Replacement of Cardinality

[joes]

It is strange that blatantly false results as the equinumerosity of
prime numbers and algebraic numbers could capture mathematics and stay
there for over a century. But by what meaningful mathematics can we
replace Cantor's wrong bijection rules?

Juste because it doesn't match your intuition doesn't mean it's not
useful.

Not all infinite sets can be compared by size, but we can establish some
useful rules

that you would like instead.

_The rule of subset_ proves that every proper subset has less elements
than its superset. So there are more natural numbers than prime numbers,
|ℕ| > |P|, and more complex numbers than real numbers. Even finitely
many exceptions from the subset-relation are admitted for infinite
subsets. Therefore there are more odd numbers than prime numbers.

What exceptions do you mean?
This immediately creates as many sizes as there are naturals, one for
each of your endsegments.

_The rule of construction_ yields the numbers of integers |Z| = 2|ℕ| + 1
and of fractions |Q| = 2|ℕ|^2 + 1 (there are less rational numbers).
Since all products of rational numbers with an irrational number are
irrational, there are many more irrational numbers than rational
numbers.

Only this leads to some contradictions depending on the construction.

_The rule of symmetry_ yields precisely the same number of reals in
every interval (n, n+1] and with at most a small error same number of
odd numbers and of even numbers in every finite interval and in the
whole real line.

How small an error?

[joes]

By your logic, if you take a set and replace every element with a
number that is twice that value, it would by the rule of construction
say they must be the same size.

That is true in potential infinity. But I assume actual infinity.

So, what part is not true?

In potential infinity there is no ω.

Neither is there in actual infinity.

Are you stating that replacing every element with another unique
distinct element something that make the set change size?

Yes they are.

In actual infinity the number of elements of any infinite set is fixed.
Doubling all elements of the set ℕ U ω = {2, 4, 6, ..., ω} yields the
set {2, 4, 6, ..., ω, ω+2, ω+4, ..., ω*2}.

I wonder how you get the second infinity. What is the preimage of all
the omegas?

[WM]

Post by joes

By your logic, if you take a set and replace every element with a
number that is twice that value, it would by the rule of construction
say they must be the same size.

That is true in potential infinity. But I assume actual infinity.

So, what part is not true?

In potential infinity there is no ω.

Neither is there in actual infinity.

That is a solitary opinion.

Post by joes

Are you stating that replacing every element with another unique
distinct element something that make the set change size?

Yes they are.

In actual infinity the number of elements of any infinite set is fixed.
Doubling all elements of the set ℕ U ω = {2, 4, 6, ..., ω} yields the

Here I made a mistake: ℕ U ω = {1, 2, 3, ..., ω}

Post by joes

set {2, 4, 6, ..., ω, ω+2, ω+4, ..., ω*2}.

I wonder how you get the second infinity. What is the preimage of all
the omegas?

See my correction.

Regards, WM

[Jim Burns]

sci.logic and sci.math

_The rule of subset_ proves that
every proper subset has
less elements than its superset.

⎛ Each non.{}.set A of ordinals holds min.A
⎜
⎜ Ordinal j = {i:i<j} set of ordinals before j
⎜
⎜ Finite ordinal j has fewer elements than j∪{j}
⎜
⎝ ℕⁿᵒᵗᐧᵂᴹ is the set of ALL finite ordinals.

ℕⁿᵒᵗᐧᵂᴹ breaks the rule of subset.

If ℕ has fewer elements than ℕ∪{ℕ}
then
|ℕ| ∈ ℕ
|ℕ∪{ℕ}| ∈ ℕ
ℕ ≠⊂ {i:i<ℕ∪{ℕ}} ⊂ ℕ
ℕ ≠⊂ ℕ
and
ℕ has fewer elements than ℕ

Because ℕ does not have fewer elements than ℕ
ℕ does not have fewer elements than ℕ∪{ℕ}
and the rule of subsets is broken.

_The rule of subset_ proves that

To make a claim
is not sufficient
to make a proof.

To make a finite sequence of claims
such that no claim is first.false
is sufficient
to make a proof.

The most that is true here is that
the rule of subset _claims_ without proof that

every proper subset has
less elements than its superset.

⎛ In English, grammatically speaking,
⎜ it is never correct to say "less <plural.noun>"
⎜
⎜ English has mass nouns (Stoffnamen)
⎜ "less rock" ...
⎜ and count nouns (zählbare Substantive)
⎜ "one rock", "fewer rocks" ...
⎜ Only count nouns have a plural.
⎜ Only mass nouns are modified by "less".
⎝ "Less rocks" and "lescs elements" are never correct.

⎛ ...as you (WM) know, since you've been told.
⎜ Why you say what you say is often more mysterious
⎜ than what you say.
⎜
⎜ Perhaps you see a mass noun as more appropriate
⎜ to your darkᵂᴹ numbers -- to number purée?
⎜ That's not the way it would be said.
⎜
⎜⎛ The smaller bucket holds fewer rocks and less rock.
⎜⎝ Every proper subset has less element than its superset.
⎜
⎜ Perhaps you intentionally insert small errors
⎜ in order to draw attention away from large errors:
⎜ dropping verbal chaff.
⎜ https://en.wikipedia.org/wiki/Chaff_(countermeasure)
⎜
⎝ Perhaps looking ignorant is your kink.

[WM]

Post by Jim Burns
If ℕ has fewer elements than ℕ∪{ℕ}
then
|ℕ| ∈ ℕ

|ℕ| = ω-1 ∈ ℕ

Post by Jim Burns
ℕ has fewer elements than ℕ

ℕ has ω-1 elements.

Post by Jim Burns
Because ℕ does not have fewer elements than ℕ
ℕ does not have fewer elements than ℕ∪{ℕ}
and the rule of subsets is broken.

ℕ = {1, 2, 3, ..., ω-1} = {1, 2, 3, ..., |ℕ|}

Post by Jim Burns

_The rule of subset_ proves that

To make a claim
is not sufficient
to make a proof.
To make a finite sequence of claims
such that no claim is first.false
is sufficient
to make a proof.

First false is your claim that |ℕ| is larger than all elements of ℕ.
ℕ counts its elements.

Post by Jim Burns
The most that is true here is that
the rule of subset _claims_ without proof that

every proper subset has
less elements than its superset.

The proof is easy. Since the superset has at least one more element than
its proper subset, it has more elements than its proper subset.

Post by Jim Burns
⎛ In English, grammatically speaking,
⎜ it is never correct to say "less <plural.noun>"
⎜
⎜ English has mass nouns (Stoffnamen)
⎜ "less rock" ...
⎜ and count nouns (zählbare Substantive)
⎜ "one rock", "fewer rocks" ...
⎜ Only count nouns have a plural.
⎜ Only mass nouns are modified by "less".
⎝ "Less rocks" and "lescs elements" are never correct.

Thank you, I will try to remember it.

Regards, WM

[Jim Burns]

Post by WM

Post by Jim Burns

_The rule of subset_ proves that
every proper subset has less elements than its superset.

If ℕ has fewer elements than ℕ∪{ℕ}
then
|ℕ| ∈ ℕ

|ℕ| = ω-1 ∈ ℕ

⎛ Each non.{}.set A of ordinals holds min.A
⎜
⎜ Ordinal j = {i:i<j} set of ordinals before j
⎜
⎜ Finite ordinal j has fewer elements than j∪{j}
⎜
⎝ ℕⁿᵒᵗᐧᵂᴹ is the set of ALL finite ordinals.

No finite.ordinal is last.finite,
no visibleᵂᴹ finite.ordinal,
no darkᵂᴹ finite.ordinal.
In particular, no finite.ordinal is ω-1

Also, no before.first infinite.ordinal is
before the first infinite.ordinal ω
In particular, no infinite.ordinal is ω-1

----
Consider ordinals i j k such that
i∪{i} = j and j∪{j} = k

Obviously, their order is i < j < k

Either they're all finite
|i| < |j| < |k|
or they're all infinite
|i| = |j| = |k|

No finite.to.infinite step exists.
no visibleᵂᴹ finite.to.infinite step,
no darkᵂᴹ finite.to.infinite step.

Defining declares the meaning of one's words.
'Defining into existence' that which doesn't exist
makes nonsense of whatever meaning one's words have.

⎛ if
⎜ g: j∪{j}→i∪{i}: 1.to.1
⎜ then
⎜ f(x) := (g(x)=i ? g(j) : g(x))
⎜ (Perl ternary conditional operator)
⎜ f: j→i: 1.to.1
⎜
⎜ if
⎜ f: j→i: 1.to.1
⎜ then
⎜ g(x) := (x=j ? i : f(x))
⎝ g: j∪{j}→i∪{i}: 1.to.1

Therefore,
i has fewer than j iff j has fewer than k

Post by WM

Post by Jim Burns
ℕ has fewer elements than ℕ

ℕ has ω-1 elements.

ℕⁿᵒᵗᐧᵂᴹ holds all finite ordinals.

Finite doesn't need to be small.
ℕⁿᵒᵗᐧᵂᴹ holds ordinals which
are big compared to Avogadroᴬᵛᵒᵍᵃᵈʳᵒ,
but those big ordinals have an immediate predecessor,
and each non.0.ordinal before them has
an immediate predecessor.
That makes them finite, but not necessarily small.

Post by WM

Post by Jim Burns
Because ℕ does not have fewer elements than ℕ
ℕ does not have fewer elements than ℕ∪{ℕ}
and the rule of subsets is broken.

ℕ = {1, 2, 3, ..., ω-1} = {1, 2, 3, ..., |ℕ|}

∀j ∈ ℕⁿᵒᵗᐧᵂᴹ:
∃k ∈ ℕⁿᵒᵗᐧᵂᴹ\{0}:
k = j+1 ∧ ¬∃kₓ≠k: kₓ=j+1

'+1': ℕⁿᵒᵗᐧᵂᴹ→ℕⁿᵒᵗᐧᵂᴹ\{0}: 1.to.1
and the rule of subset is broken.

