All numbers that you can apply have ℵo successors.

What does that mean?

If the followers of k are finitely-many,
they can be totally-ordered such that,
for each split, there is its crossing-pair,
and there are two ends.

If the followers of k are NOT finitely-many,
then they cannot be totally-ordered that way.

For each FISON-ender k, the FISON-enders which follow k
cannot be ordered that way, with crossing-pairs and ends.
This is what I mean by ~( E(k) < ℵo )

If you supplement FISON-enders with dark numbers,
they still cannot be ordered that way.
If you deny that the set E(k) exists,
the followers of k still cannot be ordered that way.

> You cannot apply any other natnumber.
> Now, according to your arguing, the successors
> would also have a minimum M.

Right.

| ∀M ⊆ ℕ,
| 0 < |M| <->
| ∃i ∈ M, ∀j ∈ M (i =< j)

Recall that, by definition, ℕ contains only FISON-enders.
Suppose 0 < |M|
Then m ∈ M and FISON {1,...,m} exists
min(M) (that is, i ) provably exists in {1,...,m}

Unless M is empty.

> But you cannot find it.

Some things we cannot find because they do not exist.
The minimum of the intersection of all end segments
is such a thing.

> Therefore, you say that in
> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> the set M cannot exist.

That's not what I say.
What _you and I_ say is

| ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

No _infinite_ intersection M exists.
An empty one exists, though.

> So all E(n) and their subset M are actually infinite.

All E(n) and their subset are infinite or empty.
That's a consequence of being subset of ℕ without maximum.
~( 0 < |M| < ℵo) <->
~( 0 < |M| ) ∨ ~( |M| < ℵo )

Each end segment has a minimum, it is not empty.
The only choice left is infinite.

There is no single minimum of all end segments.
No minimum for their intersection: empty.

[Dan Christensen]

On Sunday, November 14, 2021 at 1:54:40 PM UTC-5, Mostowski Collapse wrote:

> Omega is just an ordinal. You need first a notion of ordinal.
> Dan Christensen schrieb am Sonntag, 14. November 2021 um 17:19:19 UTC+1:
> > On Sunday, November 14, 2021 at 5:16:04 AM UTC-5, Mostowski Collapse wrote:
> > > Thats almost as hilarious as Dan Christensen thinking ZFC is
> > > an alternative to Peano.
> > Where did you get that idea, Jan Burse?

> By your idiotic comments:

> > Better that you should start with the basics. Again, you might start by constructing
> > (i.e. proving the existence of) the set of natural numbers N.
>
> Thats not the basics of ZFC. Check out the table of contents here:
>
> Table of Contents - Basic Set Theory - Levy
> https://books.google.ch/books?id=hwrmDAAAQBAJ&pg=PR11&hl=de&source=gbs_selected_pages&cad=3#v=onepage&q&f=false
>

Maybe proving the existence of N in ZFC is not as trivial as you claimed. It is almost trivial in DC Proof to construct N given an arbitrary Dedekind infinite set (supposedly the equivalent of AoI in ZFC). See: http://www.dcproof.com/ProofByInduction.html (only 247 lines).

[Mostowski Collapse]

Maybe, Maybe, Maybe.

Maybe you are a moron.

[WM]

Jim Burns schrieb am Sonntag, 14. November 2021 um 20:17:35 UTC+1:
> On 11/14/2021 5:50 AM, WM wrote:
> > Jim Burns schrieb
> > am Samstag, 13. November 2021 um 22:47:38 UTC+1:
> >> On 11/13/2021 12:33 PM, WM wrote:
>
> >>> Here we have two correct statements, according to ZF
> >>> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> >>> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
> >>
> >> ∀M ⊆ ℕ,
> >> 0 < |M| < ℵo <->
> >> ∃k ∈ M, ∀j ∈ M (j =< k)
> >
> > No. M is dark.
> When you use M in the second statement,
> |
> | M is an infinite subset of ℕ which is a subset of
> | each end segment of ℕ

Not of each endsegment but of each *infinite* endsegment. All endsegments of definable natnumbers are infinite. That is colloquially expressed by "every natural number has infinitely many successors". The intersection of a sequence of gapless infinite sets is a gapless infinite set.

> |
> is a reasonable rendering of what I take you
> to mean by "M is dark".
>
> Then that second statement is '~∃M, ( M is dark )'
> "Dark M does not exist".

In the second statement we have to consider all natural numbers. They are existing by definition (that is what I accept - it cannot be proved other than by claim or "axiom"). Under this premise there are all natural numers existing. Hence there cannot be infinite endsegments of numbers larger than all natural numbers, because also the contents of the endsegments are natural numbers. That is the difference between definable numbers and all numbers. The difference between the first and the second line above.

>
> And you claim you agree with that second statement.
> And you keep on arguing as though everyone else is
> stupid or dishonest for saying dark M does not exist.

Please distinguish: Every natural numer has ℵo successors cannot be correct for all natural numbers. ℕ has no successors between itself and ω. But experience teaches us that every named or defined or addressed natnumber has ℵo succesors.

> M either contains a maximum

M is the set of natural numbers that follows beyond every natural number. Every natural number that ever ca be named has an infinite endsegment. M is in all of them.

> My claim is about numbers with FISONs and sets of
> numbers with FISONs.

They are defined ny the FISONs. A FISON defines its last numbers (and also all predecessors).

> My claim applies to any "dark numbers" with FISONs.

There is no dark number with FISON.

> My claim DOES NOT apply to any "dark numbers"
> WITHOUT FISONs.

That is the property of being dark. All FISONs are not dark but
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
cannot be circumvented.

> > All numbers that you can apply have ℵo successors.
> What does that mean?
>
> If the followers of k are finitely-many,
> they can be totally-ordered such that,
> for each split, there is its crossing-pair,
> and there are two ends.

Dark numbers cannot be ordered. They are inaccessible. Either they exist in this form, or they don't exist at all.

>
> If the followers of k are NOT finitely-many,
> then they cannot be totally-ordered that way.
>
> For each FISON-ender k, the FISON-enders which follow k
> cannot be ordered that way, with crossing-pairs and ends.
> This is what I mean by ~( E(k) < ℵo )

No definable endsegment is smaller than ℵo. But these ℵo numbers are also natural numbers. What else should the endegments consist of. And when we consider all natural numbers, then also those sitting in the endsegmets are meant. Then nothing remains loarger than all natural numbers.

>
> > You cannot apply any other natnumber.
> > Now, according to your arguing, the successors
> > would also have a minimum M.
> Right.
> | ∀M ⊆ ℕ,
> | 0 < |M| <->
> | ∃i ∈ M, ∀j ∈ M (i =< j)
> Recall that, by definition, ℕ contains only FISON-enders.

If you claim that, then you get the potentially infinite Peano set.

>
> Each end segment has a minimum, it is not empty.
> The only choice left is infinite.
>
> There is no single minimum of all end segments.

There is no single minimum because there is not largest FISON. But all FISONs are finite (potential infinity, oo) and have an infinite complement (actual infinity, ℵo).

|ℕ| = oo + ℵo

> No minimum for their intersection: empty.

No minimum for the endsegments: empty.

Regards, WM

[Jim Burns]

On 11/14/2021 5:02 PM, WM wrote:
> Jim Burns schrieb

> am Sonntag, 14. November 2021 um 20:17:35 UTC+1:

>> | ∀M ⊆ ℕ,
>> | 0 < |M| <->
>> | ∃i ∈ M, ∀j ∈ M (i =< j)
>>
>> Recall that, by definition, ℕ contains only FISON-enders.
>
> If you claim that,

I claim that.

If k ∈ ℕ then k is a FISON-ender.
If k is a FISON-ender, then k ∈ ℕ.

> then you get the potentially infinite Peano set.

We get ℕ, for which
if k ∈ ℕ then k is a FISON-ender, and
if k is a FISON-ender, then k ∈ ℕ.

One consequence of this definition is
| ∀B ⊆ ℕ,
| 0 < |B| <->
| ∃i ∈ B, ∀j ∈ B (i =< j)

Another consequence is
| ∀B ⊆ ℕ,
| 0 < |B| < ℵo <->
| ∃k ∈ B, ∀j ∈ B (j =< k)

( I have change the _name_ of the subset of ℕ.
( Using 'M' made it difficult for you to
( understand me, for some unfathomable reason.
( You're welcome.

Define
0 < |B| < ℵo <->
a total order of B exists such that,
for each split of B, there is a crossing-pair,
and there is an upper end and a lower end in B.

