Le 17/01/2024 à 13:55, Richard Damon a écrit :
> On 1/17/24 6:49 AM, WM wrote:

>> NUF(x) cannot grow anywhere from 0 to ℵo without passing finite values.
>
> Why not?

Because of mathemtics. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.

Regards, WM

On 1/16/2024 11:39 PM, Ross Finlayson wrote:
> On Tuesday, January 16, 2024
> at 2:29:07 PM UTC-8, Ross Finlayson wrote:

>> [...]
>
> In proof theory, it's provable, that
> in classical logic,

proof.theory ≠ set.theory
proof ≠ set

logic < (non.logical) set.theory
set.theory = logic+set.description

set.description = infinity∨¬infinity
set.description = foundation∨¬foundation

different.description ⟺ different.models

> one must always assert
> Empty, Infinity, and well-foundedness
> for the inference of the existence of any set,
> for otherwise inference may arrive at
> Russell's set via quantification.

Russell's ∃{x|x∉x} ⟸ unrestricted.comprehension

s/comprehension/specification+replacement

¬∃{x|x∉x} ⟸ classical
On the other hand,
revision.theory.of.truth+more?

> So,
> in classical logic,
> ZF's axioms aren't independent.

Did you (RF) intend to make
that abrupt change of topic there?

That's not how I'd use "so"
Cha​cun à son goût.

Because models exist for
Infinity
¬Infinity
Foundation
¬Foundation
Choice
¬Choice
a (hypothetical) proof of
necessary.Infinity
necessary.Foundation
necessary.Choice
would be wrong.

tl;dr
Infinity, Foundation, Choice are
independent.

> Of course, these days we know that
> that's "quasi-modal" logic.

Uhm?
You mean?
https://philpapers.org/rec/GOLQEO-2
Quasi-Modal Equivalence of Canonical Structures

> Yet, it's not the point here that
> "classical" logic isn't monotonic.

Good to hear that it's not the point..
Classical is monotonic.

> Yet, the idea that
> "there's isn't a standard model of the integers",
> and, also,
> "there are multiple non-standard models of integers",
> in set theory, and as well in number theory,
> has a lot going on with
> "there's an infinite integer in
> some infinitudes of integers".

"has a lot going on with" = ?

Consider a finite set,
each element in its place.

A place is removed,
and not all elements have places.
We say:
the set's cardinality is finite.

Consider only all finite cardinals,
the ones which
cannot fit in a place removed.

For each finite cardinal,
inserting another element makes
a set which won't fit in its places,
one with a different cardinality.
Because finite.

Insert more and it still won't fit.
Insert all the cardinals and it still won't fit.

For each finite cardinal,
the cardinality of only all finite cardinals
is not its cardinality.

<drum.roll>
the cardinality of only all finite cardinals
is not a finite cardinality.
It is an infinite cardinality.
<cymbal.crash>

Check:
It isn't any cardinality which
cannot fit in a place removed.
It should fit in a place removed.

0@[1] 1@[2] 2@[3] 3@[4] ...
And check.

Weird as it is to fit in a place removed,
it must be so, or else
all the finite cardinals aren't
all the finite cardinals.

On 1/17/2024 6:49 AM, WM wrote:
> Le 16/01/2024 à 21:19, Dieter Heidorn a écrit :

>> [...]
>
> For the less-than relation
> there is no quantifier magic.

For anti.symmetric relations, such as less.than,
there is quantifier anti.magic.

| Assume P(x,y) ⇔ ¬P(y,x)
| Assume ∀x:∃y:P(x,y)
| | ∀x:∃y:P(x,y)
| if and only if
| ¬∃x:∀y:¬P(x,y)
| if and only if
| ¬∃x:∀y:P(y,x)
| relabel 'x''z','y''x','z''y'
| ¬∃y:∀x:P(x,y)

Therefore,
P(x,y) ⇔ ¬P(y,x)
∀x:∃y:P(x,y)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
¬∃y:∀x:P(x,y)
anti.magic

∀x:∃y≠x: x<y
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
¬∃y:∀y≠x: x<y
anti.magic

On 1/17/2024 7:55 AM, Richard Damon wrote:
> On 1/17/24 6:53 AM, WM wrote:
>> Le 16/01/2024 à 18:11, Jim Burns a écrit :

>>> (i)
>>> (∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
>>> true
>>
>> Yes, that is correct.