[Ross Finlayson]

Post by Jim Burns

Post by WM

Post by Jim Burns

_The rule of subset_ proves that
every proper subset has less elements than its superset.

If ℕ has fewer elements than ℕ∪{ℕ}
then
|ℕ| ∈ ℕ

|ℕ| = ω-1 ∈ ℕ

⎛ Each non.{}.set A of ordinals holds min.A
⎜
⎜ Ordinal j = {i:i<j} set of ordinals before j
⎜
⎜ Finite ordinal j has fewer elements than j∪{j}
⎜
⎝ ℕⁿᵒᵗᐧᵂᴹ is the set of ALL finite ordinals.
No finite.ordinal is last.finite,
no visibleᵂᴹ finite.ordinal,
no darkᵂᴹ finite.ordinal.
In particular, no finite.ordinal is ω-1
Also, no before.first infinite.ordinal is
before the first infinite.ordinal ω
In particular, no infinite.ordinal is ω-1
----
Consider ordinals i j k such that
i∪{i} = j and j∪{j} = k
Obviously, their order is i < j < k
Either they're all finite
|i| < |j| < |k|
or they're all infinite
|i| = |j| = |k|
No finite.to.infinite step exists.
no visibleᵂᴹ finite.to.infinite step,
no darkᵂᴹ finite.to.infinite step.
Defining declares the meaning of one's words.
'Defining into existence' that which doesn't exist
makes nonsense of whatever meaning one's words have.
⎛ if
⎜ g: j∪{j}→i∪{i}: 1.to.1
⎜ then
⎜ f(x) := (g(x)=i ? g(j) : g(x))
⎜ (Perl ternary conditional operator)
⎜ f: j→i: 1.to.1
⎜
⎜ if
⎜ f: j→i: 1.to.1
⎜ then
⎜ g(x) := (x=j ? i : f(x))
⎝ g: j∪{j}→i∪{i}: 1.to.1
Therefore,
i has fewer than j iff j has fewer than k

Post by WM

Post by Jim Burns
ℕ has fewer elements than ℕ

ℕ has ω-1 elements.

ℕⁿᵒᵗᐧᵂᴹ holds all finite ordinals.
Finite doesn't need to be small.
ℕⁿᵒᵗᐧᵂᴹ holds ordinals which
are big compared to Avogadroᴬᵛᵒᵍᵃᵈʳᵒ,
but those big ordinals have an immediate predecessor,
and each non.0.ordinal before them has
an immediate predecessor.
That makes them finite, but not necessarily small.

Post by WM

Post by Jim Burns
Because ℕ does not have fewer elements than ℕ
ℕ does not have fewer elements than ℕ∪{ℕ}
and the rule of subsets is broken.

ℕ = {1, 2, 3, ..., ω-1} = {1, 2, 3, ..., |ℕ|}

k = j+1 ∧ ¬∃kₓ≠k: kₓ=j+1
'+1': ℕⁿᵒᵗᐧᵂᴹ→ℕⁿᵒᵗᐧᵂᴹ\{0}: 1.to.1
and the rule of subset is broken.

That's, ..., nice and all, yet, are you,
"preaching to the choir", or,
"reaching to the higher", the higher ground.

I.e., here it's not saying much.

Where's the "extra"-ordinary.

It's a matter of deductive inference there is one,
while the naive nicely arrives at it directly.

[Ross Finlayson]

Post by Ross Finlayson

Post by Jim Burns

Post by WM

Post by Jim Burns

_The rule of subset_ proves that
every proper subset has less elements than its superset.

If ℕ has fewer elements than ℕ∪{ℕ}
then
|ℕ| ∈ ℕ

|ℕ| = ω-1 ∈ ℕ

⎛ Each non.{}.set A of ordinals holds min.A
⎜
⎜ Ordinal j = {i:i<j} set of ordinals before j
⎜
⎜ Finite ordinal j has fewer elements than j∪{j}
⎜
⎝ ℕⁿᵒᵗᐧᵂᴹ is the set of ALL finite ordinals.
No finite.ordinal is last.finite,
no visibleᵂᴹ finite.ordinal,
no darkᵂᴹ finite.ordinal.
In particular, no finite.ordinal is ω-1
Also, no before.first infinite.ordinal is
before the first infinite.ordinal ω
In particular, no infinite.ordinal is ω-1
----
Consider ordinals i j k such that
i∪{i} = j and j∪{j} = k
Obviously, their order is i < j < k
Either they're all finite
|i| < |j| < |k|
or they're all infinite
|i| = |j| = |k|
No finite.to.infinite step exists.
no visibleᵂᴹ finite.to.infinite step,
no darkᵂᴹ finite.to.infinite step.
Defining declares the meaning of one's words.
'Defining into existence' that which doesn't exist
makes nonsense of whatever meaning one's words have.
⎛ if
⎜ g: j∪{j}→i∪{i}: 1.to.1
⎜ then
⎜ f(x) := (g(x)=i ? g(j) : g(x))
⎜ (Perl ternary conditional operator)
⎜ f: j→i: 1.to.1
⎜
⎜ if
⎜ f: j→i: 1.to.1
⎜ then
⎜ g(x) := (x=j ? i : f(x))
⎝ g: j∪{j}→i∪{i}: 1.to.1
Therefore,
i has fewer than j iff j has fewer than k

Post by WM

Post by Jim Burns
ℕ has fewer elements than ℕ

ℕ has ω-1 elements.

ℕⁿᵒᵗᐧᵂᴹ holds all finite ordinals.
Finite doesn't need to be small.
ℕⁿᵒᵗᐧᵂᴹ holds ordinals which
are big compared to Avogadroᴬᵛᵒᵍᵃᵈʳᵒ,
but those big ordinals have an immediate predecessor,
and each non.0.ordinal before them has
an immediate predecessor.
That makes them finite, but not necessarily small.

Post by WM

Post by Jim Burns
Because ℕ does not have fewer elements than ℕ
ℕ does not have fewer elements than ℕ∪{ℕ}
and the rule of subsets is broken.

ℕ = {1, 2, 3, ..., ω-1} = {1, 2, 3, ..., |ℕ|}

k = j+1 ∧ ¬∃kₓ≠k: kₓ=j+1
'+1': ℕⁿᵒᵗᐧᵂᴹ→ℕⁿᵒᵗᐧᵂᴹ\{0}: 1.to.1
and the rule of subset is broken.

That's, ..., nice and all, yet, are you,
"preaching to the choir", or,
"reaching to the higher", the higher ground.
I.e., here it's not saying much.
Where's the "extra"-ordinary.
It's a matter of deductive inference there is one,
while the naive nicely arrives at it directly.

Foundations is more than a field.

[Ross Finlayson]

Post by Ross Finlayson

Post by Ross Finlayson

Post by Jim Burns

Post by WM

Post by Jim Burns

_The rule of subset_ proves that
every proper subset has less elements than its superset.

If ℕ has fewer elements than ℕ∪{ℕ}
then
|ℕ| ∈ ℕ

|ℕ| = ω-1 ∈ ℕ

⎛ Each non.{}.set A of ordinals holds min.A
⎜
⎜ Ordinal j = {i:i<j} set of ordinals before j
⎜
⎜ Finite ordinal j has fewer elements than j∪{j}
⎜
⎝ ℕⁿᵒᵗᐧᵂᴹ is the set of ALL finite ordinals.
No finite.ordinal is last.finite,
no visibleᵂᴹ finite.ordinal,
no darkᵂᴹ finite.ordinal.
In particular, no finite.ordinal is ω-1
Also, no before.first infinite.ordinal is
before the first infinite.ordinal ω
In particular, no infinite.ordinal is ω-1
----
Consider ordinals i j k such that
i∪{i} = j and j∪{j} = k
Obviously, their order is i < j < k
Either they're all finite
|i| < |j| < |k|
or they're all infinite
|i| = |j| = |k|
No finite.to.infinite step exists.
no visibleᵂᴹ finite.to.infinite step,
no darkᵂᴹ finite.to.infinite step.
Defining declares the meaning of one's words.
'Defining into existence' that which doesn't exist
makes nonsense of whatever meaning one's words have.
⎛ if
⎜ g: j∪{j}→i∪{i}: 1.to.1
⎜ then
⎜ f(x) := (g(x)=i ? g(j) : g(x))
⎜ (Perl ternary conditional operator)
⎜ f: j→i: 1.to.1
⎜
⎜ if
⎜ f: j→i: 1.to.1
⎜ then
⎜ g(x) := (x=j ? i : f(x))
⎝ g: j∪{j}→i∪{i}: 1.to.1
Therefore,
i has fewer than j iff j has fewer than k

Post by WM

Post by Jim Burns
ℕ has fewer elements than ℕ

ℕ has ω-1 elements.

ℕⁿᵒᵗᐧᵂᴹ holds all finite ordinals.
Finite doesn't need to be small.
ℕⁿᵒᵗᐧᵂᴹ holds ordinals which
are big compared to Avogadroᴬᵛᵒᵍᵃᵈʳᵒ,
but those big ordinals have an immediate predecessor,
and each non.0.ordinal before them has
an immediate predecessor.
That makes them finite, but not necessarily small.

Post by WM

Post by Jim Burns
Because ℕ does not have fewer elements than ℕ
ℕ does not have fewer elements than ℕ∪{ℕ}
and the rule of subsets is broken.

ℕ = {1, 2, 3, ..., ω-1} = {1, 2, 3, ..., |ℕ|}

k = j+1 ∧ ¬∃kₓ≠k: kₓ=j+1
'+1': ℕⁿᵒᵗᐧᵂᴹ→ℕⁿᵒᵗᐧᵂᴹ\{0}: 1.to.1
and the rule of subset is broken.