(No set with upper or lower end is empty.)

Define
0 < |B| <->
something exists in B.

>> No minimum for their intersection: empty.
>
> No minimum for the endsegments: empty.

An end segment has a minimum.

By definition, an end segment is not empty.
Some k is in the end segment.
That k is in ℕ.
{1,...,k} exists.
The split {not-in end segment},{in end-segment}
of {1,...k} has a crossing-pair i,j.
j is the minimum of the end segment.

[WM]

Jim Burns schrieb am Montag, 15. November 2021 um 06:07:34 UTC+1:
> On 11/14/2021 5:02 PM, WM wrote:
> > Jim Burns schrieb

Spare your attempts! If you really claim that

~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo

is not implying
~(∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo)
then we have no common logical basis.

> > No minimum for the endsegments: empty.
> An end segment has a minimum.

But the set of all infinite endsegments has no minimum. We only know that no finite endsegment belongs to it. It is same with all unit fractions. Whatever 1/n you choose, the interval (0, 1/n] contains ℵo unit fractions. There is no minimum. But if you consider the set of all unit fractions, then none remains outside. That is the difference between dark and visible.

Regards, WM

[Jim Burns]

On 11/15/2021 9:08 AM, WM wrote:
> Jim Burns schrieb
> am Montag, 15. November 2021 um 06:07:34 UTC+1:
>> On 11/14/2021 5:02 PM, WM wrote:

> Spare your attempts! If you really claim that
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> is not implying
> ~(∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo)
> then we have no common logical basis.

[This is where your argument should go.]

| ~∃ M⊂ℕ ∀ n∈ℕ : Claim(n,M)
does not imply
| ~( ∀ n∈ℕ ∃ M⊂ℕ: Claim(n,M) )

That does not mean that there is
no Claim(n,M) for which both statements are true.

Rather, that means that, for some Claim(n,M),
the first is true and the second is false.

That means that going from the first to the second
_can't be used in an argument_
Even if we already know the first, we don't know
the second. Knowing the second would be the point of
including that step, first to second.

It's certainly possible for us to know both the
first and the second in some instances, but, for
us to know if this is one such instance, we would
need to know what Claim(n,M) represents.

Suppose, in one instance, Claim(n,M) stood for
goodClaim(n,M) and we know by other means that
| ~∃ M⊂ℕ ∀ n∈ℕ : goodClaim(n,M)
and
| ~( ∀ n∈ℕ ∃ M⊂ℕ: goodClaim(n,M) )

Then we would know both the first and the second
statements _but by those other means_ NOT by that
quantifier exchange.

Anything we would know after using it, we would
have already known before we used it.

----
This is why we disallow that exchange in an argument,
as long as there are any conceivable claims giving
the wrong answer, even if the wrongly-answered
claim seems thoroughly irrelevant to our current
circumstances. We know we can't rely on it.

----
| ~∃ M⊂ℕ ∀ n∈ℕ : Claim(n,M)
does not imply
| ~( ∀ n∈ℕ ∃ M⊂ℕ: Claim(n,M) )

As I explained, one counter-example is enough here,
even one that seems irrelevant.

Let n refer to one of the members of a theater audience.
Let M refer to a number.
Let Claim(n,M) == "member n is thinking of M"
I'll write ThinkingOf(n,M)

Suppose a professional psychic/magician asks each
member to think of a number.
After a moment, we have situation in which
it is _false_ that there is some number M which
each audience member n is thinking of,
| ~∃ M⊂ℕ ∀ n∈ℕ : ThinkingOf(n,M)
but it is _true_ that each audience member
is thinking of some number
| ∀ n∈ℕ ∃ M⊂ℕ: ThinkingOf(n,M)

This is enough reason to disallow an inference from
| ~∃ M⊂ℕ ∀ n∈ℕ : Claim(n,M)
to
| ~( ∀ n∈ℕ ∃ M⊂ℕ: Claim(n,M) )

because Claim(n,M) might be ThinkingOf(n,M)

---
On the other hand, we CAN make the inference
in the other direction,from
| ~( ∀ n∈ℕ ∃ M⊂ℕ: ThinkingOf(n,M) )
to
| ~∃ M⊂ℕ ∀ n∈ℕ : ThinkingOf(n,M)

Returning to our example,
suppose not all the audience members obeyed.
Maybe someone was thinking about the Lovely Assistant.
| ~( ∀ n∈ℕ ∃ M⊂ℕ: ThinkingOf(n,M) )

Not all are thinking of a number.

If we know that, somehow, then we know
no number exists which all are thinking of.
| ~∃ M⊂ℕ ∀ n∈ℕ : ThinkingOf(n,M)

| ~( ∀ n∈ℕ ∃ M⊂ℕ: ThinkingOf(n,M) )
implies
| ~∃ M⊂ℕ ∀ n∈ℕ : ThinkingOf(n,M)

----
The example with the Lovely Assistant is not a _proof_
that this way always gives the right answer.
But there _is_ a proof. For always.

So, even though

| ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo

does not imply

| ~( ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo )

| ~( ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo )

does imply

| ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo

This seems to be saying roughly
if not all end segments are infinite,
the intersection of all end segments is not infinite.

This is not as valuable as it might look at first glance.
We must avoid the fallacy of affirming the consequent,
which keeps us away from some really amazing conclusions.

https://en.wikipedia.org/wiki/Affirming_the_consequent

[Jim Burns]

On 11/15/2021 9:08 AM, WM wrote:

> Spare your attempts! If you really claim that
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> is not implying
> ~(∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo)
> then we have no common logical basis.

https://en.wikipedia.org/wiki/Quantifier_shift
Quantifier shift
|
| A quantifier shift is a logical fallacy in which the
| quantifiers of a statement are erroneously transposed.
| The change in the logical nature of the statement
| may not be obvious when it is stated in a natural
| language like English.
|
| Definition
|
| The fallacious deduction is that:
| For every A, there is a B, such that C.
| Therefore, there is a B, such that for every A, C.
| ∀x ∃y Rxy ⊢ ∃y ∀x Rxy
|
| However, an inverse switching:
| ∃y ∀x Rxy ⊢ ∀x ∃y Rxy
|
| is logically valid.

[WM]

Jim Burns schrieb am Montag, 15. November 2021 um 17:32:59 UTC+1:
> On 11/15/2021 9:08 AM, WM wrote:

> | ~∃ M⊂ℕ ∀ n∈ℕ : Claim(n,M)
> does not imply
> | ~( ∀ n∈ℕ ∃ M⊂ℕ: Claim(n,M) )

Don't try to confuse the clear argument! If it is true that there is no nonempty set M between all elements of ℕ and ω
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0
then it is also true that there is no infinite set M between all elements and ω
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
Then it is wrong that alle elements of ℕ have infinitely many elements between themselves and ω
~(∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo).

You seem to follow D.K. Davis advice: "If Cantor's work is invalid, modern mathematics goes up in smoke. The investment is too great – if something's wrong we'll just change logic." I assure you that my stidents will not follow your tricks, and also other students will refuse it, once they have learned what the are expected and commited to defend in matheology.

> That does not mean that there is
> no Claim(n,M) for which both statements are true.

Therefore stop your diversions and adhere to the above claim. Other claims are off topic.

>
> Rather, that means that, for some Claim(n,M),
> the first is true and the second is false.
>
> That means that going from the first to the second
> _can't be used in an argument_

It can be used in that case where I use it.

> Even if we already know the first, we don't know
> the second. Knowing the second would be the point of
> including that step, first to second.
>
> It's certainly possible for us to know both the
> first and the second in some instances, but, for
> us to know if this is one such instance, we would
> need to know what Claim(n,M) represents.

We know it. I have explained it.
>
> Suppose,

M. Dejess : "The pseudo idea of infinity has been replaced by smart 'mathematricksters' into the opposite". Obviously any frauds of your sort are around up to mischief.

> This is why we disallow that exchange in an argument,

There is no exchange but a simple logical step from
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
to
~(∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo).

The first is true in ZF. Even
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0 .
And
∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
is also true in ZF.

> Let n refer to one of the members of a theater audience.

Whatever you are trying to prove, it is of no interest. If there is no infiite distance between the set ℕ and ω, then there are members of this set which prevent this result.