> Except we show it isn't for y = x/2

Well, for y = x/2, the antecedent is false,
thus the implication is true.
But WM probably isn't thinking of that.

(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
if and only if
0 < y ⇒ (∃x ∈ (0,1]: x ≤ y)

Fritz Feldhase schrieb am Mittwoch, 17. Januar 2024 um 20:15:47 UTC+1:
> On Wednesday, January 17, 2024 at 12:49:48 PM UTC+1, WM wrote:
>
> > For the less-than relation there is no quantifier magic.
> Was immer Deine "quantifier magic" auch sein soll; aber die (richtige)
> Reihenfolge der Quantoren ist wesentlich:
>
> An e IN: Em e IN: n < m (true)
>
> Em e IN: An e IN: n < m (false)

Both are true, but the average person cannot look into the infinite.
Therefore I have invented the simpler example with unit fractions.
If for every x ∈ (0, 1] there are ℵ smaller unit fractions, then ℵ
unit fractions are smaller than every x ∈ (0, 1] and lie left-hand side
of the whole interval. Otherwise a first one would appear within the
interval.

Please note: Without ℵ unit fractions left-hand side of (0, 1], the
above true statement cannot be true.

Regards, WM

On Wednesday, January 17, 2024 at 8:29:31 AM UTC-8, Jim Burns wrote:
> On 1/16/2024 11:39 PM, Ross Finlayson wrote:
> > On Tuesday, January 16, 2024
> > at 2:29:07 PM UTC-8, Ross Finlayson wrote:
> >> [...]
> >
> > In proof theory, it's provable, that
> > in classical logic,
> proof.theory ≠ set.theory
> proof ≠ set
>
> logic < (non.logical) set.theory
> set.theory = logic+set.description
>
> set.description = infinity∨¬infinity
> set.description = foundation∨¬foundation
>
> different.description ⟺ different.models
> > one must always assert
> > Empty, Infinity, and well-foundedness
> > for the inference of the existence of any set,
> > for otherwise inference may arrive at
> > Russell's set via quantification.
> Russell's ∃{x|x∉x} ⟸ unrestricted.comprehension
>
> s/comprehension/specification+replacement
>
> ¬∃{x|x∉x} ⟸ classical
> On the other hand,
> revision.theory.of.truth+more?
> > So,
> > in classical logic,
> > ZF's axioms aren't independent.
> Did you (RF) intend to make
> that abrupt change of topic there?
>
> That's not how I'd use "so"
> Cha​cun à son goût.
>
> Because models exist for
> Infinity
> ¬Infinity
> Foundation
> ¬Foundation
> Choice
> ¬Choice
> a (hypothetical) proof of
> necessary.Infinity
> necessary.Foundation
> necessary.Choice
> would be wrong.
>
> tl;dr
> Infinity, Foundation, Choice are
> independent.
>
> > Of course, these days we know that
> > that's "quasi-modal" logic.
> Uhm?
> You mean?
> https://philpapers.org/rec/GOLQEO-2
> Quasi-Modal Equivalence of Canonical Structures
> > Yet, it's not the point here that
> > "classical" logic isn't monotonic.
> Good to hear that it's not the point..
> Classical is monotonic.
> > Yet, the idea that
> > "there's isn't a standard model of the integers",
> > and, also,
> > "there are multiple non-standard models of integers",
> > in set theory, and as well in number theory,
> > has a lot going on with
> > "there's an infinite integer in
> > some infinitudes of integers".
> "has a lot going on with" = ?
>
>
> Consider a finite set,
> each element in its place.
>
> A place is removed,
> and not all elements have places.
> We say:
> the set's cardinality is finite.
>
>
> Consider only all finite cardinals,
> the ones which
> cannot fit in a place removed.
>
> For each finite cardinal,
> inserting another element makes
> a set which won't fit in its places,
> one with a different cardinality.
> Because finite.
>
> Insert more and it still won't fit.
> Insert all the cardinals and it still won't fit.
>
> For each finite cardinal,
> the cardinality of only all finite cardinals
> is not its cardinality.
>
> <drum.roll>
> the cardinality of only all finite cardinals
> is not a finite cardinality.
> It is an infinite cardinality.
> <cymbal.crash>
>
> Check:
> It isn't any cardinality which
> cannot fit in a place removed.
> It should fit in a place removed.
>
> 0@[1] 1@[2] 2@[3] 3@[4] ...
> And check.
>
> Weird as it is to fit in a place removed,
> it must be so, or else
> all the finite cardinals aren't
> all the finite cardinals.