That's, ..., nice and all, yet, are you,
"preaching to the choir", or,
"reaching to the higher", the higher ground.
I.e., here it's not saying much.
Where's the "extra"-ordinary.
It's a matter of deductive inference there is one,
while the naive nicely arrives at it directly.

Foundations is more than a field.

Now, if there is something as relevant as Cardinality,
as primary, for mathematical foundations, it's: Continuity,
that Continuity, is so essentially primary, fundamental,
central, and ubiquitous, makes for the Cardinality as
next to Ordinality for counting vis-a-vis Numbering,
in where there are various (and perhaps, nowhere only
"standard") models of integers, where Cohen for the
Independence of the Continuum Hypothesis in Cardinals
makes an extra-ordinary bit of model there courtesy
a pretty simple induction about Ordinals vis-a-vis Cardinals
in a theory with numbering vis-a-vis counting that there
is: the extra-ordinary, about ubiquitous ordinals
in any old theory.

That there is one at all, ....

[Jim Burns]

Post by Ross Finlayson

[...]

[...]

[...]

about ubiquitous ordinals

What are ubiquitous ordinal?

[Ross Finlayson]

Post by Jim Burns

Post by Ross Finlayson

[...]

[...]

[...]

about ubiquitous ordinals

What are ubiquitous ordinal?

Well, you know that ORD, is, the order type of ordinals,
and so it's an ordinal, of all the ordinals.

The "ubiquitous ordinals", sort of recalls Kronecker's
"G-d made the integers, the rest is the work of Man",
that the Integer Continuum, is the model and ground
model, of any sort of language of finite words,
like set theory.

It's like the universe of set theory, then as that
there's _always_ an arithmetization, or as with regards
to ordering and numbering as a bit weaker property than
collecting and counting, so that "ubiquitous ordinals"
is what you get from a discrete world.

Then there's that according to the set-theoretic Powerset
theorem of Cantor, that when the putative function is
successor, in ubiquitous ordinals where order type is
powerset is successor, then there's no missing element.

So, "ubiquitous ordinals" is exactly what it says.

[Ross Finlayson]

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

[...]

[...]

[...]

about ubiquitous ordinals

What are ubiquitous ordinal?

Well, you know that ORD, is, the order type of ordinals,
and so it's an ordinal, of all the ordinals.
The "ubiquitous ordinals", sort of recalls Kronecker's
"G-d made the integers, the rest is the work of Man",
that the Integer Continuum, is the model and ground
model, of any sort of language of finite words,
like set theory.
It's like the universe of set theory, then as that
there's _always_ an arithmetization, or as with regards
to ordering and numbering as a bit weaker property than
collecting and counting, so that "ubiquitous ordinals"
is what you get from a discrete world.
Then there's that according to the set-theoretic Powerset
theorem of Cantor, that when the putative function is
successor, in ubiquitous ordinals where order type is
powerset is successor, then there's no missing element.
So, "ubiquitous ordinals" is exactly what it says.

Or, you know, "infinity plus one".

[Jim Burns]

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
about ubiquitous ordinals

What are ubiquitous ordinal?

Well, you know that ORD, is, the order type of ordinals,
and so it's an ordinal, of all the ordinals.

If a ubiquitous ordinal is an ordinal,
then I recommend referring to as an ordinal.

Post by Ross Finlayson
The "ubiquitous ordinals", sort of recalls Kronecker's
"G-d made the integers, the rest is the work of Man",
that the Integer Continuum, is the model and ground
model, of any sort of language of finite words,
like set theory.

If a ubiquitous ordinal is
a finite ordinal ==
a natural number ==
a non.negative integer,
then
(I bet you see where I'm headed here)
I recommend that you refer to it as
a finite ordinal or
a natural number or
a non.negative integer.

Post by Ross Finlayson
It's like the universe of set theory,

Do you and I mean the same by "universe of set theory"?

I am most familiar with theories of
well.founded sets without urelements.

In the von Neumann hierarchy of hereditary well.founded sets
V[0] = {}
V[β+1] = 𝒫(V[β])
V[γ] = ⋃[β<γ] V[β]

V[ω] is the universe of hereditarily finite sets.

For the first inaccessible ordinal κ
V[κ] is a model of ZF+Choice.

For first inaccessible ordinal κ
[0,κ) holds an uncountable ordinal and
is closed under cardinal arithmetic.

Post by Ross Finlayson
then as that there's _always_ an arithmetization, or
as with regards to ordering and numbering
as a bit weaker property than collecting and counting,
so that "ubiquitous ordinals" is
what you get from a discrete world.

Is a ubiquitous ordinal a finite ordinal?
I would appreciate a "yes" or a "no" in your response.

Post by Ross Finlayson
Then there's that
according to the set-theoretic Powerset theorem of Cantor,
that when the putative function is successor,
in ubiquitous ordinals
where order type is powerset is successor,
then there's no missing element.
So, "ubiquitous ordinals" is exactly what it says.

I find it concerning that you (Ross Finlayson) think that
"what it says" answers "What does it say?" in any useful way.

[Ross Finlayson]

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
about ubiquitous ordinals

What are ubiquitous ordinal?

Well, you know that ORD, is, the order type of ordinals,
and so it's an ordinal, of all the ordinals.

If a ubiquitous ordinal is an ordinal,
then I recommend referring to as an ordinal.

Post by Ross Finlayson
The "ubiquitous ordinals", sort of recalls Kronecker's
"G-d made the integers, the rest is the work of Man",
that the Integer Continuum, is the model and ground
model, of any sort of language of finite words,
like set theory.

If a ubiquitous ordinal is
a finite ordinal ==
a natural number ==
a non.negative integer,
then
(I bet you see where I'm headed here)
I recommend that you refer to it as
a finite ordinal or
a natural number or
a non.negative integer.

Post by Ross Finlayson
It's like the universe of set theory,

Do you and I mean the same by "universe of set theory"?
I am most familiar with theories of
well.founded sets without urelements.
In the von Neumann hierarchy of hereditary well.founded sets
V[0] = {}
V[β+1] = 𝒫(V[β])
V[γ] = ⋃[β<γ] V[β]
V[ω] is the universe of hereditarily finite sets.
For the first inaccessible ordinal κ
V[κ] is a model of ZF+Choice.
For first inaccessible ordinal κ
[0,κ) holds an uncountable ordinal and
is closed under cardinal arithmetic.

Post by Ross Finlayson
then as that there's _always_ an arithmetization, or
as with regards to ordering and numbering
as a bit weaker property than collecting and counting,
so that "ubiquitous ordinals" is
what you get from a discrete world.

Is a ubiquitous ordinal a finite ordinal?
I would appreciate a "yes" or a "no" in your response.

Post by Ross Finlayson
Then there's that
according to the set-theoretic Powerset theorem of Cantor,
that when the putative function is successor,
in ubiquitous ordinals
where order type is powerset is successor,
then there's no missing element.
So, "ubiquitous ordinals" is exactly what it says.

I find it concerning that you (Ross Finlayson) think that
"what it says" answers "What does it say?" in any useful way.

The ubiquitous ordinals are, for example, a theory where
the primary elements are ordinals, for ordering theory,
and numbering theory, which may be more fundamental,
than set theory, with regards to a theory of one relation.

The "template theory" is as of a Comenius language, a
language of only truisms its elements, one of which is
the Liar paradox, only a truism as prototyping a fallacy,
here relating that to ORD, that ORD exists is a truism
in the theory.

The "universals" in language invoke "the universals",
that's exactly what it says, that's its name.


Usual ordinary well-founded set theory is a nice little theory.

Giving it an axiom of "ordinary infinity" is a sort of
restriction of comprehension, sort of like it's a false axiom,
or rather, hypocritical, in the sense that "Russell's retro-thesis",
rejects what's otherwise a matter of deductive resolution of
the paradoxes of the quantifier ambiguity and impredicativity,
that you claim Russell claims don't exist.

What it says is what's its name is what it is.

There is no universe in ZFC, don't be saying otherwise.
It simply doesn't exist and isn't available. Then, if
you get into class/set distinction and these kinds of things,
then it's automatically extra-ordinary and your meta-theory
doesn't admit Russell's retro-thesis except as a conditional
property of a sort of sub-class of propositions in tertium non datur.

ZFC with classes, ..., "proper" classes, or "ultimate" as Quine puts it.


Of course, some have integers as primary, while others have
continuity as primary, that the integers and modular just fall
out in the middle, sort of downward-Kronecker invoking Skolem and
Louwenheim and Levy, if you've heard of them.


So, if you've heard of Skolem and Louwenheim and Levy,
then you've probably heard about the generic extension,
and if you've heard of the Independence of the Continuum Hypothesis
you've probably heard about the generic extension, and collapse,
to either a higher or lower cardinality an ordinal still keeping
a model, like, ..., a model of ubiquitous ordinals.

Add it up, it's where the numbers come from, not what you make of them.

[Ross Finlayson]

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
about ubiquitous ordinals

What are ubiquitous ordinal?

Well, you know that ORD, is, the order type of ordinals,
and so it's an ordinal, of all the ordinals.

If a ubiquitous ordinal is an ordinal,
then I recommend referring to as an ordinal.

Post by Ross Finlayson
The "ubiquitous ordinals", sort of recalls Kronecker's
"G-d made the integers, the rest is the work of Man",
that the Integer Continuum, is the model and ground
model, of any sort of language of finite words,
like set theory.

If a ubiquitous ordinal is
a finite ordinal ==
a natural number ==
a non.negative integer,
then
(I bet you see where I'm headed here)
I recommend that you refer to it as
a finite ordinal or
a natural number or
a non.negative integer.

