You only make it worse!

There are roundabout arguments that, for example,
the FINITE ORDINALS, as a set, consequently contain
themselves, as an element. This is a direct
compactness result.

If you want to have ordinals that contain themselves,
you need to mention an encoding. Because per se,
we understand by ordinal an order type.

There ware various encodings for finite ordinals around:
1) von Neuman encoding, based on succ(X) = X u {X} and 0 = {}
2) Zermelo encoding, bsaed on succ(X) = {X} and 0 = {}
3) Your Powerset idea, based on succ(X) = P(X) and 0 = {}

All 3 have the property that:

/* provable */
n in n+1 and n is finite

Proof:
case 1): n+1 = n u {n}, n in n+1 because n in {n}.
further succ(X) sendes an already finite set into a finite set.
case 2): n+1 = {n}, n in n+1 because n in {n}.
further succ(X) sendes an already finite set into a finite set.
case 3): n+1 = P(n), n in n+1 because n in P(n).
further succ(X) sendes an already finite set into a finite set.
Q.E.D.

But none has the property that omega = { n } contains
itself, the proof of contradiction applies irrelevant
of the encoding, it only makes use of the

notion finite and infinite:

/* provable */
~(omega in omega) & (Y in omega => Y finite)

Proof:
(Y in omega => Y finite) follows by the claim that
omega = { n }, i.e. the least set that contains all finite
ordinals in the corresponding encoding. If it would
contain something infinite it would not be the least

set that contains all finite ordinals, would have some
extra in it. Violating the very construction of omega from
the finite ordinals.
Q.E.D.


Ross Finlayson schrieb:

On 02/19/2024 05:01 PM, Mild Shock wrote:

The contradiction is very easy:

Lets say X is the set of all finite ordinals.

- observe that X is an infinite ordinal.
- observe that if Y in X, then Y is a finite ordinal.
- hence if X in X it would be an infinite and finite ordinal at the
same time.
- an X cannot be infinite and finite at the same time.
Q.E.D:

Ross Finlayson schrieb am Dienstag, 20. Februar 2024 um 00:04:06 UTC+1: >>>>>> There are roundabout arguments that, for example, the finite

ordinals,
as a set, consequently contain themselves, as an element. This is a >>>>>> direct compactness result.

(Maybe that's just me.)

Imagine if ordinals' proper model was that the successor
was powerset, instead of just any old ordered pair.

So, those together are the "sets that don't contain themselves",
the sets of ordinals.

Quantifying over those, results the "Russell set the ordinal",
it contains itself.

So here Y isn't necessarily a finite ordinal.

Q.E.R.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

You could use an encoding of finite ordinals
into infinite objects, like:

0 = omega, 1 = omega+1, etc..

Then my proof doesn't work so easily. You can then
use the regularity axiom, to show:

/* provable */
~(omega in omega)

Axiom of regularity
https://en.wikipedia.org/wiki/Axiom_of_regularity

Is this your A "paradox" is not a set in ZF?
In non-ZF you could aim at making omega a Quine atom: https://en.wikipedia.org/wiki/Urelement#Quine_atoms

Or any other construction and encoding where you
would sneak in a set into itself.

Mild Shock schrieb:

You only make it worse!

There are roundabout arguments that, for example,
the FINITE ORDINALS, as a set, consequently contain
themselves, as an element. This is a direct
compactness result.

If you want to have ordinals that contain themselves,
you need to mention an encoding. Because per se,
we understand by ordinal an order type.

There ware various encodings for finite ordinals around:
1) von Neuman encoding, based on succ(X) = X u {X} and 0 = {}
2) Zermelo encoding, bsaed on succ(X) = {X} and 0 = {}
3) Your Powerset idea, based on succ(X) = P(X) and 0 = {}

All 3 have the property that:

/* provable */
n in n+1 and n is finite

Proof:
case 1): n+1 = n u {n}, n in n+1 because n in {n}.
further succ(X) sendes an already finite set into a finite set.
case 2): n+1 = {n}, n in n+1 because n in {n}.
further succ(X) sendes an already finite set into a finite set.
case 3): n+1 = P(n), n in n+1 because n in P(n).
further succ(X) sendes an already finite set into a finite set.
Q.E.D.

But none has the property that omega = { n } contains
itself, the proof of contradiction applies irrelevant
of the encoding, it only makes use of the

notion finite and infinite:

/* provable */
~(omega in omega) & (Y in omega => Y finite)

Proof:
(Y in omega => Y finite) follows by the claim that
omega = { n }, i.e. the least set that contains all finite
ordinals in the corresponding encoding. If it would
contain something infinite it would not be the least

set that contains all finite ordinals, would have some
extra in it. Violating the very construction of omega from
the finite ordinals.
Q.E.D.

Ross Finlayson schrieb:

On 02/19/2024 05:01 PM, Mild Shock wrote:

The contradiction is very easy:

Lets say X is the set of all finite ordinals.

- observe that X is an infinite ordinal.
- observe that if Y in X, then Y is a finite ordinal.
- hence if X in X it would be an infinite and finite ordinal at the
same time.
- an X cannot be infinite and finite at the same time.
Q.E.D:

Ross Finlayson schrieb am Dienstag, 20. Februar 2024 um 00:04:06 UTC+1: >>>>>>> There are roundabout arguments that, for example, the finite

ordinals,
as a set, consequently contain themselves, as an element. This is a >>>>>>> direct compactness result.

(Maybe that's just me.)

Imagine if ordinals' proper model was that the successor
was powerset, instead of just any old ordered pair.

So, those together are the "sets that don't contain themselves",
the sets of ordinals.

Quantifying over those, results the "Russell set the ordinal",
it contains itself.

So here Y isn't necessarily a finite ordinal.

Q.E.R.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

The phrase the "The set of all ordinals" is meaningless if ordinals
are not sets itself. At the time of Burali-Forti set theory was
not that evolved. And the proof at that time didn't use regularity axiom.

So the set of all ordinals was a notion of naive set theory, and not
formulated in modern set theory ordinal terminology, but as a
question about transfinite numbers:

Una questione sui numeri transfiniti https://zenodo.org/records/2362091/files/article.pdf

As one can see from the paper the proof proceeded by establishing:

Ω + 1 > Ω and Ω + 1 < Ωmarkus...@gmail.com schrieb:

No set can contain itself. The set of all ordinals would be a new ordinal

and thus contain itself. Ergo, there cannot be a set of all ordinals.


--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 2/20/2024 3:15 PM, Ross Finlayson wrote:

On 02/20/2024 11:35 AM, markus...@gmail.com wrote:

No set can contain itself.
The set of all ordinals would be a new ordinal,
and thus contain itself.
Ergo, there cannot be a set of all ordinals.

"... in set-theories like ZF
that are ordinary/well-founded,
according to an axiom like Regularity
of restriction of comprehension."

There are others, ..., "Mengenlehre(n)".

However,
whatever sets might be,
ordinals would not be ordinals
if they weren't well.ordered by ∈

In any theory in which ordinals are ordinals,
at least the ordinals have finite.descent,
whatever might be true of other sets.

A proposed set.of.all.ordinals which
held itself would not have finite descent.


Ordinals are well.ordered.
Well.ordered.ness can be re.phrased as
transfinite.induction.ness.
(∀α:(∀β<α:P(β))⇒P(α)) ⟹ ∀γ:P(γ)

FD(γ) == "γ has finite descent"

| Assume each ordinal β < α has finite descent.
| ∀β<α:FD(β)
|
| ⟨ α β δ ε ... ⟩ is a strictly.descending sequence
| α > β
| β has finite descent.
| ⟨ β δ ε ... ⟩ is finite
| ⟨ α β δ ε ... ⟩ is finite
| Generalizing over sequences,
| α has finite descent.

Therefore, generalizing over ordinals,
∀α:(∀β<α:FD(β))⇒FD(α)

By transfinite.induction (by well.order),
∀γ:FD(γ)
Each ordinal has finite descent.