Again your reasoning shows you in a good light,
I'll look to it.

About the "so, ..., axioms of restriction aren't independent...",
it's the inference as of what is directly above, or, "no, not non sequitur"..

It's pretty interesting and fun that model theory and proof theory
are equi-interpretable, intermixable, and not allowed to contradict each other.

Also free comprehension always exists.

On Wednesday, January 17, 2024 at 12:27:13 PM UTC-8, Fritz Feldhase wrote:
> On Wednesday, January 17, 2024 at 8:53:56 PM UTC+1, WM wrote:
> > Fritz Feldhase schrieb am Mittwoch, 17. Januar 2024 um 20:15:47 UTC+1:
>
> > > die (richtige) Reihenfolge der Quantoren ist wesentlich:
> > >
> > > An e IN: Em e IN: n < m (true)
> > >
> > > Em e IN: An e IN: n < m (false)
> > >
> > Both are true,
> Nope. "Em e IN: An e IN: n < m" is FALSE, otherwise there would have to be a natural number which is bigger than itself.
> > but the average person cannot look into the infinite.
> Yeah whatever, Mückenheim.

Yeah, that's wrong and Calvinistic, and rejected,
Duns Scotus and Spinoza give us a mathematical infinity,
that not only we can know, but according to Zeno, must.

That's not the same as "Absolute Infinity" and "the G-dhead",
which anyways are two things, that are named concepts with
universals in their definition.

Go away MW, Hodges won't be adding you to his hopeless.
An editor recalls some papers, ....

On 1/17/2024 11:53 AM, WM wrote:
> Fritz Feldhase schrieb am Mittwoch, 17. Januar 2024 um 20:15:47 UTC+1:
>> On Wednesday, January 17, 2024 at 12:49:48 PM UTC+1, WM wrote:
>> > For the less-than relation there is no quantifier magic.
>> Was immer Deine "quantifier magic" auch sein soll; aber die (richtige)
>> Reihenfolge der Quantoren ist wesentlich:
>> An e IN: Em e IN: n < m (true)
>> Em e IN: An e IN: n < m (false)
>
> Both are true, but the average person cannot look into the infinite.

You are the ultra finitist here... Pot Kettle?

> Therefore I have invented the simpler example with unit fractions.
> If for every x ∈ (0, 1] there are ℵ smaller unit fractions, then ℵ unit
> fractions are smaller than every x ∈ (0, 1] and lie left-hand side of
> the whole interval. Otherwise a first one would appear within the interval.
>
> Please note: Without ℵ unit fractions left-hand side of (0, 1], the
> above true statement cannot be true.
>
> Regards, WM
>
>

On 1/17/24 2:35 PM, Jim Burns wrote:
> On 1/17/2024 7:55 AM, Richard Damon wrote:
>> On 1/17/24 6:53 AM, WM wrote:
>>> Le 16/01/2024 à 18:11, Jim Burns a écrit :
>
>>>> (i)
>>>> (∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
>>>> true
>>>
>>> Yes, that is correct.
>
>> Except we show it isn't for y = x/2
>
> Well, for y = x/2, the antecedent is false,
> thus the implication is true.
> But WM probably isn't thinking of that.
>
>

What part of the antecedent (∀x ∈ (0,1]: y < x) is false?

x is such that x ∈ (0,1], and y is such that y < x

> (∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
>  if and only if
> 0 < y  ⇒  (∃x ∈ (0,1]: x ≤ y)
>
>

On 1/17/24 9:09 AM, WM wrote:
> Le 17/01/2024 à 13:55, Richard Damon a écrit :
>> On 1/17/24 6:49 AM, WM wrote:
>
>>> NUF(x) cannot grow anywhere from 0 to ℵo without passing finite values.
>>
>> Why not?
>
> Because of mathemtics. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
> Regards, WM

And how does that say that NUF(x) can't grow to ℵo without passing
finite values.