Regards, WM

[Jim Burns]

On 11/15/2021 4:42 PM, WM wrote:
> Jim Burns schrieb
> am Montag, 15. November 2021 um 17:32:59 UTC+1:

>> That does not mean that there is
>> no Claim(n,M) for which both statements are true.
>
> Therefore stop your diversions and adhere to the above
> claim. Other claims are off topic.

A counter-example to an inference you use is on-topic.

>> Rather, that means that, for some Claim(n,M),
>> the first is true and the second is false.
>>
>> That means that going from the first to the second
>> _can't be used in an argument_
>
> It can be used in that case where I use it.

It is a faulty inference. That's what the counter-example
shows.

"It can be used" in the same sense that you can claim
that you are Oberon, king of the fairies. Yes, you have
the physical ability to claim that. But no one has
any reason to believe you. Your insistence on making
that claim is not such a reason.

>> Even if we already know the first, we don't know
>> the second. Knowing the second would be the point of
>> including that step, first to second.
>>
>> It's certainly possible for us to know both the
>> first and the second in some instances, but, for
>> us to know if this is one such instance, we would
>> need to know what Claim(n,M) represents.
>
> We know it. I have explained it.

Have you? Then you should be able to explain it again
without mentioning the faulty WM-exchange this time.
Go ahead.

>> This is why we disallow that exchange in an argument,
>
> There is no exchange but a simple logical step from
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
> to
> ~(∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo).

"No exchange."
https://en.wikipedia.org/wiki/Gaslighting

Maybe this will help you.
https://en.wikipedia.org/wiki/Quantifier_shift
Quantifier shift

[WM]

Jim Burns schrieb am Montag, 15. November 2021 um 23:49:05 UTC+1:
> On 11/15/2021 4:42 PM, WM wrote:

> > Therefore stop your diversions and adhere to the above
> > claim. Other claims are off topic.
> A counter-example to an inference you use is on-topic.

The inference is divided into two cases, correct and incorrect results. The correctness depends on special circumstances. For instance your magician or simply the well known
∀ n∈ℕ ∃ m∈ℕ: m > n does not imply ∃ m∈ℕ ∀ n∈ℕ : m > n .

But if I was the magician I would aks the audience to think of endsegments, and I would be able to conclude that the endsegments of all imagined numbers have a common set M. Splendid magic, isn't it? Siegfried and Roy appear like amateurs compared to that achievement.

So we can be sure: If there are infinite endsegments, then they have infinitely many numbers in common.

> > It can be used in that case where I use it.
> It is a faulty inference. That's what the counter-example
> shows.

Your claim that all inferences of this kind are unreliable is wrong. By the way, that's why I have been talking about FISONs and endsegments for many years now.

> "It can be used" in the same sense that you can claim
> that you are Oberon, king of the fairies. Yes, you have
> the physical ability to claim that. But no one has
> any reason to believe you. Your insistence on making
> that claim is not such a reason.

But this is the reason: According to ZFC this statement is true
~∃ M⊂ℕ ∀ n∈ℕ : M ⊂ E(n) ∧ |M| > 0 .
It is also true by common sense because there is nothing beyond ℕ and therefore between ω and all elements of ℕ.
This implies by mathematics

~∃ M⊂ℕ ∀ n∈ℕ : M ⊂ E(n) ∧ |M| = ℵo .

There is no infinite distance between ℕ, that is all elements n of ℕ, and ω.

> >> Even if we already know the first, we don't know
> >> the second. Knowing the second would be the point of
> >> including that step, first to second.

This statement is true according to ZFC (M may differ but is infinite in any case)
∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
It has the negation
~(∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo)
which is the same as
∃ n∈ℕ ~∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo.
There is no infinite distance between all elements n of ℕ and ω.

> > We know it. I have explained it.
> Have you? Then you should be able to explain it again
> without mentioning the faulty WM-exchange this time.
> Go ahead.

See above. For ℕ_def = ℕ there is a contradiction.

Regards, WM

[Mostowski Collapse]

Your track record concerning ZFC is very weak:
- Only a few days ago you proved for the first time ALL(x):~(x e x).
- Then you cited a shitty wikipedia comment claiming axiom of regularity unnecessary.
- But you never verified all Peano axioms for ZFC's 0={} and succ(x)=x u {x}.

Now you even dont know whether the axiom of regularity is
really unnecessary or not for Peano inside ZFC.

LMAO!

[Dan Christensen]

On Tuesday, November 16, 2021 at 4:31:38 AM UTC-5, Mostowski Collapse wrote:

> Dan Christensen schrieb am Sonntag, 14. November 2021 um 20:35:26 UTC+1:
> > On Sunday, November 14, 2021 at 1:54:40 PM UTC-5, Mostowski Collapse wrote:
> >
> > > Omega is just an ordinal. You need first a notion of ordinal.
> > > Dan Christensen schrieb am Sonntag, 14. November 2021 um 17:19:19 UTC+1:
> > > > On Sunday, November 14, 2021 at 5:16:04 AM UTC-5, Mostowski Collapse wrote:
> > > > > Thats almost as hilarious as Dan Christensen thinking ZFC is
> > > > > an alternative to Peano.
> > > > Where did you get that idea, Jan Burse?
> >
> > > By your idiotic comments:
> >
> > > > Better that you should start with the basics. Again, you might start by constructing
> > > > (i.e. proving the existence of) the set of natural numbers N.
> > >
> > > Thats not the basics of ZFC. Check out the table of contents here:
> > >
> > > Table of Contents - Basic Set Theory - Levy
> > > https://books.google.ch/books?id=hwrmDAAAQBAJ&pg=PR11&hl=de&source=gbs_selected_pages&cad=3#v=onepage&q&f=false
> > >
> > Maybe proving the existence of N in ZFC is not as trivial as you claimed. It is almost trivial in DC Proof to construct N given an arbitrary Dedekind infinite set (supposedly the equivalent of AoI in ZFC). See: http://www.dcproof.com/ProofByInduction.html (only 247 lines).

.

> Your track record concerning ZFC is very weak:
> - Only a few days ago you proved for the first time ALL(x):~(x e x).
> - Then you cited a shitty wikipedia comment claiming axiom of regularity unnecessary.

.
Not ENTIRELY unnecessary. That quote again:
.
"Virtually all results in the branches of mathematics based on set theory hold even in the absence of regularity."
https://en.wikipedia.org/wiki/Axiom_of_regularity

.
> - But you never verified all Peano axioms for ZFC's 0={} and succ(x)=x u {x}.
>

I left that as a exercise for you, Jan Burse. I have very little interest in ZFC as it just doesn't come up much in most math textbooks, e.g. in algebra or analysis.

.
> Now you even dont know whether the axiom of regularity is
> really unnecessary or not for Peano inside ZFC.

.
Again, I left any further proofs in ZFC as an exercise for you. I just demonstrated for you how it is possible to do ZFC proofs in DC Proof (e.g. proving ALL(a):~a in a). You just specify the ZFC axioms at the beginning of your proof. They are not hard-coded in DC Proof. That would only confuse the intended users.
.
Dan
.

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

.

[Jim Burns]

On 11/16/2021 4:27 AM, WM wrote:
> Jim Burns schrieb
> am Montag, 15. November 2021 um 23:49:05 UTC+1:
>> On 11/15/2021 4:42 PM, WM wrote:

>>> Therefore stop your diversions and adhere to the above
>>> claim. Other claims are off topic.
>>
>> A counter-example to an inference you use is on-topic.
>
> The inference

... the WM-exchange, quantifier shift, one of
[?] ∀x ∃y Rxy ⊢ ∃y ∀x Rxy
or
[?] ~∃y ∀x Rxy ⊢ ~∀x ∃y Rxy

> The inference is divided into two cases,
> correct and incorrect results.
> The correctness depends on special circumstances.

>>> It can be used in that case where I use it.
>>
>> It is a faulty inference. That's what the counter-example
>> shows.
>
> Your claim that all inferences of this kind are unreliable
> is wrong.

What I mean by "unreliable" is that
the inference is divided into two cases,
correct and incorrect results.

What I mean by "unreliable" is that
the correctness depends on special circumstances.

[?] ∀x ∃y Rxy ⊢ ∃y ∀x Rxy
and
[?] ~∃y ∀x Rxy ⊢ ~∀x ∃y Rxy

are unreliable.

----
Compare that to a _reliable_ inference.
| {P, P->Q} ⊢ Q

Under whatever circumstances P is true and P->Q is true,
Q is true. Reliably.

We don't need to know more about P and P->Q than
that they are true, and we will know that Q is true.
There are no special circumstances.