Therefore,
the ordinal(?) holding all(?) ordinals
does not hold itself.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Le 20/02/2024 à 23:02, Jim Burns a écrit :

In any theory in which ordinals are ordinals,
at least the ordinals have finite.descent,

That proves finite ascend too, because otherwise every ordinal could be ascended and then the way upstairs could be gone back downstairs. Finite
ascend and descend prove that most ordinals are dark.

Ordinals are well.ordered.

Only those which can be specified.

Regards, WM

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 2/21/2024 3:33 AM, WM wrote:

Le 20/02/2024 à 23:02, Jim Burns a écrit :

Ordinals are well.ordered.

Only those which can be specified.

No.
All of them are well.ordered.
Anything else wouldn't be the ordinals.

Anything else would be like declaring
that only specifiableᵂᴹ right.triangles
have three corners.

In any theory in which ordinals are ordinals,
at least the ordinals have finite.descent,

That proves finite ascend too,
because
otherwise
every ordinal could be ascended
and then
the way upstairs could be gone back downstairs.

The ordinals' descents and ascents are not the same.

If any of an ordinal's descents is infinite,
the ordinal doesn't have finite.descent.

If any of an ordinal's ascents is infinite,
the ordinal doesn't have finite.ascent.


Each ordinal α has a successor α+1
α+1 has α+2, etc.

For ordinal a
⟨ α α+1 α+2 α+3 ... ⟩ is an infinite ascent.
α doesn't have finite.ascent.

Generalizing over ordinals,
no ordinal a has finite.ascent.


For each ordinal ψ
if ψ has any infinite descent,
then, because well.order,
an ordinal χ exists first with any infinite descent.

However,
one step down from χ to any ordinal β < χ is to
β with only finite descents,
and finite plus one is finite.
First χ doesn't have any infinite descent.
Contradiction.

ψ doesn't have an infinite descent.

Generalizing over ordinals,
each ordinal ψ has finite.descent.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

In the philosophy of mathematics, specifically the philosophical
foundations of set theory, limitation of size is a concept developed by
Philip Jourdain and/or Georg Cantor to avoid Cantor's paradox. It
identifies certain "inconsistent multiplicities", in Cantor's
terminology, that cannot be sets because they are "too large". In modern terminology these are called proper classes. https://en.wikipedia.org/wiki/Limitation_of_size

You might like this book:

Cantor's ideas formed the basis for set theory and also for the
mathematical treatment of the concept of infinity. The philosophical and heuristic framework he developed had a lasting effect on modern
mathematics, and is the recurrent theme of this volume. Hallett explores Cantor's ideas and, in particular, their ramifications for
Zermelo-Frankel set theory. https://academic.oup.com/pq/article-abstract/36/144/429/1567519

Mild Shock schrieb:

Seriously, you don't know what classes are?

The membership relation is the same
for members of classes and for members of sets.
Since members of classes are sets just like

the members of sets are sets, in ZF. And there is
only one membership relation ∈ between sets. The
distinction between classes and sets was described

in the past as:

sets: includes collections of sizes from the numbers to
  the transfinite numbers
classes: includes collections that Cantor called
  NCONSISTENT MULTIPLICITIES

You had them somewhere in one of your random posts:

Ross Finlayson schrieb:

Of course, the goal is "there are no paradoxes at all",
then what seem "inconsistent multiplicities", just don't relate.

But this below is awful gibberish:

Ross Finlayson schrieb:

If ORD involves class/set distinction,
and a set-theory can also be written as a part-theory,
then what's part/particle distinction/

If set theory's relation is "elt", element-of, "in"
and class theory's relation is "members", "contains", "has",
then, is :
class/set theory
set/part theory?

Here that "numbering" and "counting" are two different things,
one for ordering theory the other for collection,
ordinals and sets, numbering and counting,
what about
set/class distinction and
set/part distinction and
part/class distinction?

See, this is among reasons why
I've been way both ahead of
and on top of this for a long time,
and trying to tell you so all the time.

I told you, ..., I told you.

Mostly is for understanding that
"numbering" and "counting" are
two different things, and they
involve each other in their resources.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Seriously, you don't know what classes are?

The membership relation is the same
for members of classes and for members of sets.
Since members of classes are sets just like

the members of sets are sets, in ZF. And there is
only one membership relation ∈ between sets. The
distinction between classes and sets was described

in the past as:

sets: includes collections of sizes from the numbers to
the transfinite numbers
classes: includes collections that Cantor called
NCONSISTENT MULTIPLICITIES

You had them somewhere in one of your random posts:

Ross Finlayson schrieb:

Of course, the goal is "there are no paradoxes at all",
then what seem "inconsistent multiplicities", just don't relate.

But this below is awful gibberish:

Ross Finlayson schrieb:

If ORD involves class/set distinction,
and a set-theory can also be written as a part-theory,
then what's part/particle distinction/

If set theory's relation is "elt", element-of, "in"
and class theory's relation is "members", "contains", "has",
then, is :
class/set theory
set/part theory?

Here that "numbering" and "counting" are two different things,
one for ordering theory the other for collection,
ordinals and sets, numbering and counting,
what about
set/class distinction and
set/part distinction and
part/class distinction?

See, this is among reasons why
I've been way both ahead of
and on top of this for a long time,
and trying to tell you so all the time.

I told you, ..., I told you.

Mostly is for understanding that
"numbering" and "counting" are
two different things, and they
involve each other in their resources.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Le 21/02/2024 à 18:59, Jim Burns a écrit :

On 2/21/2024 3:33 AM, WM wrote:

Le 20/02/2024 à 23:02, Jim Burns a écrit :

Ordinals are well.ordered.

Only those which can be specified.

No.
All of them are well.ordered.

How do you know?

Anything else wouldn't be the ordinals.

In fact, not these ordinals.

Anything else would be like declaring
that only specifiableᵂᴹ right.triangles
have three corners.

That is too drastic. Natnumbers keep almost all of their properties.

The ordinals' descents and ascents are not the same.

Every way up can be reversed. That proves that also the ascents are
finite.

For each ordinal ψ
if ψ has any infinite descent,
then, because well.order,
an ordinal χ exists first with any infinite descent.

However,
one step down from χ to any ordinal β < χ is to
β with only finite descents,
and finite plus one is finite.
First χ doesn't have any infinite descent.
Contradiction.

ψ doesn't have an infinite descent.

And one step upwards is finite too. Finite plus one is finite.a

ψ doesn't have an infinite ascent (for every visible predecessor).

Generalizing over ordinals,
each ordinal ψ has finite.descent.

Each ordinal has finite ascent.

Regards, WM

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 2/22/2024 8:00 AM, WM wrote:

Le 21/02/2024 à 18:59, Jim Burns a écrit :

On 2/21/2024 3:33 AM, WM wrote:

Le 20/02/2024 à 23:02, Jim Burns a écrit :

Ordinals are well.ordered.

Only those which can be specified.

No.
All of them are well.ordered.

How do you know?

In the same way that I know
that right.triangles have three corners.
Not by looking at ordinals.
By knowing what "ordinal" means.

Anything else wouldn't be the ordinals.

In fact, not these ordinals.

Then "these ordinals" are like
four.cornered right.triangles.

Anything else would be like declaring
that only specifiableᵂᴹ right.triangles
have three corners.

That is too drastic.
Natnumbers keep almost all of their properties.

Four.cornered right.triangles share
many properties with other plane figures.
They aren't right.triangles.

The ordinals' descents and ascents are not the same.

Every way up can be reversed.

An infinite way up isn't _to_ any ordinal.
Reversed, it isn't _from_ any ordinal.

That proves that also the ascents are finite.

By excluding all infinite ascents,
we can prove that all _remaining_ ascents
are finite.

...which doesn't deny that
⟨ α α+1 α+2 α+3 ... ⟩ is an infinite ascent.

In contrast,
there is no first ordinal with
an infinite descent, so
there is no ordinal with
an infinite descent.

For each ordinal ψ
if ψ has any infinite descent,
then, because well.order,
an ordinal χ exists first with any infinite descent.