That equation in fact proves that there can not be a smallest 1/n as the
'level gap' below 1/n is only 1/(n+1) of the distance between 0 and 1/n,
so there is room for at least n+1 more unit fractions below it.

You need to find a point where the gap is as big as 1/n, and it never is.

On 1/17/2024 8:36 PM, Richard Damon wrote:
> On 1/17/24 2:35 PM, Jim Burns wrote:
>> On 1/17/2024 7:55 AM, Richard Damon wrote:
>>> On 1/17/24 6:53 AM, WM wrote:
>>>> Le 16/01/2024 à 18:11, Jim Burns a écrit :

>>>>> (i)
>>>>> (∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
>>>>> true
>>>>
>>>> Yes, that is correct.
>>
>>> Except we show it isn't for y = x/2
>>
>> Well, for y = x/2, the antecedent is false,
>> thus the implication is true.
>> But WM probably isn't thinking of that.
>
> What part of the antecedent
> (∀x ∈ (0,1]: y < x) is false?

Excuse me, I was thinking of something else.

As I'm sure you know, if
x ∈ (0,1] ⇒ y < x
is false for any value of x,
then all of
(∀x ∈ (0,1]: y < x)
is false,
and all of
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
is true.

For y > 0 and x = min{y/2,1}
x ∈ (0,1] ⇒ y < x
is false
(∀x ∈ (0,1]: y < x)
is false, and
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
is true.

On the other hand,
for y ≤ 0
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
is true,
whatever the case is for
∀x ∈ (0,1]: y < x

In sum,
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
is true, but
its truth doesn't tell us about y

I confess that I don't know
what significance that formula has to WM.

It's possible that, for WM,
talking about
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
replaces talking about
∀eps > 0: NUF(eps) ∉ ℕ

Fritz Feldhase schrieb am Mittwoch, 17. Januar 2024 um 21:27:13 UTC+1:
> On Wednesday, January 17, 2024 at 8:53:56 PM UTC+1, WM wrote:
> > Fritz Feldhase schrieb am Mittwoch, 17. Januar 2024 um 20:15:47 UTC+1:
>
> > > die (richtige) Reihenfolge der Quantoren ist wesentlich:
> > >
> > > An e IN: Em e IN: n < m (true)
> > >
> > > Em e IN: An e IN: n < m (false)
> > >
> > Both are true,
> Nope.

Yes, a mistake. Both are wrong. But the first statement is true for
*definable* numbers.

Regards, WM

Le 18/01/2024 à 02:36, Richard Damon a écrit :
> On 1/17/24 9:09 AM, WM wrote:
>> Le 17/01/2024 à 13:55, Richard Damon a écrit :
>>> On 1/17/24 6:49 AM, WM wrote:
>>
>>>> NUF(x) cannot grow anywhere from 0 to ℵo without passing finite values.
>>>
>>> Why not?
>>
>> Because of mathemtics. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.

> And how does that say that NUF(x) can't grow to ℵo without passing
> finite values.

Because after every unit fraction the function NUF(x) is constant over d_n
> 0.
>
> That equation in fact proves that there can not be a smallest 1/n as the
> 'level gap' below 1/n is only 1/(n+1) of the distance between 0 and 1/n,
> so there is room for at least n+1 more unit fractions below it.

Nevertheless ***all*** unit fractions have gaps between each other. There
is no exception.
>
> You need to find a point where the gap is as big as 1/n, and it never is.

We cannot investigate individuals within the dark domain.