There is no division into two cases, {P, P->Q, Q)
and {P, P->Q, ~Q}. The second class, which would
make {P, P->Q} ⊢ Q unreliable, is empty.

I'm not calling Q reliable.
I'm calling _taking that step_ reliable, the '⊢'
in {P, P->Q} ⊢ Q.

A proof is a finite sequence of _reliable_ steps
from assumptions to a conclusion. We know we can
rely on the conclusion as much as we can rely on
the assumptions, because of that sequence of reliable
-- _perfectly_ reliable -- steps.

That inference, the quantifier shift, is NOT
perfectly reliable. Sometimes { ∀x ∃y Rxy, ∃y ∀x Rxy }
but sometimes { ∀x ∃y Rxy, ~∃y ∀x Rxy }
This is why we disallow the quantifier shift as
one of the steps in a proof.

That doesn't mean we can't prove ∃y ∀x Rxy.
It only means we can prove ∃y ∀x Rxy _that way_
by quantifier shift.

If you find some different route from ∀x ∃y Rxy
to ∃y ∀x Rxy that takes only perfectly reliable
steps, then I have no objection.

( Compare to:
( If you instruct me to cross a river by driving over
( an unreliable bridge, I'll reject your instructions.
( But that's not an objection to crossing the river.
( If you direct me to a different bridge, one that is
( reliable, I'll accept your instructions.

[Jim Burns]

On 11/16/2021 1:36 PM, Jim Burns wrote:

> That doesn't mean we can't prove  ∃y ∀x Rxy.

> It only means we [can't] prove  ∃y ∀x Rxy  _that way_

> by quantifier shift.
>
> If you find some different route from  ∀x ∃y Rxy
> to  ∃y ∀x Rxy that takes only perfectly reliable
> steps, then I have no objection.

I am reminded of the proof of Fermat's Last Theorem
by Andrew Wiles. When he discovered an error in his
proof, a less-than-perfectly-reliable step, he
looked for and found a different route to the desired
conclusion, without less-than-perfectly-reliable steps.

Morals of the story.
(i)
For _everyone_ no matter how well-regarded, a proof
is not a proof until every step is perfectly reliable.
(ii)
Even if a less-than-perfectly-reliable step is
discovered, all is not _necessarily_ lost. A different
route to the same destination might be found.

Note: "found" is not "conjectured to exist".

[Python]

Excellent point.
Andrew Wiles deserve to be called a mathematician because
he cares about checking and fixing mistakes, and also he
has 1) intellectual integrity 2) rational thinking.

Crank Wolfgang Mückenheim, from Hochschule Augsburg has
NOTHING of, and NEVER had, these qualities.

[Jim Burns]

Another example, possibly even better, is when Edward
Nelson very publicly _withdrew_ his claim to have proven
that arithmetic is inconsistent.

It was in on-line discussion with Terence Tao. Tao
made a subtle, perhaps-counter-intuitive point that
a _sub-theory_ can have a greater complexity than the
theory it's contained in.

Good for Terence Tao for understanding and explaining
that point. But it seems to me that the actual hero here
is Edward Nelson. It had to have stung him to withdraw
his claim. But that willingness to accept correction is
the bones within mathematics that makes it different from
literary criticism or whatever. Without _heroes_
(typically unsung) like Edward Nelson, there would be
no mathematics.

http://m-phi.blogspot.com/2011/10/nelson-withdraws-his-claim.html

[Python]

Thanks for the reference, words below should be engraved on
crank Wolfgang Mückenheim's skin while he's still alived:

“This seems to be a good example of what J. Azzouni has described as
the 'uniqueness' of mathematics as a social practice: in just a few
days, a consensus has emerged as to what was wrong with Nelson's
purported proof, including Nelson himself. I cannot think of any
other field of inquiry where consensus on a substantive, serious
issue/challenge would emerge with the same swiftness. There really i
is something special about mathematics...“

This is exactly what impressed me when I was studying math at univ.
Even complete asshole mathematicians have this behaviour.

Maybe, then, he would understand why he is in no way a mathematician
(not event a scientist btw), has never been, and never will... While,
moreover, being still an asshole.

[WM]

Jim Burns schrieb am Dienstag, 16. November 2021 um 19:36:19 UTC+1:
> On 11/16/2021 4:27 AM, WM wrote:

> > The inference is divided into two cases,
> > correct and incorrect results.

> > Your claim that all inferences of this kind are unreliable
> > is wrong.
> What I mean by "unreliable" is that
> the inference is divided into two cases,
> correct and incorrect results.

Therefore we have to investigate the particular example, not the general case.

>
> What I mean by "unreliable" is that
> the correctness depends on special circumstances.

Therefore we have to investigate the particular example, not the general case.

Diversion concerning general case of implication deleted.

>
> If you find some different route from ∀x ∃y Rxy
> to ∃y ∀x Rxy that takes only perfectly reliable
> steps, then I have no objection.

I have shown you the different route. ZFC has the theorem
∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo . (*)
Its negation is

~(∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo)
which is the same as

∃ n∈ℕ ~∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo. (**)
This is not a quantifier shift but simple logic. If not all, then there is at least one exception:
~∀ n∈ℕ: P(n) ==> ∃ n∈ℕ: ~P(n).
It exists an n with not infinite subset M of its endsegment.

ZFC has the theorem

~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0 .

There exists no nonempty subset M of all endsegments, let alone an infinite M:

~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo.

And this is the bridge you can safely cross: If not all endsegments have infinite subsets M, then at least one endsegment has only smaller subsets:

∃ n∈ℕ ~∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

same as (**), the negation of (*).

Regards, WM

[WM]

Jim Burns schrieb am Dienstag, 16. November 2021 um 22:37:42 UTC+1:
> But that willingness to accept correction is
> the bones within mathematics that makes it different from
> literary criticism or whatever. Without _heroes_
> (typically unsung) like Edward Nelson, there would be
> no mathematics.

Then try to become a hero. If every defined n has an infinite distance from ω but no distance larger than zero can be found between ω and all natnumbers, then the definable natnumbers are not all natnumbers.

But I am afraid you will join D.K. Davis: If Cantor's work is invalid, modern mathematics goes up in smoke. The investment is too great – if something's wrong we'll just change logic. [D.K. Davis in "Cantor's transfinite numbers", sci.math (31 Oct 1996)]

In fact there are comparable attempts introducing non-measurable sets where it may happen that almost all objects are red and almost all objects are small, but no object is small and red.

I cannot forbid this way of "arguing", I can only try to drag this nonsense into the public eye.

Regards, WM

[Jim Burns]

On 11/17/2021 5:34 AM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 16. November 2021 um 19:36:19 UTC+1:
>> On 11/16/2021 4:27 AM, WM wrote:

>>> The inference is divided into two cases,
>>> correct and incorrect results.
>>> Your claim that all inferences of this kind
>>> are unreliable is wrong.
>>
>> What I mean by "unreliable" is that
>> the inference is divided into two cases,
>> correct and incorrect results.
>
> Therefore we have to investigate the particular
> example, not the general case.

And we have to use only reliable steps.
Truth-preserving steps.
More commonly, valid inferences.

>> If you find some different route from ∀x ∃y Rxy
>> to ∃y ∀x Rxy that takes only perfectly reliable
>> steps, then I have no objection.
>
> I have shown you the different route.
> ZFC has the theorem
> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo . (*)

That's fine. {*} is true.

> Its negation is
> ~(∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo)
> which is the same as
> ∃ n∈ℕ ~∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo. (**)

Negation is not a reliable step. (**) is false.

Actually, we could call it an anti-reliable step,
if we were doing this often enough to need a name.
Negation never gives a true second statement from
a true first statement.

But maybe that's okay, given where you're going with this.
I can think of proofs which use the _falseness_
of a certain statement to establish the truth of another.

> This is not a quantifier shift but simple logic.
> If not all, then there is at least one exception:
> ~∀ n∈ℕ: P(n) ==> ∃ n∈ℕ: ~P(n).
> It exists an n with not infinite subset M of
> its endsegment.

And that is the negation of a true statement. False.

> ZFC has the theorem
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0 .
> There exists no nonempty subset M of all endsegments,
> let alone an infinite M:
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo.

That may have been mentioned once or twice in these
threads. True.

> And this is the bridge you can safely cross:
> If not all endsegments have infinite subsets M,
> then at least one endsegment has only smaller subsets:
> ∃ n∈ℕ ~∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> same as (**), the negation of (*).