However,
one step down from χ to any ordinal β < χ is to
β with only finite descents,
and finite plus one is finite.
First χ doesn't have any infinite descent.
Contradiction.

ψ doesn't have an infinite descent.

And one step upwards is finite too.
Finite plus one is finite.a

Finite ascents exist.
Finite descents exist too.

Infinite ascents exist,
⟨ α α+1 α+2 α+3 ... ⟩
A first infinite descent is
a contradiction.

ψ doesn't have an infinite ascent
(for every visible predecessor).

Daek numbers wouldn't make
⟨ ψ ψ+1 ψ+2 ψ+3 ... ⟩ less infinite.

Generalizing over ordinals,
each ordinal ψ has finite.descent.

Each ordinal has finite ascent.

Know what "finite ascent" means:
no infinite ascents.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Doesn't make any sense at all.

Not a single mention of proper classes here: https://plato.stanford.edu/ENTRIES/mereology/

Ross Finlayson schrieb:

Actually, for class/set distinction,
I just introduced set/part distinction,
and part/particle distinction,
and set/particle distinction.

set:class::part:particle

set:part::class:particle

This is a usual form that A:B::C:D is
that A relates to B as C relates to D,
"set is to class as part is to particle", and
"set is to part as class is to particle".

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

I guess we have reached your intellectual
boundaries, inherent in your squirell brain
sized, that of a walnut, cerebrum and cerebrellum.

Mild Shock schrieb:

Doesn't make any sense at all.

Not a single mention of proper classes here: https://plato.stanford.edu/ENTRIES/mereology/

Ross Finlayson schrieb:

Actually, for class/set distinction,
I just introduced set/part distinction,
and part/particle distinction,
and set/particle distinction.

set:class::part:particle

set:part::class:particle

This is a usual form that A:B::C:D is
that A relates to B as C relates to D,
"set is to class as part is to particle", and
"set is to part as class is to particle".

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

What we can say is the following:

i) Every set is a class
ii) Not every class is a set

So there is a hypernym / hyponym relationship
between the two. Here is are proof of i) and ii):

Proof i): Let s be a set. Then we can form
the class { x | x e s }. So there is an injection
from the sets to the classes.

Proof ii): Let V be the class { x | true },
this is the universal class which is provably
not a set. So there is no surjection from
the sets to the classes.

Hope this helps. Injection is usually taken
as indicative that two sets are in the
less than or equal relation ship, i.e. ⊆.
And lack of surjection indicates that there
is no bijection, i.e. ≠, so we have:

Sets ⊆ Classes and Sets ≠ Classes

Or together:

Sets ⊂ Classes

The difference Class \ Sets, those things
that are classes but not sets, are called
proper classes.

Mild Shock schrieb:

I guess we have reached your intellectual
boundaries, inherent in your squirell brain
sized, that of a walnut, cerebrum and cerebrellum.

Mild Shock schrieb:

Doesn't make any sense at all.

Not a single mention of proper classes here:
https://plato.stanford.edu/ENTRIES/mereology/

Ross Finlayson schrieb:

Actually, for class/set distinction,
I just introduced set/part distinction,
and part/particle distinction,
and set/particle distinction.

set:class::part:particle

set:part::class:particle

This is a usual form that A:B::C:D is
that A relates to B as C relates to D,
"set is to class as part is to particle", and
"set is to part as class is to particle".

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Are you pulling a John Gabriel? Back
to greek ratios. Euclids general form:

A : B = C : D

What about this form:

A : B : C = E : F : G http://aleph0.clarku.edu/~djoyce/elements/bookV/defV3.html

I would especially recommend equations of the form:

cornet:walnut:pistachio
= cup:banana:mango
= hotday:ingest:cooling


Ross Finlayson schrieb:

Hm.  "squirrel:brain::walnut:cerebrum".

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Better symbolism would be:

Sets' ⫋ Classes

Where Sets' results from Sets by the injection
{ x | x e s } for each x e Sets. This gives a little
transfer principle. If you can prove, i.e. that

a property holds for all classes:

∀X P(X)

Then it follows, that the property holds for all sets.

∀x P(x)

Proof: In higher order logic one would probably
write λy.(x y) for { y | y e x }, by eta reduction
we have λy.(x y)= x, so one can prove:

∀X P(X)
------------ (∀ elim)
P(λy.(x y))
------------ (η-reduction)
P(x)
------------ (∀ Intro)
∀x P(x)

Q.E.D.

η-reduction expresses the idea of extensionality https://en.wikipedia.org/wiki/Lambda_calculus#%CE%B7-reduction

Wao! Now I did a lot of cheating, sweeping a lot of
details under the rug. I guess this is not
the standard way to do these things.

Better have a look here:

Basic Set Theory - Azriel Levy
https://www.amazon.com/dp/0486420795

Mild Shock schrieb:

What we can say is the following:

i) Every set is a class
ii) Not every class is a set

So there is a hypernym / hyponym relationship
between the two. Here is are proof of i) and ii):

Proof i): Let s be a set. Then we can form
the class { x | x e s }. So there is an injection
from the sets to the classes.

Proof ii): Let V be the class { x | true },
this is the universal class which is provably
not a set. So there is no surjection from
the sets to the classes.

Hope this helps. Injection is usually taken
as indicative that two sets are in the
less than or equal relation ship, i.e. ⊆.
And lack of surjection indicates that there
is no bijection, i.e. ≠, so we have:

Sets ⊆ Classes and Sets ≠ Classes

Or together:

Sets ⊂ Classes

The difference Class \ Sets, those things
that are classes but not sets, are called
proper classes.

Mild Shock schrieb:

I guess we have reached your intellectual
boundaries, inherent in your squirell brain
sized, that of a walnut, cerebrum and cerebrellum.

Mild Shock schrieb:

Doesn't make any sense at all.

Not a single mention of proper classes here:
https://plato.stanford.edu/ENTRIES/mereology/

Ross Finlayson schrieb:

Actually, for class/set distinction,
I just introduced set/part distinction,
and part/particle distinction,
and set/particle distinction.

set:class::part:particle

set:part::class:particle

This is a usual form that A:B::C:D is
that A relates to B as C relates to D,
"set is to class as part is to particle", and
"set is to part as class is to particle".

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

If you follow this thinking long enough,
you will proof Mückenheims identifiable numbers.
And suddently have finite ascent in mathematics.

LoL

Ross Finlayson schrieb:

I.e., the universe of mathematical objects
"is what it is", it's got numbers in it, and on it,
and counting is an act, two different things.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Wolfgang Mückenheim copy cats are rather boring.

Mild Shock schrieb:

If you follow this thinking long enough,
you will proof Mückenheims identifiable numbers.
And suddently have finite ascent in mathematics.

LoL

Ross Finlayson schrieb:

I.e., the universe of mathematical objects
"is what it is", it's got numbers in it, and on it,
and counting is an act, two different things.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 2/22/2024 3:03 AM, Mild Shock wrote:

Ross Finlayson schrieb:

[...]

Seriously, you don't know what classes are?

The membership relation is the same
for members of classes and for members of sets.
Since members of classes are sets just like

the members of sets are sets, in ZF. And there is
only one membership relation ∈ between sets. The
distinction between classes and sets was described

in the past as:

sets: includes collections of sizes from the numbers to
  the transfinite numbers
classes: includes collections that Cantor called
  NCONSISTENT MULTIPLICITIES

I once had a lecturer in math who would refer to
proper classes as "syntactic sugar",
which I found out today is a _programming_ term.

https://en.wikipedia.org/wiki/Syntactic_sugar
| In computer science, syntactic sugar is syntax within
| a programming language that is designed to make things
| easier to read or to express. It makes the language
| "sweeter" for human use: things can be expressed
| more clearly, more concisely, or in an alternative style
| that some may prefer.