Regards, WM

Le 18/01/2024 à 02:36, Richard Damon a écrit :
> On 1/17/24 2:35 PM, Jim Burns wrote:
>> On 1/17/2024 7:55 AM, Richard Damon wrote:
>>> On 1/17/24 6:53 AM, WM wrote:
>>>> Le 16/01/2024 à 18:11, Jim Burns a écrit :
>>
>>>>> (i)
>>>>> (∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0
>>>>> true
>>>>
>>>> Yes, that is correct.
>>
>>> Except we show it isn't for y = x/2
>>
>> Well, for y = x/2, the antecedent is false,
>> thus the implication is true.
>> But WM probably isn't thinking of that.
>>
>>
>
> What part of the antecedent (∀x ∈ (0,1]: y < x) is false?
>
> x is such that x ∈ (0,1], and y is such that y < x

Only if y is less than all x ∈ (0,1], the implication holds. If the
antecedent is violated, the implication is true nevertheless. But that is
irrelevant.
>
>> (∀x ∈ (0,1]: y < x)  ⇒  y ≤ 0

Regards, WM

Le 18/01/2024 à 06:15, Jim Burns a écrit :

> I confess that I don't know
> what significance that formula has to WM.

If ∀x ∈ (0, 1]: NUF(x) = ℵo was true, it would prove negative unit
fractions y.

Regards, WM

On 1/18/2024 4:15 AM, WM wrote:
> Le 18/01/2024 à 06:15, Jim Burns a écrit :

>> In sum,
>> (∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
>> is true, but
>> its truth doesn't tell us about y

>> I confess that I don't know
>> what significance that formula has to WM.
>
> If
> ∀x ∈ (0, 1]: NUF(x) = ℵo
> was true,
> it would prove negative unit fractions y.

How does
(∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
and
∀x ∈ (0,1]: NUF(x) = ℵ₀
prove negative unit fractions y?

----
In arithmetic,
∀k <ℵ₀: ∀j <ℵ₀: j+k <ℵ₀

∀k <ℵ₀: ∀i <ℵ₀: i < |{j+k | j <ℵ₀}|

∀k <ℵ₀: ¬∃i <ℵ₀: i = |{j+k | j <ℵ₀}|

∀k <ℵ₀: ¬( |{j+k | j <ℵ₀}| <ℵ₀ )

----
In arithmetic,
∀k ∈ℕ₁: ∀j ∈ℕ: 0 < ⅟(j+k) ≤ 1

∀k ∈ℕ₁: {⅟(j+k) | j ∈ℕ} ⊆ (0,1] ∧
∀i ∈ℕ: i < |{⅟(j+k) | j ∈ℕ}|

∀k ∈ℕ₁: {⅟(j+k) | j ∈ℕ} ⊆ (0,1] ∧
¬∃i ∈ℕ: i = |{⅟(j+k) | j ∈ℕ}|

∀k ∈ℕ₁: {⅟(j+k) | j ∈ℕ} ⊆ (0,1] ∧
¬( |{⅟(j+k) | j ∈ℕ}| < |ℕ| )

At what formula, if any, have we stopped
using arithmetic?

Keep in mind that
_those formulas_ aren't dark.
You can see them on your screen.

WM explained :
> Le 18/01/2024 à 02:36, Richard Damon a écrit :
>> On 1/17/24 9:09 AM, WM wrote:
>>> Le 17/01/2024 à 13:55, Richard Damon a écrit :
>>>> On 1/17/24 6:49 AM, WM wrote:
>>>
>>>>> NUF(x) cannot grow anywhere from 0 to ℵo without passing finite values.
>>>>
>>>> Why not?
>>>
>>> Because of mathemtics. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
>
>> And how does that say that NUF(x) can't grow to ℵo without passing finite
>> values.
>
> Because after every unit fraction the function NUF(x) is constant over d_n
>> 0.
>>
>> That equation in fact proves that there can not be a smallest 1/n as the
>> 'level gap' below 1/n is only 1/(n+1) of the distance between 0 and 1/n, so
>> there is room for at least n+1 more unit fractions below it.
>
> Nevertheless ***all*** unit fractions have gaps between each other. There is
> no exception.

Yes, there are gaps in Q+ with respect to the positive reals fractional
parts. Unless "***all*** unit fractions" means the set of unit
fractions, in which case you just babble as usual.