I agree that this is a reliable step.
But a reliable step from where?

"not all endsegments have infinite subsets M"

was not arrived at by _a sequence of reliable steps_
but by negation.

You still need a safe bridge _to get to_
this last safe bridge and then cross it..

[WM]

Jim Burns schrieb am Mittwoch, 17. November 2021 um 12:49:09 UTC+1:
> On 11/17/2021 5:34 AM, WM wrote:

> > Therefore we have to investigate the particular
> > example, not the general case.
> And we have to use only reliable steps.
> Truth-preserving steps.
> More commonly, valid inferences.

It is valid to say : If not all endsegments have infinite subsets M, then at least one endsegment has only smaller subsets:

∃ n∈ℕ ~∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

> > ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo . (*)
> That's fine. {*} is true.
> > Its negation is
> > ~(∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo)
> > which is the same as
> > ∃ n∈ℕ ~∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo. (**)
> Negation is not a reliable step. (**) is false.

Of course.

>
> Actually, we could call it an anti-reliable step,
> if we were doing this often enough to need a name.
> Negation never gives a true second statement from
> a true first statement.

No, I have only writen it in the same form as below, where it is a reliable truth. We'll obtain a contradiction you know.

>
> And that is the negation of a true statement. False.
> > ZFC has the theorem
> > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0 .
> > There exists no nonempty subset M of all endsegments,
> > let alone an infinite M:
> > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo.
> That may have been mentioned once or twice in these
> threads. True.
> > And this is the bridge you can safely cross:
> > If not all endsegments have infinite subsets M,
> > then at least one endsegment has only smaller subsets:
> > ∃ n∈ℕ ~∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> > same as (**), the negation of (*).
> I agree that this is a reliable step.
> But a reliable step from where?

From ZFC! See above:

~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0 .
There exists no nonempty subset M of all endsegments, let alone an infinite M:
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo.

> "not all endsegments have infinite subsets M"
> was not arrived at by _a sequence of reliable steps_
> but by negation.

No, not by the negation. The last derivation was from ZFC: There exists no nonempty subset M of all endsegments, let alone an infinite M.

Regards, WM

[Jim Burns]

On 11/17/2021 1:14 PM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 17. November 2021 um 12:49:09 UTC+1:
>> On 11/17/2021 5:34 AM, WM wrote:

>>> And this is the bridge you can safely cross:
>>> If not all endsegments have infinite subsets M,

If ~∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

>>> then at least one endsegment has only smaller subsets:
>>> ∃ n∈ℕ ~∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
>>> same as (**), the negation of (*).
>>
>> I agree that this is a reliable step.
>> But a reliable step from where?

I'm asking, how do we get reliably to
| ~∀ n∈ℕ ∃ M⊂ℕ : M⊂E(n) ∧ |M| = ℵo

in order to cross the bridge you are pointing to?

> From ZFC! See above:
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0 .
> There exists no nonempty subset M of all endsegments,
> let alone an infinite M:

| ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo

is not
| ~∀ n∈ℕ ∃ M⊂ℕ : M⊂E(n) ∧ |M| = ℵo

That step is a quantifier shift.

_All_ the steps need to be reliable.

How do we get reliably to
| ~∀ n∈ℕ ∃ M⊂ℕ : M⊂E(n) ∧ |M| = ℵo

[Serg io]

On 11/17/2021 4:45 AM, WM wrote:
> Jim Burns schrieb am Dienstag, 16. November 2021 um 22:37:42 UTC+1:
>> But that willingness to accept correction is
>> the bones within mathematics that makes it different from
>> literary criticism or whatever. Without _heroes_
>> (typically unsung) like Edward Nelson, there would be
>> no mathematics.
>
> Then try to become a hero. If every defined n has an infinite distance from ω but no distance larger than zero can be found between ω and all natnumbers, then the definable natnumbers are not all natnumbers.

"If every defined n"

Wrong, your "defined" daffynition is a joke, it shows you as a fraud.

> has an infinite distance from ω

OK

> but no distance larger than zero can be found between ω and all natnumbers,

wrong, in all cases.

>then the definable natnumbers


wrong, "definalbe" is poop

> are not all natnumbers.

Wrong all natnumbers are natnumbers

> In fact there are comparable attempts introducing non-measurable sets where it may happen that almost all objects are red and almost all objects are small, but no object is small and red.

that is out of left field, provide a full example and why it relats to this claim

>
> I cannot forbid this way of "arguing", I can only try to drag this nonsense into the public eye.

please drag your nonsense over into alt.fake.math


>
> Regards, WM
>

[WM]

Jim Burns schrieb am Mittwoch, 17. November 2021 um 20:02:27 UTC+1:

> How do we get reliably to
> | ~∀ n∈ℕ ∃ M⊂ℕ : M⊂E(n) ∧ |M| = ℵo

Start with ZFC:

~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0

and ask yourself: Why does not even a set of card 1 exist for all n∈ℕ?
Any idea for a correct answer?

Regards, WM

[Fritz Feldhase]

On Thursday, November 18, 2021 at 10:37:43 AM UTC+1, WM wrote:

> ~∃M⊂ℕ ∀n∈ℕ : M⊂E(n) ∧ |M| > 0
>

> Why does not even a set [M⊂ℕ] of card 1 exist [such that] for all n ∈ ℕ [M⊂E(n)]?

> Any idea for a correct answer?

Hint: |M| > 0 mans that M is not empty. Now any nonempty set of natural numbers (condition M⊂ℕ) has a least element. Assume that n_0 is this least element of M. But since n_0 !e E(n_0+1), M isn't a subset of E(n_0+1). Hence M isn't a subset of ALL endsegments (i. e. ∀n∈ℕ : M⊂E(n) doesn't hold).

[Jim Burns]

<JB<WM>>
>
> Define all natural numers individually
> such that none remains.
> Fail.

Say something known about a natural number k
even when it's not known which natural number
is referred to.

{0,...,k} exists.
Succeed.

</JB<WM>>

Subject: Re: Synopsis
Date: Mon, 15 Nov 2021 07:23:51 -0500


Returning to the "bridge" analogy,
we have a successful method for crossing rivers
in the Land of Infinitely-Many.
The method is to reason truth-preserving-ly,
_reliably_ from something we already know about one of
the infinitely-many. Something like "{0,...,k} exists"

Your argument for the unreliability of our bridge is
to build a bridge yourself, one that fails,
one that is intended to fail, I think.
Since yours fails, how could ours succeed?
That seems to be the question you implicitly ask.

I can't think of any other circumstance in which
that would be an effective argument.

----

> Why does not even a set of card 1 exist for all n∈ℕ?

For each n ∈ ℕ, FISON {0,...,n} exists.

∀j,k ∈ ℕ ( j ∈ {0,...,k} ∨ k ∈ {0,...,j} )

∀j,k ∈ ℕ ( k ∈ E(j) <-> j ∈ {0,...,k} )

∃k ∈ ℕ, ∀j ∈ ℕ ( k ∈ E(j) ) <->
∃k ∈ ℕ, ∀j ∈ ℕ ( j ∈ {0,...,k} )

~∃k ∈ ℕ ( k+1 ∈ {0,...,k} )

~∃k ∈ ℕ, ∀j ∈ ℕ ( j ∈ {0,...,k} )

~∃k ∈ ℕ, ∀j ∈ ℕ ( k ∈ E(j) )

[WM]

Fritz Feldhase schrieb am Donnerstag, 18. November 2021 um 18:21:18 UTC+1:
> On Thursday, November 18, 2021 at 10:37:43 AM UTC+1, WM wrote:
>
> > ~∃M⊂ℕ ∀n∈ℕ : M⊂E(n) ∧ |M| > 0
> >
> > Why does not even a set [M⊂ℕ] of card 1 exist [such that] for all n ∈ ℕ [M⊂E(n)]?
> > Any idea for a correct answer?
> Hint: |M| > 0 mans that M is not empty. Now any nonempty set of natural numbers (condition M⊂ℕ) has a least element.

No. Every natnumber has infinitely many successors. There is no least element of this imfinite set that never will vanish. Nevertheless it is existing because all endsegments are infinite, according to ZF.

Fact is: All *definable* natnumbers disappear in the course of "n is lost in E(n+1)". But all endsegments are infinite. There is a contents which does never get lost. Do you recognize a difference?

> Assume that n_0 is this least element of M.

If n_0 is definable, then it has ℵo successors. I am talking about an M contained within the infinite set of successors.