I take that to mean that
| α ∈ ORD
is a "sweeter" way to say something like
| α is a regular transitive set of transitive sets
| Reg(α) ∧ Trans(α) ∧ ∀β ∈ α: Trans(β)

Reg(α) ⟺
α ≠ ∅ ⟹ ∃β ∈ α: β∩α = ∅

Trans(α) ⟺
∀β,γ: γ ∈ β ∧ β ∈ α ⟹ γ ∈ α


On the one hand,
questions surrounding the existence of ORD
have a lower bar to clear,
but
some things which a set can do
proper class ORD can't do.

Or that is my impression.
Caveat lector.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

This is the school that tries to avoid
higher order logic (HOL). And then sneaks in
classes as a kind of syntactic sugar into

FOL, with this rule:

x e { y | p(y) } <=> p(x)

If you are not wearing this chastity belt, its
just some HOL juggling. Like here:

Ross Finlayson schrieb:

https://leanprover-community.github.io/mathlib_docs/set_theory/zfc/basic.html#Class


But I suggest to study simple types first. And
then maybe dependent types. To understand the
type theoretic capture of set theory.

See also:

Should Type Theory Replace Set Theory as
the Foundation of Mathematics?

13 February 2023 - Thorsten Altenkirch https://link.springer.com/article/10.1007/s10516-023-09676-0

Jim Burns schrieb:

I once had a lecturer in math who would refer to
proper classes as "syntactic sugar",
which I found out today is a _programming_ term.

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Le 22/02/2024 à 17:13, Jim Burns a écrit :

On 2/22/2024 8:00 AM, WM wrote:

Le 21/02/2024 à 18:59, Jim Burns a écrit :

On 2/21/2024 3:33 AM, WM wrote:

Le 20/02/2024 à 23:02, Jim Burns a écrit :

Ordinals are well.ordered.

Only those which can be specified.

No.
All of them are well.ordered.

How do you know?

In the same way that I know
that right.triangles have three corners.

Yes, you are right. I exaggerated. Having three corners is essential for triangles. Being well-ordered is essential for ordinals. What I meant is
that we cannot follow the well-order into the dark realm. In particular
Peano ceases.

Regards, WM

--- SoupGate-Win32 v1.05
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Le 27/02/2024 à 20:25, Jim Burns a écrit :

On 2/23/2024 3:47 AM, WM wrote:

We know that Peano induction not.ceases in
the Peano (final) ordinals.

We know that every visible step is reversible.

Regardes, WM

--- SoupGate-Win32 v1.05
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On 2/23/2024 3:47 AM, WM wrote:

Le 22/02/2024 à 17:13, Jim Burns a écrit :

...]

Having three corners is essential for triangles.
Being well-ordered is essential for ordinals.
What I meant is that
we cannot follow the well-order into the dark realm.
In particular Peano ceases.

"Following the well.order" is a metaphor.

I am not a Form..
I've never met you. Still, I'm confident that
you aren't a Form, either.
We aren't in the same Realm as the ordinals.
Un.metaphorically, in our Realm,
we have no opportunity to follow or to not.follow
the well.order.

What I gather that you (WM) are asserting by metaphor
is a wrong assertion.

Having three corners is essential for triangles.
Being well-ordered is essential for ordinals.
What I meant is that
we cannot follow the well-order into the dark realm.
In particular Peano ceases.

We know that Peano induction not.ceases in
the Peano (final) ordinals.

It is a mistake to expect more than
Peano induction not.ceasing in
the Peano (final) ordinals.

We know that transfinite induction not.ceases in
the transfinite ordinals.

It is a mistake to expect more than
transfinite induction not.ceasing in
the transfinite ordinals.

We know that
being well.ordered not.ceases in
the transfinite ordinals.
Even beyond the first _inaccessible_ ordinal κ
being well.ordered not.ceases.

https://en.wikipedia.org/wiki/Inaccessible_cardinal
| In set theory, an uncountable cardinal is inaccessible
| if it cannot be obtained from smaller cardinals by
| the usual operations of cardinal arithmetic.

We know that because
transfinite induction is well.order in drag.
Thus, just like well.order,
transfinite induction not.ceases in the ordinals.

----
Well.ordering not.ceases in the ordinals.

If exists ordinal γ: p(γ)
then exists first ordinal β: p(β)

∃ᵒʳᵈγ:p(γ) ⟹ ∃#1ᵒʳᵈβ:p(β)

∃ᵒʳᵈγ:p(γ) ⟹ ∃ᵒʳᵈβ:(p(β) ∧ ¬∃ᵒʳᵈα<β:p(α))

¬∃ᵒʳᵈβ:(p(β) ∧ ¬∃ᵒʳᵈα<β:p(α)) ⟹ ¬∃ᵒʳᵈγ:p(γ)

∀ᵒʳᵈβ:(¬p(β) ∨ ¬∀ᵒʳᵈα<β:¬p(α)) ⟹ ∀ᵒʳᵈγ:¬p(γ)

∀ᵒʳᵈβ:(​̅p(β) ∨ ¬∀ᵒʳᵈα<β:​̅p(α)) ⟹ ∀ᵒʳᵈγ:​̅p(γ)

∀ᵒʳᵈβ:(∀ᵒʳᵈα<β:​̅p(α) ⇒ ​̅p(β)) ⟹ ∀ᵒʳᵈγ:​̅p(γ)

∀ᵒʳᵈβ:(​̅pᣔ[0,β) ⇒ ​̅pᣔ[0,β⁺¹)) ⟹ ∀ᵒʳᵈγ:​̅p(γ)

where
​̅pᣔ[0,β) ⟺ ∀ᵒʳᵈα<β:​̅p(α)
​̅pᣔ[0,β) ∧ ​̅p(β) ⟺ ​̅pᣔ[0,β⁺¹)

If, for each ordinal β
​̅p.before.β implies ​̅p.before.β⁺¹
then, for each ordinal γ ​̅p

Transfinite induction not.ceases in the ordinals.

--- SoupGate-Win32 v1.05
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On 2/27/2024 3:05 PM, WM wrote:

Le 27/02/2024 à 20:25, Jim Burns a écrit :

On 2/23/2024 3:47 AM, WM wrote:

We know that Peano induction not.ceases in
the Peano (final) ordinals.

We know that every visible step is reversible.

A one.ended ascent is reversible,
but is not a descent.


A one.ended ascent starts and not.stops.

A reverse one.ended ascent stops and not.starts.

A descent starts.

A reverse one.ended ascent isn't a descent.

----
No ordinal is first to start a one.ended descent.

No ordinal starts a one.ended descent.

----
ℕ​̲⇊ is the set of final ordinals.

For each ordinal ξ
{ξ+n: n ∈ ℕ​̲⇊} is
a darkᵂᴹfree one.ended ascent from ξ

--- SoupGate-Win32 v1.05
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Le 27/02/2024 à 23:24, Jim Burns a écrit :

On 2/27/2024 3:05 PM, WM wrote:

Le 27/02/2024 à 20:25, Jim Burns a écrit :

On 2/23/2024 3:47 AM, WM wrote:

We know that Peano induction not.ceases in
the Peano (final) ordinals.

We know that every visible step is reversible.

A one.ended ascent is reversible,
but is not a descent.

Every reversion of an ascent is a descent.

Regards, WM

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 2/28/2024 4:48 AM, WM wrote:

Le 27/02/2024 à 23:24, Jim Burns a écrit :

On 2/27/2024 3:05 PM, WM wrote:

Le 27/02/2024 à 20:25, Jim Burns a écrit :

On 2/23/2024 3:47 AM, WM wrote:

We know that Peano induction not.ceases in
the Peano (final) ordinals.

We know that every visible step is reversible.

A one.ended ascent is reversible,
but is not a descent.

Every reversion of an ascent is a descent.

A one.ended ascent starts and not.stops.

A reverse one.ended ascent stops and not.starts.

A descent starts.

A reverse one.ended ascent isn't a descent.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Le 28/02/2024 à 12:52, Jim Burns a écrit :

On 2/28/2024 4:48 AM, WM wrote:

Le 27/02/2024 à 23:24, Jim Burns a écrit :

On 2/27/2024 3:05 PM, WM wrote:

Le 27/02/2024 à 20:25, Jim Burns a écrit :

On 2/23/2024 3:47 AM, WM wrote:

We know that Peano induction not.ceases in
the Peano (final) ordinals.