Post by Ross Finlayson
It's like the universe of set theory,

Do you and I mean the same by "universe of set theory"?
I am most familiar with theories of
well.founded sets without urelements.
In the von Neumann hierarchy of hereditary well.founded sets
V[0] = {}
V[β+1] = 𝒫(V[β])
V[γ] = ⋃[β<γ] V[β]
V[ω] is the universe of hereditarily finite sets.
For the first inaccessible ordinal κ
V[κ] is a model of ZF+Choice.
For first inaccessible ordinal κ
[0,κ) holds an uncountable ordinal and
is closed under cardinal arithmetic.

Post by Ross Finlayson
then as that there's _always_ an arithmetization, or
as with regards to ordering and numbering
as a bit weaker property than collecting and counting,
so that "ubiquitous ordinals" is
what you get from a discrete world.

Is a ubiquitous ordinal a finite ordinal?
I would appreciate a "yes" or a "no" in your response.

Post by Ross Finlayson
Then there's that
according to the set-theoretic Powerset theorem of Cantor,
that when the putative function is successor,
in ubiquitous ordinals
where order type is powerset is successor,
then there's no missing element.
So, "ubiquitous ordinals" is exactly what it says.

I find it concerning that you (Ross Finlayson) think that
"what it says" answers "What does it say?" in any useful way.

The ubiquitous ordinals are, for example, a theory where
the primary elements are ordinals, for ordering theory,
and numbering theory, which may be more fundamental,
than set theory, with regards to a theory of one relation.
The "template theory" is as of a Comenius language, a
language of only truisms its elements, one of which is
the Liar paradox, only a truism as prototyping a fallacy,
here relating that to ORD, that ORD exists is a truism
in the theory.
The "universals" in language invoke "the universals",
that's exactly what it says, that's its name.
Usual ordinary well-founded set theory is a nice little theory.
Giving it an axiom of "ordinary infinity" is a sort of
restriction of comprehension, sort of like it's a false axiom,
or rather, hypocritical, in the sense that "Russell's retro-thesis",
rejects what's otherwise a matter of deductive resolution of
the paradoxes of the quantifier ambiguity and impredicativity,
that you claim Russell claims don't exist.
What it says is what's its name is what it is.
There is no universe in ZFC, don't be saying otherwise.
It simply doesn't exist and isn't available. Then, if
you get into class/set distinction and these kinds of things,
then it's automatically extra-ordinary and your meta-theory
doesn't admit Russell's retro-thesis except as a conditional
property of a sort of sub-class of propositions in tertium non datur.
ZFC with classes, ..., "proper" classes, or "ultimate" as Quine puts it.
Of course, some have integers as primary, while others have
continuity as primary, that the integers and modular just fall
out in the middle, sort of downward-Kronecker invoking Skolem and
Louwenheim and Levy, if you've heard of them.
So, if you've heard of Skolem and Louwenheim and Levy,
then you've probably heard about the generic extension,
and if you've heard of the Independence of the Continuum Hypothesis
you've probably heard about the generic extension, and collapse,
to either a higher or lower cardinality an ordinal still keeping
a model, like, ..., a model of ubiquitous ordinals.
Add it up, it's where the numbers come from, not what you make of them.

And von Neumann was a belligerent drunk.

[Jim Burns]

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
about ubiquitous ordinals

What are ubiquitous ordinal?

Well, you know that ORD, is, the order type of ordinals,
and so it's an ordinal, of all the ordinals.

Is a ubiquitous ordinal a finite ordinal?
I would appreciate a "yes" or a "no" in your response.

The ubiquitous ordinals are, for example,
a theory where the primary elements are ordinals,
for ordering theory, and numbering theory,
which may be more fundamental, than set theory,
with regards to a theory of one relation.

Apparently,
what you mean by ubiquitous ordinals are ordinals,
without further qualification.

Ordinals can be represented as sets, and are,
most often by the von Neumann scheme, λ = [0,λ)ᴼʳᵈ

Apparently,
ubiquitous ordinals are
what are represented by the [0,λ)ᴼʳᵈ

Ordinals are well.ordered.
The [0,λ)ᴼʳᵈ are well.ordered.
That is entirely non.accidental.
There isn't much reason to choose between
the von Neumann ordinal.representations and
the raw, unfiltered "ubiquitous" ordinals.

The one advantage which
representations have over the ubiquitous(?) is that
they are are objects in a theory of sets which
we have great confidence isn't contradictory,
and that extends our great confidence to
the non.contradictoriness of theorems about ordinals.

Unless we are considering their existence,
which is to say, their non.contradictoriness,
ordinals.of.unspecified.origin are well.ordered,
and that's an end to their description.

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
It's like the universe of set theory,

Do you and I mean the same by "universe of set theory"?
I am most familiar with theories of
well.founded sets without urelements.
In the von Neumann hierarchy of hereditary well.founded sets
V[0] = {}
V[β+1] = 𝒫(V[β])
V[γ] = ⋃[β<γ] V[β]
V[ω] is the universe of hereditarily finite sets.
For the first inaccessible ordinal κ
V[κ] is a model of ZF+Choice.
For first inaccessible ordinal κ
[0,κ) holds an uncountable ordinal and
is closed under cardinal arithmetic.

There is no universe in ZFC, don't be saying otherwise.

I will continue not.saying that
there is a universe _IN_ ZFC.

There are universes _OF_ ZFC which are also called domains.
V[κ] is one of the domains of ZFC.

None of the sets described by the theory is
the universal set, holding all sets IN the domain.


Not precisely to your point, but interesting:
Some of the sets IN ZFC satisfy all its axioms,
which make them also domains OF ZFC

Compare that to the way in which
each end.segment of ℕ is also a model of ℕ

[Ross Finlayson]

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
about ubiquitous ordinals

What are ubiquitous ordinal?

Well, you know that ORD, is, the order type of ordinals,
and so it's an ordinal, of all the ordinals.

Is a ubiquitous ordinal a finite ordinal?
I would appreciate a "yes" or a "no" in your response.

The ubiquitous ordinals are, for example,
a theory where the primary elements are ordinals,
for ordering theory, and numbering theory,
which may be more fundamental, than set theory,
with regards to a theory of one relation.

Apparently,
what you mean by ubiquitous ordinals are ordinals,
without further qualification.
Ordinals can be represented as sets, and are,
most often by the von Neumann scheme, λ = [0,λ)ᴼʳᵈ
Apparently,
ubiquitous ordinals are
what are represented by the [0,λ)ᴼʳᵈ
Ordinals are well.ordered.
The [0,λ)ᴼʳᵈ are well.ordered.
That is entirely non.accidental.
There isn't much reason to choose between
the von Neumann ordinal.representations and
the raw, unfiltered "ubiquitous" ordinals.
The one advantage which
representations have over the ubiquitous(?) is that
they are are objects in a theory of sets which
we have great confidence isn't contradictory,
and that extends our great confidence to
the non.contradictoriness of theorems about ordinals.
Unless we are considering their existence,
which is to say, their non.contradictoriness,
ordinals.of.unspecified.origin are well.ordered,
and that's an end to their description.

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
It's like the universe of set theory,

Do you and I mean the same by "universe of set theory"?
I am most familiar with theories of
well.founded sets without urelements.
In the von Neumann hierarchy of hereditary well.founded sets
V[0] = {}
V[β+1] = 𝒫(V[β])
V[γ] = ⋃[β<γ] V[β]
V[ω] is the universe of hereditarily finite sets.
For the first inaccessible ordinal κ
V[κ] is a model of ZF+Choice.
For first inaccessible ordinal κ
[0,κ) holds an uncountable ordinal and
is closed under cardinal arithmetic.

There is no universe in ZFC, don't be saying otherwise.

I will continue not.saying that
there is a universe _IN_ ZFC.
There are universes _OF_ ZFC which are also called domains.
V[κ] is one of the domains of ZFC.
None of the sets described by the theory is
the universal set, holding all sets IN the domain.
Some of the sets IN ZFC satisfy all its axioms,
which make them also domains OF ZFC
Compare that to the way in which
each end.segment of ℕ is also a model of ℕ

Well, thanks for asking, I guess.

Starting with the axioms of ZFC before getting
to an actual axiom of infinity, then the embodiment
of Russell's paradox is those finite sets you've found,
all the sets that don't contain themselves, then you
invoke the the restriction of comprehension of the
axiom of infinity (you mean it doesn't say there's
an infinite, it says there's not an extra-ordinary infinite?)
those fulfilling Russell's retro-thesis for him.

"Because it would immediately lead to a contradiction."

If infinite end-segments are models of N,
they're also models of co-finite with non-empty
complements.

I'm not sure you're aware of usual operations "direct sum"
and "direct product", of sets, and as with regards to that the
direct sum of infinitely many copies of a set like as
modeled by finite natural integers, is a matter of definition
and it sticks out that according to the usual definition,
it's one way, and according to the usual imposed definition
on the case of the infinite, it's the opposite. This is
about the direct sum of the infinitely-many copies of N
and whether it's empty or full, that the imposed definition
is the other way than how it is for the finite sets.

"Because it would immediately lead to a contradiction."

So, these are examples, in ZF set theory,
some that you know you have and have to remember
to invoke or as result usual sorts fundamental theorems
that are always assumed, and more examples that you
don't know you have, about consequences of comprehension
and the fact that it's natural that the infinite is extra-ordinary
and then when there's an Integer Continuum then the model of
words is on a substrate of "ubiquitous ordinals".

Then, for talk about the universe and the constructible universe
and V = L and with regards to whether V = L, according to Goedel
it's not, that a gist of his incompleteness theorems, which of
course follow his more naive though entirely ordinary completeness
theorems, for which you can thank Frege.

That V = L has there's a universe already and it constructs.