On 1/18/24 4:09 AM, WM wrote:
> Le 18/01/2024 à 02:36, Richard Damon a écrit :
>> On 1/17/24 9:09 AM, WM wrote:
>>> Le 17/01/2024 à 13:55, Richard Damon a écrit :
>>>> On 1/17/24 6:49 AM, WM wrote:
>>>
>>>>> NUF(x) cannot grow anywhere from 0 to ℵo without passing finite
>>>>> values.
>>>>
>>>> Why not?
>>>
>>> Because of mathemtics. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
>
>> And how does that say that NUF(x) can't grow to ℵo without passing
>> finite values.
>
> Because after every unit fraction the function NUF(x) is constant over d_n

So? That doesn't mean that NUF(x) can't instantly grow to infinity
between 0 and the range (0,1]

>> 0.
>>
>> That equation in fact proves that there can not be a smallest 1/n as
>> the 'level gap' below 1/n is only 1/(n+1) of the distance between 0
>> and 1/n, so there is room for at least n+1 more unit fractions below it.
>
> Nevertheless ***all*** unit fractions have gaps between each other.
> There is no exception.

And again, we aren't talking BETWEEN unit fractions, but between 0 and
(0,1].

>>
>> You need to find a point where the gap is as big as 1/n, and it never is.
>
> We cannot investigate individuals within the dark domain.

You can not investigate the dark domain because it doesn't exist.

>
> Regards, WM
>
>

Le 18/01/2024 à 13:24, Jim Burns a écrit :
> On 1/18/2024 4:15 AM, WM wrote:
>> Le 18/01/2024 à 06:15, Jim Burns a écrit :
>
>>> In sum,
>>> (∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
>>> is true, but
>>> its truth doesn't tell us about y
>
>>> I confess that I don't know
>>> what significance that formula has to WM.
>>
>> If
>> ∀x ∈ (0, 1]: NUF(x) = ℵo
>> was true,
>> it would prove negative unit fractions y.
>
> How does
> (∀x ∈ (0,1]: y < x) ⇒ y ≤ 0
> and
> ∀x ∈ (0,1]: NUF(x) = ℵ₀
> prove negative unit fractions y?

For every point x > 0 there are ℵ smaller unit fractions.
==>
There exists no point x > 0 without ℵ smaller unit fractions.
==>
There exist ℵ negative unit fractions.

Regards, WM

Le 18/01/2024 à 13:44, FromTheRafters a écrit :

> Yes, there are gaps in Q+ with respect to the positive reals fractional
> parts.

That means NUF(x) does not increase by more than 1 without stopping
afterwards. It starts with 0 and not with ℵ.

Regards, WM

Le 18/01/2024 à 13:51, Richard Damon a écrit :
> On 1/18/24 4:09 AM, WM wrote:

>> Because after every unit fraction the function NUF(x) is constant over d_n
>
> So? That doesn't mean that NUF(x) can't instantly grow to infinity
> between 0 and the range (0,1]

It does mean exactly this. Otherwise:

If for every point x > 0 there are ℵ smaller unit fractions,
then there exists no point x > 0 without ℵ smaller unit fractions.
Then there exist ℵ negative unit fractions.

Regards, WM

WM wrote :
> Le 18/01/2024 à 13:44, FromTheRafters a écrit :
>
>> Yes, there are gaps in Q+ with respect to the positive reals fractional
>> parts.
>
> That means NUF(x) does not increase by more than 1 without stopping
> afterwards.

No, it doesn't.

> It starts with 0 and not with ℵ.

It is not an action.