We know that every visible step is reversible.

A one.ended ascent is reversible,
but is not a descent.

Every reversion of an ascent is a descent.

A one.ended ascent starts and not.stops.

As long as it runs through visible numbers it is finite and reversible.

Regards, WM

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 2/28/24 12:24 PM, WM wrote:

Le 28/02/2024 à 12:52, Jim Burns a écrit :

On 2/28/2024 4:48 AM, WM wrote:

Le 27/02/2024 à 23:24, Jim Burns a écrit :

On 2/27/2024 3:05 PM, WM wrote:

Le 27/02/2024 à 20:25, Jim Burns a écrit :

On 2/23/2024 3:47 AM, WM wrote:

We know that Peano induction not.ceases in
the Peano (final) ordinals.

We know that every visible step is reversible.

A one.ended ascent is reversible,
but is not a descent.

Every reversion of an ascent is a descent.

A one.ended ascent starts and not.stops.

As long as it runs through visible numbers it is finite and reversible.

Regards, WM

In other words, your definition of "Visible Numbers" are finite sub-sets
of the actual set of values.

And this is because your logic can only handle finite sets.

Your "Dark" numbers are just the numbers that you can not handle with
your restricted finite limited logic.

There is actually no problem with those numbers, except you incorrectly
limited logic can't deal with them.

They are a product of YOUR limitations, not of the inability of the
proper logic to deal with them.

--- SoupGate-Win32 v1.05
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Le 28/02/2024 à 23:07, Richard Damon a écrit :

On 2/28/24 12:24 PM, WM wrote:

Le 28/02/2024 à 12:52, Jim Burns a écrit :

On 2/28/2024 4:48 AM, WM wrote:

Le 27/02/2024 à 23:24, Jim Burns a écrit :

On 2/27/2024 3:05 PM, WM wrote:

Le 27/02/2024 à 20:25, Jim Burns a écrit :

On 2/23/2024 3:47 AM, WM wrote:

We know that Peano induction not.ceases in
the Peano (final) ordinals.

We know that every visible step is reversible.

A one.ended ascent is reversible,
but is not a descent.

Every reversion of an ascent is a descent.

A one.ended ascent starts and not.stops.

As long as it runs through visible numbers it is finite and reversible.

In other words, your definition of "Visible Numbers" are finite sub-sets
of the actual set of values.

Visible narural numbers are FISONs {1, 2, 3, ..., n}.

And this is because your logic can only handle finite sets.

This because there is no infinite natural number.

Your "Dark" numbers are

the only possibility to have completed infinity. Note: "Going on and on"
is not completed infinity but potential infinity.

There is actually no problem with those numbers,

What numbers? Infinite natural numbers?

Regards, WM

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 2/29/24 3:23 AM, WM wrote:

Le 28/02/2024 à 23:07, Richard Damon a écrit :

On 2/28/24 12:24 PM, WM wrote:

Le 28/02/2024 à 12:52, Jim Burns a écrit :

On 2/28/2024 4:48 AM, WM wrote:

Le 27/02/2024 à 23:24, Jim Burns a écrit :

On 2/27/2024 3:05 PM, WM wrote:

Le 27/02/2024 à 20:25, Jim Burns a écrit :

On 2/23/2024 3:47 AM, WM wrote:

We know that Peano induction not.ceases in
the Peano (final) ordinals.

We know that every visible step is reversible.

A one.ended ascent is reversible,
but is not a descent.

Every reversion of an ascent is a descent.

A one.ended ascent starts and not.stops.

As long as it runs through visible numbers it is finite and reversible.

In other words, your definition of "Visible Numbers" are finite
sub-sets of the actual set of values.

Visible narural numbers are FISONs {1, 2, 3, ..., n}.

And so SUBSETS of the Natural Number

And this is because your logic can only handle finite sets.

This because there is no infinite natural number.

No, but there are an infinte number of them.

You logic can't handle sets of infinite/unbounded size.

Your "Dark" numbers are

the only possibility to have completed infinity. Note: "Going on and on"
is not completed infinity but potential infinity.

Only because your logic can't handle it.

There is actually no problem with those numbers,

What numbers? Infinite natural numbers?

The unbounded set of Natural Numbers that go on and on and on.

Regards, WM

You are just proving your logic is broken and your brain is two sizes
too small for what you are trying to do.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Le 29/02/2024 à 13:35, Richard Damon a écrit :

On 2/29/24 3:23 AM, WM wrote:

Visible narural numbers are FISONs {1, 2, 3, ..., n}.

And so SUBSETS of the Natural Number

Of course. What else? Every natural number belongs to the set ℕ.

There is actually no problem with those numbers,

What numbers? Infinite natural numbers?

The unbounded set of Natural Numbers that go on and on and on.

Visible numbers do never complete infinity.

Regards, WM

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 2/29/24 2:25 PM, WM wrote:

Le 29/02/2024 à 13:35, Richard Damon a écrit :

On 2/29/24 3:23 AM, WM wrote:

Visible narural numbers are FISONs {1, 2, 3, ..., n}.

And so SUBSETS of the Natural Number

Of course. What else? Every natural number belongs to the set ℕ.

And evvery natural number is finite and thus namable and thus visible.

There is actually no problem with those numbers,

What numbers? Infinite natural numbers?

The unbounded set of Natural Numbers that go on and on and on.

Visible numbers do never complete infinity.

Right, but no individual Natural Number does either, so they all can be visible.

It is the SET that "completes infinity", and the set isn't any of the individual number.

Regards, WM

--- SoupGate-Win32 v1.05
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Le 01/03/2024 à 04:12, Richard Damon a écrit :

And evvery natural number is finite and thus namable and thus visible.

That concerns potential infinity only.

It is the SET that "completes infinity", and the set isn't any of the individual number.

The set is nothig but the collection of its elements. The complete set
requires that no element is missing. That proves, via ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0, the existence of a smallest unit fraction.

Regards, WM

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 3/1/24 3:47 AM, WM wrote:

Le 01/03/2024 à 04:12, Richard Damon a écrit :

And evvery natural number is finite and thus namable and thus visible.

That concerns potential infinity only.

No, it applies to ALL Finite numbers, which of course, being finte,
never actally REACH infinite, so if you want to invent a term for that
as "Potential Infinity", so be it.

But ALL

It is the SET that "completes infinity", and the set isn't any of the
individual number.

The set is nothig but the collection of its elements.

But it is, it is the COLLECTION of ALL its elements AT ONCE.

Thus, it converts the potential infinity of the individal Natural
Numbers, none of which are themselves infinite, into the ACTUAL infinity
of the set itself, and its size.

So, I guess your "dark" numbers, the ones that can only be talked about
as sets, are to collectively is the final collection (and its size).

Thus, your "NUF(x)" function is "dark" and you can't use its values individually, so NUF(x) = 1 is an invalid statement.

The complete set
requires that no element is missing. That proves, via ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0, the existence of a smallest unit fraction.

Why does it require a smallest unit fraction?

Your equation proves to opposite.

Given that for all n 1/n - 1/(n+1) > 0

we can say that for all n (by addin 1/(n+1)

1/n > 1/(n+1)

Since for all n, n+1 exists (from the definition of the natural numbers)

Thus, there can not be a smallest 1/n

Smallest here only applies to finite sub-sets, not the final actually
infinite set.

Regards, WM

--- SoupGate-Win32 v1.05
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Le 01/03/2024 à 15:44, Richard Damon a écrit :

On 3/1/24 3:47 AM, WM wrote:

Le 01/03/2024 à 04:12, Richard Damon a écrit :

And evvery natural number is finite and thus namable and thus visible.

That concerns potential infinity only.

No, it applies to ALL Finite numbers, which of course, being finte,
never actally REACH infinite, so if you want to invent a term for that
as "Potential Infinity", so be it.

But here we assume finished infinity.