> Hence M isn't a subset of ALL endsegments (i. e. ∀n∈ℕ : M⊂E(n) doesn't hold).

Not for any definable number n_0. That is not new. The question remains however, what are the numbers in all infinite endsegments which never get lost because *all* endsegments are infinite? Any idea?

Please agree to these statements: The intersection of endsegments does not depend on their order. The intersection of endsegments does not depend on there being or not being a last endsegment. The intersection of endsegments depends only and exclusively on the size of the smallest endsegment. If all endsegments are infinite, then also the smallest endsegment is infinite. Such sets exist, for example the fractions in the intervl (0, 1+1/n).

Regards, WM

[WM]

Jim Burns schrieb am Donnerstag, 18. November 2021 um 20:04:26 UTC+1:
> On 11/18/2021 4:37 AM, WM wrote:

> Say something known about a natural number k
> even when it's not known which natural number
> is referred to.
>
> {0,...,k} exists.

Every such k will get lost in the course of "k is not in E(k+1)". But ℵo natnumbers will remain forever (for every definable endsegment) because every definable endsegment is infinite, and you believe that there are only definable endsegmets.

> > Why does not even a set of card 1 exist for all n∈ℕ?
> For each n ∈ ℕ, FISON {0,...,n} exists.

Not for the infinitely many natnumbers remaining after every definable n, i.e., in all endsegments. You say that all are infinite. You believe that they all contain infinitely many natnumbers which will never get lost:

∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

Should not at least 100 natnumbers remain withion these infinite sets?

> ~∃k ∈ ℕ, ∀j ∈ ℕ ( k ∈ E(j) )

No definable k is in all endsegments. But infinitely many natnumbers are in all endsegments. That proves the difference between definable and dark numbers:
∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo .
The set M is dark. The set ℕ in above statement is ℕ_def.

The intersection of endsegments does not depend on their order.

The intersection of endsegments does not depend on whether there is an infinite or finite set of endsegments but it depends only on the smallest endsegment (inlusion monotony).

If all endsegments are infinite, then also the smallest endsegment is infinite.

Such sets exist, for example the fractions in the interval (0, 1+1/n).

Regards, WM

[Fritz Feldhase]

On Friday, November 19, 2021 at 8:42:12 AM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 18. November 2021 um 18:21:18 UTC+1:
> > On Thursday, November 18, 2021 at 10:37:43 AM UTC+1, WM wrote:
> > >
> > > ~∃M⊂ℕ ∀n∈ℕ : M⊂E(n) ∧ |M| > 0
> > >
> > > Why does not even a set [M⊂ℕ] of card 1 exist [such that] for all n ∈ ℕ [M⊂E(n)]?
> > > Any idea for a correct answer?
> > >

> > Hint: |M| > 0 means that M is not empty. Now each and every nonempty set of natural numbers (condition M⊂ℕ) has a least [smallest] element.

> No.

Yes, Dummkopf!

See: https://en.wikipedia.org/wiki/Well-order

> > Assume that n_0 is this least [smallest] element of M.
> >
> n_0 [...] has ℵo successors.

Indeed, since each and every natural number has ℵo successors and n_0 _is_ a natural number (since it is an element in M⊂ℕ).

> I am talking <bla>

Yeah, whatever, Mückenheim.

> > Hence M isn't a subset of ALL endsegments (i. e. ∀n∈ℕ : M⊂E(n) doesn't hold).

qed

> That is not new.

So why did you ask, in the first place?

[Serg io]

On 11/19/2021 1:42 AM, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 18. November 2021 um 18:21:18 UTC+1:
>> On Thursday, November 18, 2021 at 10:37:43 AM UTC+1, WM wrote:
>>
>>> ~∃M⊂ℕ ∀n∈ℕ : M⊂E(n) ∧ |M| > 0
>>>
>>> Why does not even a set [M⊂ℕ] of card 1 exist [such that] for all n ∈ ℕ [M⊂E(n)]?
>>> Any idea for a correct answer?
>> Hint: |M| > 0 mans that M is not empty. Now any nonempty set of natural numbers (condition M⊂ℕ) has a least element.

<snip crap>

>
> Please agree to these statements: The intersection of endsegments does not depend on their order. The intersection of endsegments does not depend on there being or not being a last endsegment. The intersection of endsegments depends only and exclusively on the size of the smallest endsegment.

wrong, there is no smallest endsegment.

> If all endsegments are infinite, then also the smallest endsegment is infinite.

WRONG, there is no smallest endsegment.

> Such sets exist, for example the fractions in the intervl (0, 1+1/n).

Wrong, those are NOT endsegments.

>
> Regards, WM
>

[Serg io]

On 11/19/2021 1:53 AM, WM wrote:
> Jim Burns schrieb am Donnerstag, 18. November 2021 um 20:04:26 UTC+1:
>> On 11/18/2021 4:37 AM, WM wrote:
>
>> Say something known about a natural number k
>> even when it's not known which natural number
>> is referred to.
>>
>> {0,...,k} exists.
>
> Every such k will get lost in the course of "k is not in E(k+1)". But ℵo natnumbers will remain forever (for every definable endsegment) because every definable endsegment is infinite, and you believe that there are only definable endsegmets.
>
>>> Why does not even a set of card 1 exist for all n∈ℕ?
>> For each n ∈ ℕ, FISON {0,...,n} exists.
>
> Not for the infinitely many natnumbers remaining after every definable n, i.e., in all endsegments. You say that all are infinite. You believe that they all contain infinitely many natnumbers which will never get lost:
> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> Should not at least 100 natnumbers remain withion these infinite sets?
>
>> ~∃k ∈ ℕ, ∀j ∈ ℕ ( k ∈ E(j) )
>
> No definable k is in all endsegments. But infinitely many natnumbers are in all endsegments. That proves the difference between definable and dark numbers:

nope. you have no proof at all, spoof yes

> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo .
> The set M is dark. The set ℕ in above statement is ℕ_def.
>
> The intersection of endsegments does not depend on their order.
> The intersection of endsegments does not depend on whether there is an infinite or finite set of endsegments but it depends only on the smallest endsegment (inlusion monotony).

wrong, there is no smallest endsegment.

> If all endsegments are infinite, then also the smallest endsegment is infinite.

wrong, there is no smallest endsegment.

> Such sets exist, for example the fractions in the interval (0, 1+1/n).

wrong, that is not an endsegment.

>
> Regards, WM
>

[Jim Burns]

On 11/19/2021 2:53 AM, WM wrote:

> The intersection of endsegments does not depend on
> their order.

You've changed what "order" means.
Originally, the order of end segments matched their
subset-inclusion.

So, we'll cut out the word you're having trouble
keeping steady. "Order" no longer, it's "subsetting"

? The intersection of endsegments does not depend on
? their subsetting.

It's easy to see that that's false.
E(j) ∩ E(k) = E(j) <-> E(j) ⊆ E(k)

That wobbly word was causing premature contradiction.
Don't you feel better now?

----------------------------------------------------
| Note to myrmecologists
|
| wobble-worded ants having premature contradictions
----------------------------------------------------

From the definition of "intersection",
there are no dark end segments D, no sets at all exist
_between_ all the (infinite) end segments E(j) and
their Intersection IA(I)ES of All (Infinite) End Segments.

~∃D, ∀j ∈ ℕ, IA(I)ES ⊂ D ⊆ E(j)

∀j ∈ ℕ, IA(I)ES ⊆ E(j)

One way to describe IA(I)ES is
as the lower bound which has no _dark end segments_

IA(I)ES makes it difficult to _reasonably_ claim that
there is some sort of logical requirement for
_dark numbers_
IA(I)ES has the same _formal_ description with respect
to (infinite) end segments that ω has to natural numbers.

> The intersection of endsegments does not depend on
> their order.

s/order/subsetting

? The intersection of endsegments does not depend on
? their subsetting.

Wrong.
The intersection depends utterly on their subsetting.

----

>> Say something known about a natural number k
>> even when it's not known which natural number
>> is referred to.
>>
>> {0,...,k} exists.
>
> Every such k will get lost in the course of
> "k is not in E(k+1)".

There is no natural number k without {0,...,k}.
Every k will get lost.

> But ℵo natnumbers will remain forever

No k with {0,...,k} will remain forever.
Empty intersection.

No k with {0,...,k} is without k+1 with {0,...,k+1}
Infinite end segments ==
No end segments with two-ended steppable total orders.