Then, your talk of domains, like "the equivalence class of
singletons the cardinal 1", and so on, and don't ask why
they don't just call that a set, in as regards to whether
cardinals are even sets at all, is another example that
there's a theory where domains are primary and elements
and it's called "domain theory".

The idea that there's one theory for all this theory,
has that otherwise there isn't and you're not talking
about any of them.

[Ross Finlayson]

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
about ubiquitous ordinals

What are ubiquitous ordinal?

Well, you know that ORD, is, the order type of ordinals,
and so it's an ordinal, of all the ordinals.

Is a ubiquitous ordinal a finite ordinal?
I would appreciate a "yes" or a "no" in your response.

The ubiquitous ordinals are, for example,
a theory where the primary elements are ordinals,
for ordering theory, and numbering theory,
which may be more fundamental, than set theory,
with regards to a theory of one relation.

Apparently,
what you mean by ubiquitous ordinals are ordinals,
without further qualification.
Ordinals can be represented as sets, and are,
most often by the von Neumann scheme, λ = [0,λ)ᴼʳᵈ
Apparently,
ubiquitous ordinals are
what are represented by the [0,λ)ᴼʳᵈ
Ordinals are well.ordered.
The [0,λ)ᴼʳᵈ are well.ordered.
That is entirely non.accidental.
There isn't much reason to choose between
the von Neumann ordinal.representations and
the raw, unfiltered "ubiquitous" ordinals.
The one advantage which
representations have over the ubiquitous(?) is that
they are are objects in a theory of sets which
we have great confidence isn't contradictory,
and that extends our great confidence to
the non.contradictoriness of theorems about ordinals.
Unless we are considering their existence,
which is to say, their non.contradictoriness,
ordinals.of.unspecified.origin are well.ordered,
and that's an end to their description.

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
It's like the universe of set theory,

Do you and I mean the same by "universe of set theory"?
I am most familiar with theories of
well.founded sets without urelements.
In the von Neumann hierarchy of hereditary well.founded sets
V[0] = {}
V[β+1] = 𝒫(V[β])
V[γ] = ⋃[β<γ] V[β]
V[ω] is the universe of hereditarily finite sets.
For the first inaccessible ordinal κ
V[κ] is a model of ZF+Choice.
For first inaccessible ordinal κ
[0,κ) holds an uncountable ordinal and
is closed under cardinal arithmetic.

There is no universe in ZFC, don't be saying otherwise.

I will continue not.saying that
there is a universe _IN_ ZFC.
There are universes _OF_ ZFC which are also called domains.
V[κ] is one of the domains of ZFC.
None of the sets described by the theory is
the universal set, holding all sets IN the domain.
Some of the sets IN ZFC satisfy all its axioms,
which make them also domains OF ZFC
Compare that to the way in which
each end.segment of ℕ is also a model of ℕ

Well, thanks for asking, I guess.
Starting with the axioms of ZFC before getting
to an actual axiom of infinity, then the embodiment
of Russell's paradox is those finite sets you've found,
all the sets that don't contain themselves, then you
invoke the the restriction of comprehension of the
axiom of infinity (you mean it doesn't say there's
an infinite, it says there's not an extra-ordinary infinite?)
those fulfilling Russell's retro-thesis for him.
"Because it would immediately lead to a contradiction."
If infinite end-segments are models of N,
they're also models of co-finite with non-empty
complements.
I'm not sure you're aware of usual operations "direct sum"
and "direct product", of sets, and as with regards to that the
direct sum of infinitely many copies of a set like as
modeled by finite natural integers, is a matter of definition
and it sticks out that according to the usual definition,
it's one way, and according to the usual imposed definition
on the case of the infinite, it's the opposite. This is
about the direct sum of the infinitely-many copies of N
and whether it's empty or full, that the imposed definition
is the other way than how it is for the finite sets.
"Because it would immediately lead to a contradiction."
So, these are examples, in ZF set theory,
some that you know you have and have to remember
to invoke or as result usual sorts fundamental theorems
that are always assumed, and more examples that you
don't know you have, about consequences of comprehension
and the fact that it's natural that the infinite is extra-ordinary
and then when there's an Integer Continuum then the model of
words is on a substrate of "ubiquitous ordinals".
Then, for talk about the universe and the constructible universe
and V = L and with regards to whether V = L, according to Goedel
it's not, that a gist of his incompleteness theorems, which of
course follow his more naive though entirely ordinary completeness
theorems, for which you can thank Frege.
That V = L has there's a universe already and it constructs.
Then, your talk of domains, like "the equivalence class of
singletons the cardinal 1", and so on, and don't ask why
they don't just call that a set, in as regards to whether
cardinals are even sets at all, is another example that
there's a theory where domains are primary and elements
and it's called "domain theory".
The idea that there's one theory for all this theory,
has that otherwise there isn't and you're not talking
about any of them.

It's a pretty simple result that standard models of ordinals
don't exist at all, only fragments and extensions.

Same goes for integers.

[Jim Burns]

Post by Ross Finlayson

[...]

The idea that there's one theory for all this theory,
has that otherwise there isn't
and you're not talking about any of them.

If I remember correctly, your (RF's) name for
not.talking about
what's outside the domain of discussion
is hypocrisyᴿꟳ.

That sounds like you're delivering a value.judgment:
that we _should not_ not.talk about
what's outside the domain of discussion,
that we _should not_ for example, not.talk about
_all_ triangles when we discuss whether
the square of its longest side equals
the sum of the squares of the two remaining sides.

However,
it is because we are hypocriticalᴿꟳ (in your sense?)
that such discussions produce results.
"Conclusions", if you like.

We make finite.length.statements which
we know are true in infinitely.many senses.

We can know they are so because
we have narrowed our attention to
those for which they are true without exception.
Stated once, finitely, for infinitely.many.

Non.hypocrisyᴿꟳ (sincerityᴿꟳ?) throws that away.

[Ross Finlayson]

Post by Jim Burns

Post by Ross Finlayson

[...]

The idea that there's one theory for all this theory,
has that otherwise there isn't
and you're not talking about any of them.

If I remember correctly, your (RF's) name for
not.talking about
what's outside the domain of discussion
is hypocrisyᴿꟳ.
that we _should not_ not.talk about
what's outside the domain of discussion,
that we _should not_ for example, not.talk about
_all_ triangles when we discuss whether
the square of its longest side equals
the sum of the squares of the two remaining sides.
However,
it is because we are hypocriticalᴿꟳ (in your sense?)
that such discussions produce results.
"Conclusions", if you like.
We make finite.length.statements which
we know are true in infinitely.many senses.
We can know they are so because
we have narrowed our attention to
those for which they are true without exception.
Stated once, finitely, for infinitely.many.
Non.hypocrisyᴿꟳ (sincerityᴿꟳ?) throws that away.

You're talking about a field, I'm talking about foundations.

... Of which there is one and a universe of it.

Then, when these examples of just carrying forward inductive
inference get out, they do. That there's a class of expressions
that are outside of tertium non datur is always a thing.

Let's talk about completions and the infinite limit and
the continuum limit, there's always an inductive counterargument
that it's not so, not complete, not the sum, not continuous.

About triangles and right triangles, and classes and sets in
an ordinary theory like ZFC with classes, now your theory has
classes that aren't sets.

That analysis is sometimes catalysis, for anaphora and cataphora,
is a thing, and two things.

In foundations, there's a universe to account for,
there's nothing outside, or vice-versa, ..., and vice-versa.


Then ubiquitous ordinals you can also find in where, for example,
"ordering theory" is fundamental instead of "set theory", it's
a theory altogether with entirely different elements, then with
the decriptive approach of model theory, making models and giving
them names, the "equi-interpretability" here has that simply
the ordering theory's a bit simpler than set theory, and in
a fundamental theory the elements are simple.

So, hypocrisy, like Russell's retro-thesis, a restriction of
comprehension that goes along with other axioms that would
build for themselves a confounding confusing conflating
consterning counter-example, doesn't just go without saying.


Yeah, my mathematical conscience demands that hypocrisy is bad.

[Jim Burns]

Post by Ross Finlayson

If I remember correctly,  your (RF's) name for
not.talking about
what's outside the domain of discussion
is hypocrisyᴿꟳ.
that we _should not_ not.talk about
what's outside the domain of discussion,
that we _should not_ for example, not.talk about
_all_ triangles when we discuss whether
the square of its longest side equals
the sum of the squares of the two remaining sides.
However,
it is because we are hypocriticalᴿꟳ (in your sense?)
that such discussions produce results.
"Conclusions", if you like.
We make finite.length.statements which
we know are true in infinitely.many senses.
We can know they are so because
we have narrowed our attention to
those for which they are true without exception.
Stated once, finitely, for infinitely.many.
Non.hypocrisyᴿꟳ (sincerityᴿꟳ?) throws that away.

You're talking about a field,
I'm talking about foundations.

I doubt that
you and I are calling the same thing a field:
a set with addition, multiplication, identities, inverses
such that
a+(b+c)=(a+b)+c a+b=b+a a+0=a a+(-a)=0
a⋅(b⋅c)=(a⋅b)⋅c a⋅b=b⋅a a⋅1=a a≠0 ⇒ a⋅⅟a=1
a⋅(b+c)=(a⋅b)+(a⋅c)
?

fieldᴿꟳ == domainⁿᵒᵗᐧᴿꟳ ?

The counterpart of a variable is its domainⁿᵒᵗᐧᴿꟳ
== those to which the variable possibly refers.

From what I can see,
both fieldsᴿꟳ and foundationsᴿꟳ are domainsⁿᵒᵗᐧᴿꟳ

I'm guessing that the distinction between
fieldsᴿꟳ and foundationsᴿꟳ is the distinction between
retail mathematics and wholesale mathematics,
issues of the day and grand unification.

In given circumstances, there may well be
excellent reasons to do retail mathematics or
to do wholesale mathematics.
I am skeptical about there ever being
logical reasons to choose one over the other.