It is the SET that "completes infinity", and the set isn't any of the
individual number.

The set is nothig but the collection of its elements.

But it is, it is the COLLECTION of ALL its elements AT ONCE.

There is nothing at once. Numbers are for counting one after the other.
Your claim shows that you need matheologial magic. But that's not maths.

Regards, WM

--- SoupGate-Win32 v1.05
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On 3/1/24 1:29 PM, WM wrote:

Le 01/03/2024 à 15:44, Richard Damon a écrit :

On 3/1/24 3:47 AM, WM wrote:

Le 01/03/2024 à 04:12, Richard Damon a écrit :

And evvery natural number is finite and thus namable and thus visible.

That concerns potential infinity only.

No, it applies to ALL Finite numbers, which of course, being finte,
never actally REACH infinite, so if you want to invent a term for that
as "Potential Infinity", so be it.

But here we assume finished infinity.

If we haven't finished the Infinity, you can't have you NUF(x), since it
starts at the infinite end.

It is the SET that "completes infinity", and the set isn't any of
the individual number.

The set is nothig but the collection of its elements.

But it is, it is the COLLECTION of ALL its elements AT ONCE.

There is nothing at once. Numbers are  for counting one after the other. Your claim shows that you need matheologial magic. But that's not maths.

And it seems your definition of "Maths" doesn't support the Natural Numbers.

But the actual Maths does, so you are using a defective version of logic
to think about it.

IF you can't support "All at Once" then you can't have the Set of
Natural Numbers to talk about, and thus we can't have your NUF.

You also can show that Achilles can't pass the Tortoise, as that
requires adding up the values "all at once" to let him catch up.

In other words, your logic is just insufficent for the task you are
using it for.

Regards, WM

--- SoupGate-Win32 v1.05
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Somehow I am quite new to the ordinal analysis
that Alan Turing started. Nice find Gödel introduces
a nice notion in his incompletness paper,

namely he says a function is primitive recursive,
to degree N, if the primitive recursion schema is
applied N times. So I guess this definition of

factorial would then have degree 3, since 'R' is used 3 times:

/* The R combinator */
natrec(_, 0, X, X) :- !. % attention, not steadfast
natrec(F, N, X, Z) :- M is N-1, natrec(F, M, X, Y), call(F, Y, Z).

plus(X, Y, Z) :- natrec(succ, X, Y, Z).

mult(X, Y, Z) :- natrec(plus(X), Y, 0, Z).

step((X,Y),(Z,T)) :- succ(X,Z), mult(Z,Y,T).

factorial(X,Y) :- natrec(step,X,(0,1),(_,Y)).

Works fine, although eats quite some computing resources,
i.e. 9_303_219 inferences, since it must form
3268800 successors:

?- factorial(10,X).
X = 3628800.

?- time(factorial(10,X)).
% 9,303,219 inferences, 0.578 CPU in 0.617 seconds
(94% CPU, 16092054 Lips)
X = 3628800.

Ross Finlayson schrieb:

... fourier my ass, what has it to do with ordinals ...

--- SoupGate-Win32 v1.05
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Know that one is the secret and source of all the cardinals.
-- Abraham ibn Ezra (1153)

But have mercy to me. So far I thought in ordinal
anlysis of programs, we would simply take the
tree of execution, and this is somehow an ordinal

for a terminating program? But whats the tree
repectively ordinal for for example 3! = 6 ?
Are all finite ordinals the same or not?

For example the ordinals 0, 1, 2, 3, with von
Neumann succeessor are:

0 = {}
1 = {{}}
2 = {{},{{}}}
3 = {{},{{}},{{{},{{}}}}}.

Or as trees, * = empty set, o = non-empty set:

0 = *

1 = o
|
*

2 = o
/ \
* o
|
*


3 = o
/ | \
* o o
| / \
* * o
|
*


But what about this tree, it has also no infinite decend,
but what property is missing to make it an ordinal?

? = o
/ | \
* o o
| |
* o
|
*

Or as a set:

? = {{},{{}},{{{}}}}.

Why is it not an ordinal?

P.S.: I tried to find an answer here, but I guess
I am too lazy to read it. Its starts with the funny
quote and has funny pictures in it:

Trees, ordinals and termination
N Dershowitz · 1993 https://link.springer.com/content/pdf/10.1007/3-540-56610-4_68.pdf

Mild Shock schrieb:

Somehow I am quite new to the ordinal analysis
that Alan Turing started. Nice find Gödel introduces
a nice notion in his incompletness paper,

namely he says a function is primitive recursive,
to degree N, if the primitive recursion schema is
applied N times. So I guess this definition of

factorial would then have degree 3, since 'R' is used 3 times:

/* The R combinator */
natrec(_, 0, X, X) :- !.    % attention, not steadfast
natrec(F, N, X, Z) :- M is N-1, natrec(F, M, X, Y), call(F, Y, Z).

plus(X, Y, Z) :- natrec(succ, X, Y, Z).

mult(X, Y, Z) :- natrec(plus(X), Y, 0, Z).

step((X,Y),(Z,T)) :- succ(X,Z), mult(Z,Y,T).

factorial(X,Y) :- natrec(step,X,(0,1),(_,Y)).

Works fine, although eats quite some computing resources,
i.e. 9_303_219 inferences, since it must form
3268800 successors:

?- factorial(10,X).
X = 3628800.

?- time(factorial(10,X)).
% 9,303,219 inferences, 0.578 CPU in 0.617 seconds
(94% CPU, 16092054 Lips)
X = 3628800.

Ross Finlayson schrieb:

... fourier my ass, what has it to do with ordinals ...

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

I think the key are these terminological definitions:

"In order theory, a partial order is called well-founded if the
corresponding strict order is a well-founded relation. If the order is a
total order then it is called a well-order." https://en.wikipedia.org/wiki/Well-founded_relation

So my tree "?" in question might have no infinite descend,
but it might not belong to the same total order, as the
other sets. But how exclude "?" ? The criteria of

transitive set is not violated:
trans(A) :<=> ∀ x , y : x ∈ A ∧ y ∈ x ⇒ y ∈ A https://de.wikipedia.org/wiki/Transitive_Menge

One can easily verify that the above is satisfied by
the set "?". So what is violated? Well this
here is violate, namely each elememt should be

transitive as well, and so on:

"hereditarily transitive sets"
h-trans(A) :<=> trans(A) & ∀x(x ∈ A => h-trans(A))

Because the third branch is not transitive:

? = o
/ | \
* o o
| |
* o
|
*

Or as a set:

? = {{},{{}},{{{}}}}.

So I remember Jim Burns when he posited a more
general approach, he said transfinite induction
must be satisfied.

Otherwise we can take this Quine atom x = {x}, https://math.stackexchange.com/a/2874533

And by a suitable interpretation of the circular
h-trans definition, a definition that is not well-defined
since it has no unique interpretation,

we might judge this Quine atom an ordinal.

LoL

Mild Shock schrieb:

Know that one is the secret and source of all the cardinals.
-- Abraham ibn Ezra (1153)

But have mercy to me. So far I thought in ordinal
anlysis of programs, we would simply take the
tree of execution, and this is somehow an ordinal

for a terminating program? But whats the tree
repectively ordinal for for example 3! = 6 ?
Are all finite ordinals the same or not?

For example the ordinals 0, 1, 2, 3, with von
Neumann succeessor are:

0 = {}
1 = {{}}
2 = {{},{{}}}
3 = {{},{{}},{{{},{{}}}}}.

Or as trees, * = empty set, o = non-empty set:

0 =            *

1 =            o
               |
               *

2 =            o
              / \
             *   o
                 |
                 *

3 =            o
             / |  \
           *   o    o
               |   / \
               *  *   o
                      |
                      *

But what about this tree, it has also no infinite decend,
but what property is missing to make it an ordinal?

? =             o
              / | \
             *  o  o
                |  |
                *  o
                   |
                   *

Or as a set:

? = {{},{{}},{{{}}}}.

Why is it not an ordinal?