>>> Why does not even a set of card 1 exist for all n∈ℕ?
>>
>> For each n ∈ ℕ, FISON {0,...,n} exists.
>
> Not for the infinitely many natnumbers remaining after
> every definable n, i.e., in all endsegments.

The ℕ for which

"not even a set of card 1 exist for all n∈ℕ"

is same as the ℕ for which,
"for each n ∈ ℕ, FISON {0,...,n} exists".

You asked a question. You got your answer.

Your (WM's) problem is your words have gone wobbly.
It's giving you premature contradiction.

[WM]

Fritz Feldhase schrieb am Freitag, 19. November 2021 um 12:57:09 UTC+1:
> On Friday, November 19, 2021 at 8:42:12 AM UTC+1, WM wrote:

> Indeed, since each and every natural number has ℵo successors

Then there are not ℵo natural numbers. ℵo natnumbers followed by ℵo successors would imply two actually infinite sets within the natural order of ℕ. That is impossible. If you disagree, find the first successor.

Each and every *definable* natnumber has ℵo successors. That proves that the definable numbers are not an actually infinite set.

Likewise each and every definable endsegment has ℵo elements. Hence there cannot be ℵo definable endsegments.

Regards, WM

[WM]

Jim Burns schrieb am Freitag, 19. November 2021 um 19:38:18 UTC+1:
> On 11/19/2021 2:53 AM, WM wrote:
>
> > The intersection of endsegments does not depend on
> > their order.
> You've changed what "order" means.
> Originally, the order of end segments matched their
> subset-inclusion.

So it is. But the intersection of this set does not depend on their order. The intersection depends on the least endsegment. If this is infinite, then the intersection is infinite.

We can sort out all infinite endsegments.
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
By the axiom of separation these endsegments can be collected in a set ℕ_def such that
|∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .

Endsegments which would reduce this intersection are not included. Hence the intersection of these endsegments is infinite.

> ? The intersection of endsegments does not depend on
> ? their subsetting.
>
> It's easy to see that that's false.
> E(j) ∩ E(k) = E(j) <-> E(j) ⊆ E(k)

It is easy to see that it is correct. The intersection of E(j) and E(k) and many more does not depend on the order defined by subsets but only on the sizes of the endsegments.

>
> That wobbly word was causing premature contradiction.
> Don't you feel better now?

My unfamiliar wording may excuse your not understanding.

>
> From the definition of "intersection",
> there are no dark end segments D, no sets at all exist
> _between_ all the (infinite) end segments E(j)

There are less than ℵ₀ infinite endsegments because these endsegments contain ℵ₀ natnumbers, but two sets of ℵ₀ elements are impossible in the natural oder of ℕ. If you can't understand try to find the remaining contents after the of set of ℵ₀ endsegments has been completed. If you can't find a contents, then there is none. Endsegments without contents are not infinite endsegments.

> The ℕ for which
> "not even a set of card 1 exist for all n∈ℕ"
> is same as the ℕ for which,
> "for each n ∈ ℕ, FISON {0,...,n} exists".

Why don't you use logic?
|ℕ \ {1, 2, 3, ..., n}| = ℵo
By the axiom of separation we can collect all n satisfying this into a set ℕ_def. Then by this condition
|ℕ \ ℕ_def| = ℵo.
Hence |ℕ_def| < ℵo.
Or do you refuse the axiom of separation?

Regards, WM

[An Own]

On Saturday, 20 November 2021 at 07:44:14 UTC-4, WM wrote:
[...]

> Why don't you use logic?
> |ℕ \ {1, 2, 3, ..., n}| = ℵo
> By the axiom of separation we can collect all n satisfying this into a set ℕ_def. Then by this condition

> |ℕ \ ℕ_def| = ℵo.

This does not follow. Seems that your "logic" is a bot wobbly this morning.

[Serg io]

On 11/20/2021 5:22 AM, WM wrote:
> Fritz Feldhase schrieb am Freitag, 19. November 2021 um 12:57:09 UTC+1:
>> On Friday, November 19, 2021 at 8:42:12 AM UTC+1, WM wrote:
>
>> Indeed, since each and every natural number has ℵo successors
>
> Then there are not ℵo natural numbers. ℵo natnumbers followed by ℵo successors would imply two actually infinite sets within the natural order of ℕ. That is impossible. If you disagree, find the first successor.

your cheese has slid off its cracker.

>
> Each and every *definable* natnumber has ℵo successors. That proves that the definable numbers are not an actually infinite set.

that is a Spoof, not a proof.

>
> Likewise each and every definable endsegment has ℵo elements. Hence there cannot be ℵo definable endsegments.

wrong. the # elements has no relationship with the # sets

You get an F, for Fail.

>
> Regards, WM
>

[Serg io]

full moon...

[Fritz Feldhase]

On Saturday, November 20, 2021 at 12:44:14 PM UTC+1, WM wrote:

> Why don't you use logic?

Well, actually set theory.

A predicate with free variable "n":

> |ℕ \ {1, 2, 3, ..., n}| = ℵo
>

> By the axiom of separation we can collect all n [in IN] satisfying this into a set ℕ_def.

Right. In other words, we can state the definition:

IN_def := {n e IN : |IN \ {1, 2, 3, ..., n}| = ℵo} .

> Then by this [definition]
>
> |ℕ \ ℕ_def| = ℵo.

Nope.

Hint: Since An e IN: |IN \ {1, 2, 3, ..., n}| = ℵo we get that {n e IN : |IN \ {1, 2, 3, ..., n}| = ℵo} = IN and hence IN = IN_def. Hence IN \ IN_def = {}, and hence |IN \ IN_def| = |{}| = 0. qed

> Hence |ℕ_def| < ℵo.

Nope.

Hint: Since IN = IN_def and |IN| = ℵo by definition, we get that |IN_def| = ℵo. qed

> Or do you refuse the axiom of separation?

Learn some logic, learn some math, Mückenheim!

[Jim Burns]

On 11/20/2021 6:44 AM, WM wrote:
> Jim Burns schrieb
> am Freitag, 19. November 2021 um 19:38:18 UTC+1:
>> On 11/19/2021 2:53 AM, WM wrote:

>>> The intersection of endsegments does not depend on
>>> their order.
>>
>> You've changed what "order" means.
>> Originally, the order of end segments matched their
>> subset-inclusion.
>
> So it is.
> But the intersection of this set does not depend on
> their order.

Your words have gone wobbly. What you means is

s/order/subsetting

| But the intersection of this set does not depend on

| their subsetting.

and that's wrong.

> The intersection depends on the least endsegment.

There are two distinct senses of "less" here.
"Set B is less than set C" can mean

(i)
B ⊆ C ∧ ~(C ⊆ B)
Each element of B is an element of C,
but not the other way.
Order by subsetting, inclusion.

(ii)
|B| =< |C| ∧ ~(|C| =< |B|)
Each element of B can be matched to
a unique element of C, but not the other way.
Order by injecting, cardinality.

> The intersection depends on the least endsegment.

(i)
By subsetting,
_no least end segment E' ⊆ ℕ exists_
~∃E', ∀E, E' ⊆ E ∧ ~(E ⊆ E')

There is no end segment subset to, E' ⊆ E,
all end segments.

(ii)
By injecting,
_no least end segment E' ⊆ ℕ exists_
~∃E', ∀E, |E'| =< |E| ∧ ~(|E| =< |E'|)

There is no end segment not injectable, ~(|E'| =< |E|),
into any other end segment.

Therefore, whichever you mean,
_no least end segment E' ⊆ ℕ exists_

> If this is infinite, then the intersection is infinite.

No least end segment exists. Ex falso quodlibet

What does "infinite" mean here? "Not finite"?
What does "finite" mean here?

----
B ⊆ ℕ is maximumless iff
B ⊆ ℕ is infinite or empty

B ⊆ ℕ is minimumless iff
B ⊆ ℕ is empty

Each end segment amd the intersection of all
is maximumless.
Each end segment amd the intersection of all
is infinite or empty.

Each end segment is not empty and infinite.

The intersection of all is minumumless and empty.

> We can sort out all infinite endsegments.

Each end segment ⊆ ℕ is not empty and closed upward,
not empty and maximumless, infinite.

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

What you should say here is

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

You're describing a property that each end segment
has xor does not have. A closed formula, with no free
variables, can't be that description.
The description is the Sorting Hat. It decides which
House a particular end segment goes to.
The Sorting Hat needs to sit onthe head of the
end segment. The Hat's free variable is where the
end segment goes.