Post by Ross Finlayson
You're talking about a field,
I'm talking about foundations.
... Of which there is one and a universe of it.

If a theory has any model of infinite cardinality,
it has models of each infinite cardinality.

That's a general result.
The empty theory (with no extralogical axioms)
has models of each infinite cardinality.

Post by Ross Finlayson

that we _should not_ not.talk about
what's outside the domain of discussion,
that we _should not_ for example, not.talk about
_all_ triangles when we discuss whether
the square of its longest side equals
the sum of the squares of the two remaining sides.

About triangles and right triangles,
and classes and sets in an ordinary theory
like ZFC with classes, now your theory has
classes that aren't sets.

Somewhere, in axioms or definitions,
there are statements we know are true
as long as they are referring to a right triangle.

I fully expect that
things other than right triangles exist.
Those other things' existence doesn't change
the truth of those statements
as long as they are referring to a right triangle.

That isn't a particularly difficult insight.
We know they're true because
we know what a right triangle is. Duh.

I think that I find myself repeating
that not.particularly.difficult insight
because it _sounds like_
teeny, tiny finite beings <waves at camera> are
somehow engaging in some sort of infinite activity.

We are not engaged in any sort of
infinite activity.
Making finitely.many finite.length statements
is not an infinite activity.
Yes, they are true _about_ infinitely.many, but
we do not "true" the statements infinitely.often
as though we're laying infinitely.many bricks.

Post by Ross Finlayson
Yeah, my mathematical conscience demands that
hypocrisy is bad.

Bad why?

[Ross Finlayson]

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns
If I remember correctly, your (RF's) name for
not.talking about
what's outside the domain of discussion
is hypocrisyᴿꟳ.
that we _should not_ not.talk about
what's outside the domain of discussion,
that we _should not_ for example, not.talk about
_all_ triangles when we discuss whether
the square of its longest side equals
the sum of the squares of the two remaining sides.
However,
it is because we are hypocriticalᴿꟳ (in your sense?)
that such discussions produce results.
"Conclusions", if you like.
We make finite.length.statements which
we know are true in infinitely.many senses.
We can know they are so because
we have narrowed our attention to
those for which they are true without exception.
Stated once, finitely, for infinitely.many.
Non.hypocrisyᴿꟳ (sincerityᴿꟳ?) throws that away.

You're talking about a field,
I'm talking about foundations.

I doubt that
a set with addition, multiplication, identities, inverses
such that
a+(b+c)=(a+b)+c a+b=b+a a+0=a a+(-a)=0
a⋅(b⋅c)=(a⋅b)⋅c a⋅b=b⋅a a⋅1=a a≠0 ⇒ a⋅⅟a=1
a⋅(b+c)=(a⋅b)+(a⋅c)
?
fieldᴿꟳ == domainⁿᵒᵗᐧᴿꟳ ?
The counterpart of a variable is its domainⁿᵒᵗᐧᴿꟳ
== those to which the variable possibly refers.
From what I can see,
both fieldsᴿꟳ and foundationsᴿꟳ are domainsⁿᵒᵗᐧᴿꟳ
I'm guessing that the distinction between
fieldsᴿꟳ and foundationsᴿꟳ is the distinction between
retail mathematics and wholesale mathematics,
issues of the day and grand unification.
In given circumstances, there may well be
excellent reasons to do retail mathematics or
to do wholesale mathematics.
I am skeptical about there ever being
logical reasons to choose one over the other.

Post by Ross Finlayson
You're talking about a field,
I'm talking about foundations.
... Of which there is one and a universe of it.

If a theory has any model of infinite cardinality,
it has models of each infinite cardinality.
That's a general result.
The empty theory (with no extralogical axioms)
has models of each infinite cardinality.

Post by Ross Finlayson

Post by Jim Burns
that we _should not_ not.talk about
what's outside the domain of discussion,
that we _should not_ for example, not.talk about
_all_ triangles when we discuss whether
the square of its longest side equals
the sum of the squares of the two remaining sides.

About triangles and right triangles,
and classes and sets in an ordinary theory
like ZFC with classes, now your theory has
classes that aren't sets.

Somewhere, in axioms or definitions,
there are statements we know are true
as long as they are referring to a right triangle.
I fully expect that
things other than right triangles exist.
Those other things' existence doesn't change
the truth of those statements
as long as they are referring to a right triangle.
That isn't a particularly difficult insight.
We know they're true because
we know what a right triangle is. Duh.
I think that I find myself repeating
that not.particularly.difficult insight
because it _sounds like_
teeny, tiny finite beings <waves at camera> are
somehow engaging in some sort of infinite activity.
We are not engaged in any sort of
infinite activity.
Making finitely.many finite.length statements
is not an infinite activity.
Yes, they are true _about_ infinitely.many, but
we do not "true" the statements infinitely.often
as though we're laying infinitely.many bricks.

Post by Ross Finlayson
Yeah, my mathematical conscience demands that
hypocrisy is bad.

Bad why?

"Wrong", ....

Yes we must disambiguate algebra's "fields" and physics' "fields"
and science's "fields" and logic's "fields, all one field, as it
were, in terms of "foundations", "the field of foundations".

Definition usually expands, when it doesn't, then there is
the conscientious disambiguation, for example as you rightly
note X^WM or X^RF definitions as you see fit to delineate in
the anaphora and cataphora, what's forward definition or for
the multi-pass parsing in definition, in terms of the "descriptive
dynamics" and "definitional dynamics".

For example, where "differential geometry" reduces the world of
functions and curves, they no longer are conscientiously free
in the wider field, encumbered instead with an ambiguity, which
is why then there's function^DG and curve^DG, for example, then
that what remains a restriction has its results attached.

The notion of function varies since the days of classical function,
since when it's delineated and disambiguated with regards to the
Cartesian and function, and that functions, for example, make
their own sort of fundamental theory.

Arithmetic, geometry, number theory, algebra, set theory, part theory,
topology, function theory: each has their own "theory" in their
own fundamental terms, thus resulting modeling each other or modeling
the disambiguation, here about "ubiquitous ordinals of a linear then
integer continuum". Each has their own "field", yet there's still
only one foundations, only one "meta-theory", all one theory.

[Jim Burns]

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

If I remember correctly,  your (RF's) name for
not.talking about
what's outside the domain of discussion
is hypocrisyᴿꟳ.
that we _should not_ not.talk about
what's outside the domain of discussion,
that we _should not_ for example, not.talk about
_all_ triangles when we discuss whether
the square of its longest side equals
the sum of the squares of the two remaining sides.

Yeah, my mathematical conscience demands that
hypocrisy is bad.

Bad why?

"Wrong", ....

It is wrong to treat claims about right triangles
as though they are claims about more than right triangles.

Post by Ross Finlayson
Definition usually expands,

The hypocrisyᴿꟳ of NOT expanding
the definition of right triangle ABC
to encompass triangles without right angles
leaves it NOT wrong that
a segment CH from right angle C
perpendicular to and meeting side AB at H
makes two more triangles ACH BCH,
which are both similar to ABC
which, as similar triangles,
have corresponding sides in the same ratio
so that
A͞H/A͞C = A͞C/A͞B
H͞B/B͞C = B͞C/A͞B
(A͞H+H͞B)⋅A͞B = A͞C² +B͞C²
and
A͞B² = A͞C² + B͞C² is NOT wrong.

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
hypocrisy is bad.

If it is, then it isn't for making things wrong,
which is something hypocrisyᴿꟳ
(not.talking about outside the domain)
doesn't do.

[Ross Finlayson]

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns
If I remember correctly, your (RF's) name for
not.talking about
what's outside the domain of discussion
is hypocrisyᴿꟳ.
that we _should not_ not.talk about
what's outside the domain of discussion,
that we _should not_ for example, not.talk about
_all_ triangles when we discuss whether
the square of its longest side equals
the sum of the squares of the two remaining sides.

Yeah, my mathematical conscience demands that
hypocrisy is bad.

Bad why?

"Wrong", ....

It is wrong to treat claims about right triangles
as though they are claims about more than right triangles.

Post by Ross Finlayson
Definition usually expands,

The hypocrisyᴿꟳ of NOT expanding
the definition of right triangle ABC
to encompass triangles without right angles
leaves it NOT wrong that
a segment CH from right angle C
perpendicular to and meeting side AB at H
makes two more triangles ACH BCH,
which are both similar to ABC
which, as similar triangles,
have corresponding sides in the same ratio
so that
A͞H/A͞C = A͞C/A͞B
H͞B/B͞C = B͞C/A͞B
(A͞H+H͞B)⋅A͞B = A͞C² +B͞C²
and
A͞B² = A͞C² + B͞C² is NOT wrong.

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
hypocrisy is bad.

If it is, then it isn't for making things wrong,
which is something hypocrisyᴿꟳ
(not.talking about outside the domain)
doesn't do.

There is no "outside" the universe.

Anything else, there is.

For a while we were having a discussion about Pythagorean Triples,
which are integer tuples that happen to be side lengths of right
triangles. The discussion then got into _completions_, that
just like the least-upper-bound not existing in rationals,
yet it's built out to be and usually with an axiomatization
for field-reals then that line-reals have their own trivial sort,
that courtesy unique prime factorization that for right triangles
with side lengths that aren't integer or rational with respect
to each other, there's a sequence of Pythagorean triples that
goes to it.

That it, ..., "goes to".


Whence you might consider our discussion on Pythagorean triples,
and these sequences of them that attain to right triangles of
not-necessarily integer proportion, then also you might recall,
there was the discussion of equi-lateral triangles, and that
un-hinging them and making them their epi-cycles as it were,
that equi-lateral triangles draw sine and cosine which is the
usual role of right triangles, so, all the properties so
accordingly, can be written altogether in terms of equi-lateral
triangles besides these right triangles.