P.S.: I tried to find an answer here, but I guess
I am too lazy to read it. Its starts with the funny
quote and has funny pictures in it:

Trees, ordinals and termination
N Dershowitz · 1993 https://link.springer.com/content/pdf/10.1007/3-540-56610-4_68.pdf

Mild Shock schrieb:

Somehow I am quite new to the ordinal analysis
that Alan Turing started. Nice find Gödel introduces
a nice notion in his incompletness paper,

namely he says a function is primitive recursive,
to degree N, if the primitive recursion schema is
applied N times. So I guess this definition of

factorial would then have degree 3, since 'R' is used 3 times:

/* The R combinator */
natrec(_, 0, X, X) :- !.    % attention, not steadfast
natrec(F, N, X, Z) :- M is N-1, natrec(F, M, X, Y), call(F, Y, Z).

plus(X, Y, Z) :- natrec(succ, X, Y, Z).

mult(X, Y, Z) :- natrec(plus(X), Y, 0, Z).

step((X,Y),(Z,T)) :- succ(X,Z), mult(Z,Y,T).

factorial(X,Y) :- natrec(step,X,(0,1),(_,Y)).

Works fine, although eats quite some computing resources,
i.e. 9_303_219 inferences, since it must form
3268800 successors:

?- factorial(10,X).
X = 3628800.

?- time(factorial(10,X)).
% 9,303,219 inferences, 0.578 CPU in 0.617 seconds
(94% CPU, 16092054 Lips)
X = 3628800.

Ross Finlayson schrieb:

... fourier my ass, what has it to do with ordinals ...

--- SoupGate-Win32 v1.05
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On 3/1/2024 3:08 PM, Mild Shock wrote:

So I remember Jim Burns when he posited a more
general approach, he said transfinite induction
must be satisfied.

In a phrase I'm a little proud of, I said
transfinite.induction is well.order in drag.
One is a simple re-write of the other.

Otherwise we can take this Quine atom x = {x}, https://math.stackexchange.com/a/2874533

And by a suitable interpretation of the circular
h-trans definition, a definition that is not well-defined
since it has no unique interpretation,

we might judge this Quine atom an ordinal.

LoL

Ah.
But an ordinal is
a _regular_ transitive set of transitive sets.
So, not x = {x}

A regular non.empty set A holds
a disjoint element B.
∃B ∈ A: A∩B = ∅

But, if A is transitive.transitive,
each element is a subset, and
∀B ∈ A: A∩B = B
Transitive.transitive A can only be regular
if one of its elements is 0

∅ ∈ A ∧ ∅ ∈ A′ ties all the ordinals together.

It's a beautiful thing.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Lets work without regularity axiom, and
examine this naive attempt, hereditary =
my ancestors satisfied it as well:

"hereditarily transitive sets"
h-trans(A) :<=> trans(A) & ∀x(x ∈ A => h-trans(A))

Otherwise when regularity is present,
this excludes Quine atom q = {q}. When regularty is
not present, we can prove:

~h-trans(q)


Jim Burns schrieb:

On 3/1/2024 3:08 PM, Mild Shock wrote:

So I remember Jim Burns when he posited a more
general approach, he said transfinite induction
must be satisfied.

In a phrase I'm a little proud of, I said
transfinite.induction is well.order in drag.
One is a simple re-write of the other.

Otherwise we can take this Quine atom x = {x},
https://math.stackexchange.com/a/2874533

And by a suitable interpretation of the circular
h-trans definition, a definition that is not well-defined
since it has no unique interpretation,

we might judge this Quine atom an ordinal.

LoL

Ah.
But an ordinal is
a _regular_ transitive set of transitive sets.
So, not x = {x}

A regular non.empty set A holds
a disjoint element B.
∃B ∈ A: A∩B = ∅

But, if A is transitive.transitive,
each element is a subset, and
∀B ∈ A: A∩B = B
Transitive.transitive A can only be regular
if one of its elements is 0

∅ ∈ A  ∧  ∅ ∈ A′ ties all the ordinals together.

It's a beautiful thing.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Corr.: we cannot prove:

~h-trans(q)

See also the remark here by Andrés E. Caicedo:

Note that in the absence of foundation (= regularity),
this is a bit peculiar. For instance, if x={x}, then x
is hereditarily finite, although it does not belong to Vω.) https://math.stackexchange.com/a/2874533

About digging into "transfinite.induction is well.order in drag"
by Jim Burns. You probably mean transfinite.induction
follows from well.order. What about the other direction?

Now my question, is assume we have no foundation,
but epsilon induction, what will happen. epsilon
induction is usually not an axiom. But what

if we stipulate it as an axiom?

Considered as an axiomatic principle, it is
called the axiom schema of set induction.
∀ x . ( ( ∀ ( y ∈ x ) . ψ ( y ) ) → ψ ( x ) ) → ∀ z . ψ ( z ) https://en.wikipedia.org/wiki/Epsilon-induction

Mild Shock schrieb:

Lets work without regularity axiom, and
examine this naive attempt, hereditary =
my ancestors satisfied it as well:

"hereditarily transitive sets"
h-trans(A) :<=> trans(A) & ∀x(x ∈ A => h-trans(A))

Otherwise when regularity is present,
this excludes Quine atom q = {q}. When regularty is
not present, we can prove:

~h-trans(q)

Jim Burns schrieb:

On 3/1/2024 3:08 PM, Mild Shock wrote:

So I remember Jim Burns when he posited a more
general approach, he said transfinite induction
must be satisfied.

In a phrase I'm a little proud of, I said
transfinite.induction is well.order in drag.
One is a simple re-write of the other.

Otherwise we can take this Quine atom x = {x},
https://math.stackexchange.com/a/2874533

And by a suitable interpretation of the circular
h-trans definition, a definition that is not well-defined
since it has no unique interpretation,

we might judge this Quine atom an ordinal.

LoL

Ah.
But an ordinal is
a _regular_ transitive set of transitive sets.
So, not x = {x}

A regular non.empty set A holds
a disjoint element B.
∃B ∈ A: A∩B = ∅

But, if A is transitive.transitive,
each element is a subset, and
∀B ∈ A: A∩B = B
Transitive.transitive A can only be regular
if one of its elements is 0

∅ ∈ A  ∧  ∅ ∈ A′ ties all the ordinals together.

It's a beautiful thing.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

My numb nut rewriting faculty, after spying
wikipeda, takes the contrapositive (i.e replace

B by ~B -> ~A) of the set induction axiom:

¬ ∀ z ψ ( z ) → ¬ ∀ x ( ( ∀ y ( y ∈ x → ψ ( y ) ) → ψ ( x ) )

Now use for ψ ( z ) the formula ¬ z ∈ u, and one gets:

∃ z z ∈ u → ∃ x ¬ ( ∀ y ( y ∈ x → ¬ y ∈ u ) → ¬ x ∈ u )

Again some contrapositive:

∃ z z ∈ u → ∃ x ¬ ( x ∈ u → ¬ ∀ y ( y ∈ x → ¬ y ∈ u ))

And hence:

∃ z z ∈ u → ∃ x ( x ∈ u ∧ ∀ y ( y ∈ x → ¬ y ∈ u ))

Some last quantifier switch, and we got the regularity axiom:

∃ z z ∈ u → ∃ x ( x ∈ u ∧ ∃ y ( y ∈ x ∧ y ∈ u ))

Usually written as:

u ≠ ∅ → ∃ x (x ∈ u ∧ x ∩ u ≠ ∅)

Mild Shock schrieb:

Corr.: we cannot prove:

~h-trans(q)

See also the remark here by Andrés E. Caicedo:

Note that in the absence of foundation (= regularity),
this is a bit peculiar. For instance, if x={x}, then x
is hereditarily finite, although it does not belong to Vω.) https://math.stackexchange.com/a/2874533

About digging into "transfinite.induction is well.order in drag"
by Jim Burns. You probably mean transfinite.induction
follows from well.order. What about the other direction?