On Wednesday, January 17, 2024 at 1:07:02 PM UTC-8, Ross Finlayson wrote:
> On Wednesday, January 17, 2024 at 8:29:31 AM UTC-8, Jim Burns wrote:
> > On 1/16/2024 11:39 PM, Ross Finlayson wrote:
> > > On Tuesday, January 16, 2024
> > > at 2:29:07 PM UTC-8, Ross Finlayson wrote:
> > >> [...]
> > >
> > > In proof theory, it's provable, that
> > > in classical logic,
> > proof.theory ≠ set.theory
> > proof ≠ set
> >
> > logic < (non.logical) set.theory
> > set.theory = logic+set.description
> >
> > set.description = infinity∨¬infinity
> > set.description = foundation∨¬foundation
> >
> > different.description ⟺ different.models
> > > one must always assert
> > > Empty, Infinity, and well-foundedness
> > > for the inference of the existence of any set,
> > > for otherwise inference may arrive at
> > > Russell's set via quantification.
> > Russell's ∃{x|x∉x} ⟸ unrestricted.comprehension
> >
> > s/comprehension/specification+replacement
> >
> > ¬∃{x|x∉x} ⟸ classical
> > On the other hand,
> > revision.theory.of.truth+more?
> > > So,
> > > in classical logic,
> > > ZF's axioms aren't independent.
> > Did you (RF) intend to make
> > that abrupt change of topic there?
> >
> > That's not how I'd use "so"
> > Cha​cun à son goût.
> >
> > Because models exist for
> > Infinity
> > ¬Infinity
> > Foundation
> > ¬Foundation
> > Choice
> > ¬Choice
> > a (hypothetical) proof of
> > necessary.Infinity
> > necessary.Foundation
> > necessary.Choice
> > would be wrong.
> >
> > tl;dr
> > Infinity, Foundation, Choice are
> > independent.
> >
> > > Of course, these days we know that
> > > that's "quasi-modal" logic.
> > Uhm?
> > You mean?
> > https://philpapers.org/rec/GOLQEO-2
> > Quasi-Modal Equivalence of Canonical Structures
> > > Yet, it's not the point here that
> > > "classical" logic isn't monotonic.
> > Good to hear that it's not the point..
> > Classical is monotonic.
> > > Yet, the idea that
> > > "there's isn't a standard model of the integers",
> > > and, also,
> > > "there are multiple non-standard models of integers",
> > > in set theory, and as well in number theory,
> > > has a lot going on with
> > > "there's an infinite integer in
> > > some infinitudes of integers".
> > "has a lot going on with" = ?
> >
> >
> > Consider a finite set,
> > each element in its place.
> >
> > A place is removed,
> > and not all elements have places.
> > We say:
> > the set's cardinality is finite.
> >
> >
> > Consider only all finite cardinals,
> > the ones which
> > cannot fit in a place removed.
> >
> > For each finite cardinal,
> > inserting another element makes
> > a set which won't fit in its places,
> > one with a different cardinality.
> > Because finite.
> >
> > Insert more and it still won't fit.
> > Insert all the cardinals and it still won't fit.
> >
> > For each finite cardinal,
> > the cardinality of only all finite cardinals
> > is not its cardinality.
> >
> > <drum.roll>
> > the cardinality of only all finite cardinals
> > is not a finite cardinality.
> > It is an infinite cardinality.
> > <cymbal.crash>
> >
> > Check:
> > It isn't any cardinality which
> > cannot fit in a place removed.
> > It should fit in a place removed.
> >
> > 0@[1] 1@[2] 2@[3] 3@[4] ...
> > And check.
> >
> > Weird as it is to fit in a place removed,
> > it must be so, or else
> > all the finite cardinals aren't
> > all the finite cardinals.
> Again your reasoning shows you in a good light,
> I'll look to it.
>
> About the "so, ..., axioms of restriction aren't independent...",
> it's the inference as of what is directly above, or, "no, not non sequitur".
>
> It's pretty interesting and fun that model theory and proof theory
> are equi-interpretable, intermixable, and not allowed to contradict each other.
>
> Also free comprehension always exists.

Hi James,

I enjoyed this talk, it helps reflect quite a thorough theoretical approach
to continuity and about the objects of mathematics sort of generally.

https://www.youtube.com/watch?v=U7KbayajRaA&list=PLb7rLSBiE7F4eHy5vT61UYFR7_BIhwcOY&index=9

There's lots of talk here about infinitesimals that don't exist. Now, maybe it will
help if we sort of delineate the classical notions, of what these things are.