So, that right-triangles and equi-lateral-triangles somehow concur,
isn't that as a fact that it's an emergent property of a proper
deconstruction of them either, both, together?

Also there was the "very-tall-triangles" bit.


It sort of seems the straw-man of you to say I'm disputing Pythagoras
when all I did was point out that Russell was more-or-less lying to you.

Then, if you recall, it was, "Pick one. Ha, I put them together,
you get both or none". Which was it?

It was "anti-diagonal and only-diagonal".

[Jim Burns]

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

If I remember correctly,  your (RF's) name for
not.talking about
what's outside the domain of discussion
is hypocrisyᴿꟳ.
that we _should not_ not.talk about
what's outside the domain of discussion,
that we _should not_ for example, not.talk about
_all_ triangles when we discuss whether
the square of its longest side equals
the sum of the squares of the two remaining sides.

Yeah, my mathematical conscience demands that
hypocrisy is bad.

Bad why?

"Wrong", ....

It is wrong to treat claims about right triangles
as though they are claims about more than right triangles.

Post by Ross Finlayson
Definition usually expands,

The hypocrisyᴿꟳ of NOT expanding
  the definition of right triangle ABC
  to encompass triangles without right angles
leaves it NOT wrong that
a segment CH from right angle C
  perpendicular to and meeting side AB at H
makes two more triangles ACH BCH,
  which are both similar to ABC
  which, as similar triangles,
  have corresponding sides in the same ratio
so that
A͞H/A͞C = A͞C/A͞B
H͞B/B͞C = B͞C/A͞B
(A͞H+H͞B)⋅A͞B = A͞C² +B͞C²
and
  A͞B² = A͞C² + B͞C²  is NOT wrong.

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
hypocrisy is bad.

If it is, then it isn't for making things wrong,
  which is something hypocrisyᴿꟳ
  (not.talking about outside the domain)
doesn't do.

There is no "outside" the universe.
Anything else, there is.

There very often are things outside
the things which we are discussing.

It is perhaps surprising how important it is that,
when we are discussing certain things,
we are not discussing other things, outside.

It is, it seems to me, a key part of a technique for
exploring the infinite by finite means.

That key part seems to me to be
what you are calling hypocrisyᴿꟳ

It strikes me as _at least_ unwise to discard
such a powerful and reliable tool.

Post by Ross Finlayson
It sort of seems the straw-man of you to say
I'm disputing Pythagoras
when all I did was point out that
Russell was more-or-less lying to you.

I picked Pythagoras as a concrete example of
my best guess at what you (RF) mean by hypocrisyᴿꟳ

If my best guess is wide of the mark,
perhaps you should tell me what you DO mean.

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

that we _should not_ not.talk about
what's outside the domain of discussion,

Yeah, my mathematical conscience demands that
hypocrisy is bad.

[FromTheRafters]

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns
If I remember correctly, your (RF's) name for
not.talking about
what's outside the domain of discussion
is hypocrisyᴿꟳ.
that we _should not_ not.talk about
what's outside the domain of discussion,
that we _should not_ for example, not.talk about
_all_ triangles when we discuss whether
the square of its longest side equals
the sum of the squares of the two remaining sides.

Yeah, my mathematical conscience demands that
hypocrisy is bad.

Bad why?

"Wrong", ....

It is wrong to treat claims about right triangles
as though they are claims about more than right triangles.

Post by Ross Finlayson
Definition usually expands,

The hypocrisyᴿꟳ of NOT expanding
the definition of right triangle ABC
to encompass triangles without right angles
leaves it NOT wrong that
a segment CH from right angle C
perpendicular to and meeting side AB at H
makes two more triangles ACH BCH,
which are both similar to ABC
which, as similar triangles,
have corresponding sides in the same ratio
so that
A͞H/A͞C = A͞C/A͞B
H͞B/B͞C = B͞C/A͞B
(A͞H+H͞B)⋅A͞B = A͞C² +B͞C²
and
A͞B² = A͞C² + B͞C² is NOT wrong.

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
hypocrisy is bad.

If it is, then it isn't for making things wrong,
which is something hypocrisyᴿꟳ
(not.talking about outside the domain)
doesn't do.

There is no "outside" the universe.

Then what *is* restricted comprehension?

[Ross Finlayson]

Post by FromTheRafters

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

Post by Jim Burns
If I remember correctly, your (RF's) name for
not.talking about
what's outside the domain of discussion
is hypocrisyᴿꟳ.
that we _should not_ not.talk about
what's outside the domain of discussion,
that we _should not_ for example, not.talk about
_all_ triangles when we discuss whether
the square of its longest side equals
the sum of the squares of the two remaining sides.

Yeah, my mathematical conscience demands that
hypocrisy is bad.

Bad why?

"Wrong", ....

It is wrong to treat claims about right triangles
as though they are claims about more than right triangles.

Post by Ross Finlayson
Definition usually expands,

The hypocrisyᴿꟳ of NOT expanding
the definition of right triangle ABC
to encompass triangles without right angles
leaves it NOT wrong that
a segment CH from right angle C
perpendicular to and meeting side AB at H
makes two more triangles ACH BCH,
which are both similar to ABC
which, as similar triangles,
have corresponding sides in the same ratio
so that
A͞H/A͞C = A͞C/A͞B
H͞B/B͞C = B͞C/A͞B
(A͞H+H͞B)⋅A͞B = A͞C² +B͞C²
and
A͞B² = A͞C² + B͞C² is NOT wrong.

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
hypocrisy is bad.

If it is, then it isn't for making things wrong,
which is something hypocrisyᴿꟳ
(not.talking about outside the domain)
doesn't do.

There is no "outside" the universe.

Then what *is* restricted comprehension?

Usually it's just the antonym of expansion of comprehension.

What I ask,
if that you surpass,
the inductive impasse,
of the infinite super-task.

[Jim Burns]

Post by Ross Finlayson

Post by FromTheRafters
[...]

What I ask,
if that you surpass,
the inductive impasse,
of the infinite super-task.

I don't see what infinite super.task or
what inductive impasse you are asking about.

I see transfinite induction being
a consequence of ordinals being
well.ordered.

I see cisfinite induction being
a consequence of finite ordinals being
well.ordered and bounding only 2.ended sets.

----
In transfinite induction,
{ξ ∈ 𝕆: ¬P(ξ)} holds
no first ordinal β:
¬P(β) ∧ ¬∃α<β: ¬P(α)

⎛ Because ordinals are well.ordered,
⎝ the set is no.first only.if {ξ ∈ 𝕆: ¬P(ξ)} = {}.

¬∃β ∈ 𝕆:( ¬P(β) ∧ ¬∃α<β: ¬P(α) )
⟹
¬∃ξ ∈ 𝕆: ¬P(ξ)

∀β ∈ 𝕆:( P(β) ⇐ ∀α<β: P(α) )
⟹
∀ξ ∈ 𝕆: P(ξ)

----
⎛ Because a finite.ordinal bounds only
⎜ 2.ended non.{}.sets
⎜ finite.ordinal β bounds only
⎜ non.{} [0,α)ᴼ with a second.end
⎝ [0,α)ᴼ = [0,α-1]ᴼ for 0 < α ≤ finite β

In cisfinite induction,
{ξ ∈ 𝕆ᶠⁱⁿ: ¬P(ξ)} doesn't hold
0 or ordinal β+1: ¬P(β+1) ∧ P(β)

...which implies
{ξ ∈ 𝕆ᶠⁱⁿ: ¬P(γ)} holds
no first ordinal β:
¬P(β) ∧ ¬∃α<β: ¬P(α)
and
no.first implies
{ξ ∈ 𝕆ᶠⁱⁿ: ¬P(ξ)} = {}.

P(0) ∧ ¬∃β ∈ 𝕆ᶠⁱⁿ: ¬P(β+1) ∧ P(β)
⟹
¬∃β ∈ 𝕆ᶠⁱⁿ:( ¬P(β) ∧ ¬∃α<β: ¬P(α) )
⟹
¬∃ξ ∈ 𝕆: ¬P(ξ)

P(0) ∧ ∀β ∈ 𝕆ᶠⁱⁿ: P(β) ⇒ P(β+1)
⟹
∀β ∈ 𝕆ᶠⁱⁿ:( P(β) ⇐ ∀α<β: P(α) )
⟹
∀ξ ∈ 𝕆ᶠⁱⁿ: P(ξ)

----

Post by Ross Finlayson

Post by FromTheRafters
Then what *is* restricted comprehension?

Usually it's just
the antonym of expansion of comprehension.

I am more familiar with unrestricted comprehension
being the antonym of restricted comprehension.

Unrestricted comprehension grants that
{x:P(x)} exists because
description P(x) of its elements exists.

Restricted comprehension grants that
{x∈A:P(x)} exists because
description P(x) and set A exist.

The existence of set A might have been granted
because of Restricted.Comprehension or Infinity or
Power.Set or Union or Replacement or Pairing,
but A would be logically prior to {x∈A:P(x)}
by some route.

[Jim Burns]

Post by Ross Finlayson

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson

If I remember correctly,  your (RF's) name for
not.talking about
what's outside the domain of discussion
is hypocrisyᴿꟳ.

Yeah, my mathematical conscience demands that
hypocrisy is bad.

Bad why?

"Wrong", ....
Definition usually expands,

It sort of seems the straw-man of you
to say I'm disputing Pythagoras
when all I did was point out that
Russell was more-or-less lying to you.

This is the heart of the matter:

Expanding the definition of right triangle
does not dispute Pythagoras.

Expanding the definition of right triangle
moves us to a second conversation.
Pythagoras is in the first conversation.
The second has no effect there.