Now my question, is assume we have no foundation,
but epsilon induction, what will happen. epsilon
induction is usually not an axiom. But what

if we stipulate it as an axiom?

Considered as an axiomatic principle, it is
called the axiom schema of set induction.
∀ x . ( ( ∀ ( y ∈ x ) . ψ ( y ) ) → ψ ( x ) ) → ∀ z . ψ ( z ) https://en.wikipedia.org/wiki/Epsilon-induction

Mild Shock schrieb:

Lets work without regularity axiom, and
examine this naive attempt, hereditary =
my ancestors satisfied it as well:

"hereditarily transitive sets"
h-trans(A) :<=> trans(A) & ∀x(x ∈ A => h-trans(A))

Otherwise when regularity is present,
this excludes Quine atom q = {q}. When regularty is
not present, we can prove:

~h-trans(q)


Jim Burns schrieb:

On 3/1/2024 3:08 PM, Mild Shock wrote:

So I remember Jim Burns when he posited a more
general approach, he said transfinite induction
must be satisfied.

In a phrase I'm a little proud of, I said
transfinite.induction is well.order in drag.
One is a simple re-write of the other.

Otherwise we can take this Quine atom x = {x},
https://math.stackexchange.com/a/2874533

And by a suitable interpretation of the circular
h-trans definition, a definition that is not well-defined
since it has no unique interpretation,

we might judge this Quine atom an ordinal.

LoL

Ah.
But an ordinal is
a _regular_ transitive set of transitive sets.
So, not x = {x}

A regular non.empty set A holds
a disjoint element B.
∃B ∈ A: A∩B = ∅

But, if A is transitive.transitive,
each element is a subset, and
∀B ∈ A: A∩B = B
Transitive.transitive A can only be regular
if one of its elements is 0

∅ ∈ A  ∧  ∅ ∈ A′ ties all the ordinals together.

It's a beautiful thing.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 3/1/2024 5:11 PM, Mild Shock wrote:

About digging into
"transfinite.induction is well.order in drag"
by Jim Burns. You probably mean
transfinite.induction follows from well.order.
What about the other direction?

Both directions.

well.order

∃ᵒʳᵈγ:p(γ) ⟹ ∃#1ᵒʳᵈβ:p(β)

∃ᵒʳᵈγ:p(γ) ⟹ ∃ᵒʳᵈβ:(p(β) ∧ ¬∃ᵒʳᵈα<β:p(α))

¬∃ᵒʳᵈβ:(p(β) ∧ ¬∃ᵒʳᵈα<β:p(α)) ⟹ ¬∃ᵒʳᵈγ:p(γ)

∀ᵒʳᵈβ:(¬p(β) ∨ ¬∀ᵒʳᵈα<β:¬p(α)) ⟹ ∀ᵒʳᵈγ:¬p(γ)

∀ᵒʳᵈβ:(​̅p(β) ∨ ¬∀ᵒʳᵈα<β:​̅p(α)) ⟹ ∀ᵒʳᵈγ:​̅p(γ)

∀ᵒʳᵈβ:(∀ᵒʳᵈα<β:​̅p(α) ⇒ ​̅p(β)) ⟹ ∀ᵒʳᵈγ:​̅p(γ)

∀ᵒʳᵈβ:(​̅p[0,β) ⇒ ​̅p(β)) ⟹ ∀ᵒʳᵈγ:​̅p(γ)

∀ᵒʳᵈβ:(​̅p[0,β) ⇒ ​̅p[0,β)∧​̅p(β)) ⟹ ∀ᵒʳᵈγ:​̅p(γ)

∀ᵒʳᵈβ:(​̅p[0,β) ⇒ ​̅p[0,β⁺¹)) ⟹ ∀ᵒʳᵈγ:​̅p(γ)

transfinite.induction

Now my question, is assume we have no foundation,
but epsilon induction, what will happen. epsilon
induction is usually not an axiom. But what
if we stipulate it as an axiom?

Considered as an axiomatic principle, it is
called the axiom schema of set induction. ∀x.((∀(y∈x).ψ(y))→ψ(x))→∀z.ψ(z) https://en.wikipedia.org/wiki/Epsilon-induction

That wiki.page assumes regular sets.
I'm not sure it does so explicitly,
but it defines ordinals as
transitive sets of transitive sets
which, without regularity, include x = {x}

--- SoupGate-Win32 v1.05
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Le 01/03/2024 à 20:00, Richard Damon a écrit :

On 3/1/24 1:29 PM, WM wrote:

Le 01/03/2024 à 15:44, Richard Damon a écrit :

On 3/1/24 3:47 AM, WM wrote:

Le 01/03/2024 à 04:12, Richard Damon a écrit :

And evvery natural number is finite and thus namable and thus visible. >>>>

That concerns potential infinity only.

No, it applies to ALL Finite numbers, which of course, being finte,
never actally REACH infinite, so if you want to invent a term for that
as "Potential Infinity", so be it.

But here we assume finished infinity.

If we haven't finished the Infinity, you can't have you NUF(x), since it starts at the infinite end.

True. NUF requires completed infinity. That is the premise.

IF you can't support "All at Once" then you can't have the Set of
Natural Numbers to talk about, and thus we can't have your NUF.

You also can show that Achilles can't pass the Tortoise, as that
requires adding up the values "all at once" to let him catch up.

Every step in half time is enough.

Regards, WM

--- SoupGate-Win32 v1.05
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On 3/2/24 7:36 AM, WM wrote:

Le 01/03/2024 à 20:00, Richard Damon a écrit :

On 3/1/24 1:29 PM, WM wrote:

Le 01/03/2024 à 15:44, Richard Damon a écrit :

On 3/1/24 3:47 AM, WM wrote:

Le 01/03/2024 à 04:12, Richard Damon a écrit :

And evvery natural number is finite and thus namable and thus
visible.

That concerns potential infinity only.

No, it applies to ALL Finite numbers, which of course, being finte,
never actally REACH infinite, so if you want to invent a term for
that as "Potential Infinity", so be it.

But here we assume finished infinity.

If we haven't finished the Infinity, you can't have you NUF(x), since
it starts at the infinite end.

True. NUF requires completed infinity. That is the premise.

And thus must use a logic that ALLOWS for a completed infinity.

One at a time only doesn't.

IF you can't support "All at Once" then you can't have the Set of
Natural Numbers to talk about, and thus we can't have your NUF.

You also can show that Achilles can't pass the Tortoise, as that
requires adding up the values "all at once" to let him catch up.

Every step in half time is enough.

I guess you don't understand the problem.

Regards, WM

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Le 02/03/2024 à 15:22, Richard Damon a écrit :

On 3/2/24 7:36 AM, WM wrote:

Le 01/03/2024 à 20:00, Richard Damon a écrit :

NUF requires completed infinity. That is the premise.

And thus must use a logic that ALLOWS for a completed infinity.

Obviously there is none.

One at a time only doesn't.

One at a time only is the basis of counting. Using half the time for every
step would be possible. Otherwise there is no provable bijection. Only if
each desired step (not all) can be verified, it is matematics.

Regards, WM

--- SoupGate-Win32 v1.05
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Jim Burns schrieb:

Considered as an axiomatic principle, it is
called the axiom schema of set induction.
∀x.((∀(y∈x).ψ(y))→ψ(x))→∀z.ψ(z)
https://en.wikipedia.org/wiki/Epsilon-induction

That wiki.page assumes regular sets.

Well I showed classicaly that the set induction
axiom schema implies the regularity axiom:

Mild Shock schrieb:

My numb nut rewriting faculty, after spying

/* set induction */
¬ ∀ z ψ ( z ) → ¬ ∀ x ( ( ∀ y ( y ∈ x → ψ ( y ) ) → ψ ( x ) )

|
|
v

/* set induction */
u ≠ ∅ → ∃ x (x ∈ u ∧ x ∩ u ≠ ∅)

The only problem with this derivation, it
might not be intuitionistically valid. I might
have used a propositional or quantifier rules,

which are not accepted intuitionistically.

--- SoupGate-Win32 v1.05
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