You mention the axiom of separation.
It's the formula _without_ '∀k ∈ ℕ' which gets used.

> By the axiom of separation these endsegments can be
> collected in a set ℕ_def such that
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .

"these end segments" =? {E(k) : k ∈ ℕ_def}
better:

THESE =
{ E ∈ ENDS | ∩{E(1),...,E} = E ∧ |E| = ℵ₀ }

ENDS =
{ E ⊆ ℕ | E closed upwards, not empty }

{E(1),...,E} =
{ E' ∈ ENDS | E' ⊇ E }

----
Here's your problem:
| ∩THESE | ≠ ℵ₀

| ∩THESE | = 0

because THESE = ENDS

> Endsegments which would reduce this intersection
> are not included. Hence the intersection
> of these endsegments is infinite.

"Reduces" from what? The choices are |{}| and |ℕ|
I'll guess you mean |ℕ|

And I see three readings of "reduce".
None of the readings make what you claim true.

For each end segment E ⊆ ℕ,

E reduces[1], defined as
~( |ℕ| =< | ∩(ENDS\{E} | )
(
( ∩(ENDS\{E} = {}
( ℕ cannot be injected into {}

E reduces[2], defined as
~( |ℕ| =< | ∩(ENDS\{E(1),...,E} | )
(
( ∩(ENDS\{E(1),...,E} = {}
( ℕ cannot be injected into {}

E does not reduce[3], defined as
|ℕ| =< | ∩(ENDS\{E,...} |
(
( ∩(ENDS\{E,...} = E\{min(E)}
( ℕ _can_ be injected into E\{min(E)}

Define
"endsegments which would reduce this intersection"
variously as

REDUCES[1] =
{ E ∈ ENDS | E reduces[1] } = ENDS

REDUCES[2] =
{ E ∈ ENDS | E reduces[2] } = ENDS

REDUCES[3] =
{ E ∈ ENDS | E reduces[3] } = {}

REDUCES[1] = ENDS

> Endsegments which would reduce this intersection
> are not included. Hence the intersection

> of these endsegments is ...

∩(ENDS\REDUCES[1]) = ∩{} = {}

∩(ENDS\REDUCES[2]) = ∩{} = {}

∩(ENDS\REDUCES[3]) = ∩ENDS = {}

> ... infinite.

No, it's not infinite. No matter what you're saying.

>> ? The intersection of endsegments does not depend on
>> ? their subsetting.
>>
>> It's easy to see that that's false.
>> E(j) ∩ E(k) = E(j) <-> E(j) ⊆ E(k)
>
> It is easy to see that it is correct.
> The intersection of E(j) and E(k) and many more
> does not depend on the order defined by subsets
> but only on the sizes of the endsegments.

| Star light, star bright,
| First star I see tonight,
| I wish I may, I wish I might
| Have the wish I wish tonight
|
| I wish ...

Sorry, no. Your wish is not granted.
Your wish never works, not even for finite sets.

∩SETS is the intersection of SETS ≠ {} iff
~∃D, ∀E ∈ SETS ( ∩SETS ⊂ D ⊆ E ) ∧
∀E ∈ SETS ( ∩SETS ⊆ E )

>> That wobbly word was causing premature contradiction.
>> Don't you feel better now?
>
> My unfamiliar wording may excuse your not understanding.

I suggest that you stop wobbling your words all around.
It's unseemly.

[Fritz Feldhase]

On Saturday, November 20, 2021 at 9:16:31 PM UTC+1, Jim Burns wrote A LOT.

I'm quite sure that Mückenheim will profit quite a lot by your slightly excessive exposition.

[ Mückenheim: HEUREKA!!! ]

[WM]

An Own schrieb am Samstag, 20. November 2021 um 14:24:32 UTC+1:
> On Saturday, 20 November 2021 at 07:44:14 UTC-4, WM wrote:
> [...]
> > Why don't you use logic?
> > |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > By the axiom of separation we can collect all n satisfying this into a set ℕ_def. Then by this condition
>
> > |ℕ \ ℕ_def| = ℵo.
> This does not follow.

No? Why does the axiom of separation fail to separate all FISONs which fail to make their set ℕ_def actually infinite?

Regards, WM

[Jim Burns]

On 11/20/2021 3:34 PM, Fritz Feldhase wrote:
> On Saturday, November 20, 2021 at 9:16:31 PM UTC+1,
> Jim Burns wrote A LOT.

> I'm quite sure that Mückenheim will profit quite a lot
> by your slightly excessive exposition.

| I would have written a shorter letter,
| but I did not have the time.
|
-- Blaise Pascal
Provincial Letters: Letter XVI (4 December 1656)

> [ Mückenheim: HEUREKA!!! ]
>

[WM]

Fritz Feldhase schrieb am Samstag, 20. November 2021 um 20:11:13 UTC+1:
> On Saturday, November 20, 2021 at 12:44:14 PM UTC+1, WM wrote:
>
> > Why don't you use logic?
> Well, actually set theory.
>
> A predicate with free variable "n":
> > |ℕ \ {1, 2, 3, ..., n}| = ℵo

> > By the axiom of separation we can collect all n [in IN] satisfying this into a set ℕ_def.
>
> Right. In other words, we can state the definition:
>
> IN_def := {n e IN : |IN \ {1, 2, 3, ..., n}| = ℵo} .

No we define the set ℕ_def by using only such FISONs which satisfy

>
> > |ℕ \ ℕ_def| = ℵo.
>
> Nope.

Proof: If the set ℕ_def contains only the first three FISONs, then
|ℕ \ ℕ_def| = ℵo.

Therefore such a set exists. We collect only such FISONs in ℕ_def. ℕ_def is defined by
|ℕ \ ℕ_def| = ℵo.

>
> Hint: Since An e IN: |IN \ {1, 2, 3, ..., n}| = ℵo

We choose not all, if they make the difference ℕ \ ℕ_def empty.

> > Hence |ℕ_def| < ℵo.
>
> Nope.
>
> Hint: Since IN = IN_def and |IN| = ℵo by definition

You misunderstand. We choose only such elemenets for ℕ_def which let |ℕ_def| < ℵo.

Regards, WM

[WM]

Jim Burns schrieb am Samstag, 20. November 2021 um 21:16:31 UTC+1:
> On 11/20/2021 6:44 AM, WM wrote:

> > The intersection depends on the least endsegment.
> There are two distinct senses of "less" here.

No. The least endsegment has least elements. It has lost most natural numbers. That is a valid measure.

> Therefore, whichever you mean,
> _no least end segment E' ⊆ ℕ exists_

The least endsegment has lost most natnumbers.

> What does "finite" mean here?

The number of elements can be measured by a natural number.

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

> You mention the axiom of separation.
> It's the formula _without_ '∀k ∈ ℕ' which gets used.

The set separated from the set of all endsegments ist defined by

∩{E(k) | k ∈ ℕ_def} = ℵ₀.

> > By the axiom of separation these endsegments can be
> > collected in a set ℕ_def such that
> > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .
> "these end segments" =? {E(k) : k ∈ ℕ_def}

No. These endsegments are defined by ∩{E(k) | k ∈ ℕ_def} = ℵ₀.

> | ∩THESE | ≠ ℵ₀
>
> | ∩THESE | = 0

Then you gave a wrong definition. I collect only those ∩{E(k) | k ∈ ℕ_def} = ℵ₀.

>
> > Endsegments which would reduce this intersection
> > are not included. Hence the intersection
> > of these endsegments is infinite.
> "Reduces" from what? The choices are |{}| and |ℕ|

No, the choice is an infinite intersection but not ℕ. For instance we could use the first three endsegments. But we will continue to collect more - as long as the intersection remains infinite.

> I'll guess you mean |ℕ|

Wrong guess.

>
> And I see three readings of "reduce".

There is only one: The set of the first n endsegments reduces the intersection to {n, n+1, n+2, ...} We continue as long as the intersection is infinite, that is it has more elements than can be counted by a natural number.

Regards, WM

[Serg io]

On 11/20/2021 4:15 PM, WM wrote:
> Jim Burns schrieb am Samstag, 20. November 2021 um 21:16:31 UTC+1:
>> On 11/20/2021 6:44 AM, WM wrote:
>
>>> The intersection depends on the least endsegment.
>> There are two distinct senses of "less" here.
>
> No. The least endsegment has least elements.

you bird brain.