The ordered field is Archimedean. It has the infinitely-many, the unbounded,
and, no infinite magnitudes. Now, its result that way, sort of has to include
how it's arrived at. Archimedes, first establishes there _are_ infinitely-many,
there are not found bounds, and, then that there are no infinite magnitudes..
This then is very Pythagorean, where, the ordered field of rationals, yet of
course has no ir-rational quantities, either. This then wouldn't suffice for
most today, of course, because we have a notion of a complete ordered field,
that's a model of a linear continuum as of our model of a linear curriculum..

Then, what seems most these people are ignorant, is Aristotle's continuum.
Now, Aristotle is most well-known for inductive inference after syllogism.
But, there's also has deductive inference, for Aristotle's reflections on Parmenides.
Here though, the notion of the Aristotle's continuum, is that the segment,
is finite, but is equi-partitioned with infinitely-many. This is the most simple
usual model of line-drawing or putting pencil to paper, drawing, and lifting it,
relating exactly this model in space according to any model in time, in exactly
the course-of-passage, in space through time.

So, it's something to be known that when someone introduces this sort of
notion, Aristotle vis-a-vis Archimedes, that there's plenty going on in their
considerations of the replete courtesy courtesy their simple relations of
the mental apparatus.

These days for example there are notions of "non-Archimedean fields". Now,
you might look at something like Groethendieck universes, which is in the
space of words for algebra, and entirely separate the notions of the space
of words, and the space of points. They end up sort of being at odds.

So, is the complete ordered field, after Eudoxus' construction of the rational
moduli and its extension about methods of exhaustion that the circle is no
different than the infinitely-sided regular polygon, in the infinite limit?
This is of course that methods of exhaustion or the infinite limit was a notion
long before Weierstrass helped bolt it down with a passel of delta-epsilonics.
So, Cauchy/Dedekind is just a patching of Eudoxus-to-Weierstrass into abstract
algebra in a space of words.

Is the complete ordered field, ..., Archimedean and Pythagorean, anymore?
The field of rationals is, yes, indeed, but, _there is more to it_, it's not so much
that "it's more than the sum of its parts" as "it IS the sum of its parts".

So, "iota-values", or "standard infinitesimals", "modern paleo-classical Aristotle's
constant monotone increasing vanishing values with a different lower and upper bound",
is ANCIENT and already long, long, long ago part of the canon.

Then, to make it modern again, is for a simple few results in function theory for line-reals,
and topology for signal-reals, resulting at least THREE of more replete definitions of
continuity and continuous domains, that exist in the universe of objects of mathematics.

Where nothing non-mathematical does.

Now I'm very much enjoying your introduced notation, and as well particularly
your attention to quantifier disambiguation, I generally frame it in this sort
of way: that the universal quantifier has various forms for-any/for-each/for-every/for-all,
that these are "four different for's", that in some conditions are same and others different.

This is about quite directly the Sorites and Heap, which otherwise you'll find
all these quite standard modern adherents to exoteric dogma tripping all over.

Le 18/01/2024 à 18:02, FromTheRafters a écrit :
> WM wrote :
>> Le 18/01/2024 à 13:44, FromTheRafters a écrit :
>>
>>> Yes, there are gaps in Q+ with respect to the positive reals fractional
>>> parts.
>>
>> That means NUF(x) does not increase by more than 1 without stopping
>> afterwards.
>
> No, it doesn't.

In mathematics it does. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
>
>> It starts with 0 and not with ℵ.
>
> It is not an action.

Hogwash. It is a function.

Regards, WM

WM formulated on Thursday :
> Le 18/01/2024 à 18:02, FromTheRafters a écrit :
>> WM wrote :
>>> Le 18/01/2024 à 13:44, FromTheRafters a écrit :
>>>
>>>> Yes, there are gaps in Q+ with respect to the positive reals fractional
>>>> parts.
>>>
>>> That means NUF(x) does not increase by more than 1 without stopping
>>> afterwards.
>>
>> No, it doesn't.
>
> In mathematics it does. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
>>
>>> It starts with 0 and not with ℵ.
>>
>> It is not an action.
>
> Hogwash. It is a function.

Then map it, done. No start here and finish there involved.