First conversation.
⎛ Consider △ABC and AC⟂BC
⎜
⎜ H in AB: CH⟂AH and CH⟂BH
⎜
⎜ △ACH ∝ △ABC
⎜ A͞H/A͞C = A͞C/A͞B
⎜
⎜ △CBH ∝ △ABC
⎜ H͞B/B͞C = B͞C/A͞B
⎜
⎜ (A͞H+H͞B)⋅A͞B = A͞C² + B͞C²
⎝ A͞B² = A͞C² + B͞C²

Second conversation.
⎛ Consider △ABC ...
⎜ Comrades!
⎜ Throw off the shackles of the past!
⎜ AC⟂BC or AC̸⟂BC
⎜ You have been lied to.
⎜ For A͞B = A͞C = B͞C
⎜ A͞B² ≠ A͞C² + B͞C²
⎝ Freedom!

I am exaggerating for effect.
If you have called the rest of us comrades, etc,
I have missed it.

Nor do I think that you dispute Pythagoras.
I hope that you do not.
I want you to come over to our side and
look at the shape of your argument once more.

Are non.standard right triangles
an objection or a change of topic?

Is non.standard analysis
an objection or a change of topic?

Would you like to know why
your objections to a vast swath of of mathematics
attract such a tiny bit of attention?
No one else perceives what you're doing
to be raising an objection.

In my own case, it took me years
to guess that's what you intend.

[WM]

Post by Jim Burns

Post by WM

Post by Jim Burns

_The rule of subset_ proves that
every proper subset has less elements than its superset.

If ℕ has fewer elements than ℕ∪{ℕ}
then
|ℕ| ∈ ℕ

|ℕ| = ω-1 ∈ ℕ

No finite.ordinal is last.finite,
no visibleᵂᴹ finite.ordinal,
no darkᵂᴹ finite.ordinal.
In particular, no finite.ordinal is ω-1

NUF(x) cannot grow by more than 1 at any x. Therefore there is a first
step. First unit fraction implies last natural number.

Post by Jim Burns
Also, no before.first infinite.ordinal is
before the first infinite.ordinal ω
In particular, no infinite.ordinal is ω-1

Don't claim. Explain how you imagine the positions of the unit fractions
such that NUF does not have a first step.

Regards, WM

[Jim Burns]

Post by WM

Post by Jim Burns

Post by WM

Post by Jim Burns

_The rule of subset_ proves that
every proper subset has less elements than its superset.

If ℕ has fewer elements than ℕ∪{ℕ}
then
|ℕ| ∈ ℕ

|ℕ| = ω-1 ∈ ℕ

No finite.ordinal is last.finite,
no visibleᵂᴹ finite.ordinal,
no darkᵂᴹ finite.ordinal.
In particular, no finite.ordinal is ω-1

NUF(x) cannot grow by more than 1 at any x.

NUF(x) = |⅟ℕ∩(0,x]|
NUF(x) cannot grow by more than 1 at any x > 0
¬(0 > 0)

⅟ℕ∩(0,x] has
each non.{}.subset maximummed
each unit.fraction down.stepped
each non.maximum up.stepped
and therefore
ℵ₀.many unit.fractions

x > 0 ⇒ NUF(x) = ℵ₀

Post by WM
Therefore there is a first step.

0 < ⅟⌈1+⅟x⌉ < ⅟⌈⅟x⌉ ≤ x
⅟⌈⅟x⌉ ∈ ⅟ℕ∩(0,x]
⅟⌈1+⅟x⌉ ∈ ⅟ℕ∩(0,x]
NUF(x) = |⅟ℕ∩(0,x]| ≠ 1

Post by WM
First unit fraction implies last natural number.

Ordinal λ = the set of ordinals < λ
λ = [0,λ)ᴼʳᵈ

( k is a finite.ordinal
⇔
( each non.{}.subset of [0,k)ᴼʳᵈ is 2.ended

_Finite doesn't need to be small_

( k is a finite.ordinal
⇔
⎛ [0,k)ᴼʳᵈ is 2.ended and
⎝ ∀j ∈ (0,k)ᴼʳᵈ: [0,j)ᴼʳᵈ is 2.ended

if
( k is a finite.ordinal
then
( k+1 is a finite ordinal

...because
if
⎛ [0,k)ᴼʳᵈ is 2.ended and
⎝ ∀j ∈ (0,k)ᴼʳᵈ: [0,j)ᴼʳᵈ is 2.ended
then
⎛ [0,k+1)ᴼʳᵈ is 2.ended and
⎝ ∀j ∈ (0,k+1)ᴼʳᵈ: [0,j)ᴼʳᵈ is 2.ended

ω is defined to be the first transfinite.ordinal.
ω is not a reallyreallyreally big finite.ordinal.

ω = [0,ω)ᴼʳᵈ

if ω-1 exists
then
⎛ ω-1 = max.[0,ω)ᴼʳᵈ and
⎜⎛ [0,ω)ᴼʳᵈ is 2.ended and
⎜⎝ ∀j ∈ (0,ω)ᴼʳᵈ: [0,j)ᴼʳᵈ is 2.ended
⎜ and
⎝ ω is finite.

However,
ω is not finite, not even a reallyreallyreally big one.
ω is transfinite (the first).
[0,ω)ᴼʳᵈ isn't 2.ended.
ω-1 doesn't exist.

Post by WM

Post by Jim Burns
Also, no before.first infinite.ordinal is
before the first infinite.ordinal ω
In particular, no infinite.ordinal is ω-1

Don't claim.

Before the first star, no stars existed.
Before the first mammal, no mammal existed.
Before the first infinite.ordinal, no infinite ordinal exists.
What ω means is the first infinite ordinal.

Post by WM
Explain how you imagine
the positions of the unit fractions such that
NUF does not have a first step.

ℚ the rationals are fractal.

For each rational p/q in [0,⅟n]
there is rational n⋅p/q in [0,1]

Zoom from [0,1] in to [0,⅟n]
[0,⅟n] looks just like [0,1]

Zoom from [0,⅟n] in to [0,⅟n²]
[0,⅟n²] looks just like [0,⅟n] and
[0,⅟n] looks just like [0,1]

No zoom.level ⅟nₓ exists such that
[0,⅟nₓ] does NOT look like [0,1]

No zoom.level ⅟nₓ exists such that
some unit.fraction.free gap opens.

[0,⅟nₓ] looks like [0,1] and
[0,1] isn't unit.fraction.free.

No zoom.level ⅟nₓ exists such that
[0,⅟nₓ] holds fewer unit.fractions than [0,1]

No x > 0 exists such that
NUF(x) ≠ NUF(1) = ℵ₀

----
A lagniappe.

ℚ the rationals are fractal.

ℝ the reals are fractal AND each split is situated in ℝ:
For each ℝ.split F′ ᵃˡˡ<ᵃˡˡ H′
some x′ ∈ ℝ is last.in.F′ or first.in.H′

With each zoom.level ⅟n
x′ is at the split F′ ᵃˡˡ<ᵃˡˡ H′

[WM]

Post by Jim Burns

Post by WM
NUF(x) cannot grow by more than 1 at any x.

NUF(x) = |⅟ℕ∩(0,x]|
NUF(x) cannot grow by more than 1 at any x > 0

Correct. At x < 0 or x = 0 NUF(x) = 0 and remains so.

Post by Jim Burns
¬(0 > 0)

Correct.

Post by Jim Burns
⅟ℕ∩(0,x] has
each non.{}.subset maximummed
each unit.fraction down.stepped
each non.maximum up.stepped
and therefore
ℵ₀.many unit.fractions
x > 0 ⇒ NUF(x) = ℵ₀

That implies a growth between [0, 1] and (0, 1].
Between [0, 1] and (0, 1] NUF(x) cannot grow because no unit fraction fits
in between.
Strange how void of logical thinking matheologians are.

Regards, WM

[Jim Burns]

Post by WM

Post by Jim Burns

Post by WM
NUF(x) cannot grow by more than 1 at any x.

NUF(x) = |⅟ℕ∩(0,x]|
NUF(x) cannot grow by more than 1 at any x > 0

Correct. At x < 0 or x = 0 NUF(x) = 0 and remains so.

Post by Jim Burns
¬(0 > 0)

Correct.

Post by Jim Burns
⅟ℕ∩(0,x] has
each non.{}.subset maximummed
each unit.fraction down.stepped
each non.maximum up.stepped
and therefore
ℵ₀.many unit.fractions
x > 0  ⇒  NUF(x) = ℵ₀

That implies a growth between [0, 1] and (0, 1].

No.
x > 0 ⇒ NUF(x) = ℵ₀
does not imply a growth between [0,1] and (0,1]

(Check your work for a quantifier shift.)

NUF(x) = |⅟ℕ∩(0,x]| = ℵ₀
does not imply any unit fraction outside (0,x]

The unit.fractions in (0,x] are
each non.{}.subset maximummed
each unit.fraction down.stepped
each non.maximum up.stepped.

Things with that order.type are ℵ₀.many.

Finite doesn't need to be small.
Infinite is beyond all big.

[Moebius]

[...] between [0, 1] and (0, 1].

Could you please explain to me the meaning of the phrase "between [0, 1]
and (0, 1]"?

What IS "between [0, 1] and (0, 1]". (Which real numbers are "between
[0, 1] and (0, 1]"? 0? So why not just talk about 0 in this case?)

Thank you.

(Actually, it sounds like meaningless mumbo-jumbo to me. Something a
psychotic asshole full of shit might utter.)

[olcott]

Post by Moebius

[...] between [0, 1] and (0, 1].

Could you please explain to me the meaning of the phrase "between [0, 1]
and (0, 1]"?

One can construe them as line segments that differ in length
by a single geometric point on the number line. I came up
with that a few years ago with my infinitesimal number system
that seems to map integers to a contiguous set of immediately
adjacent geometric points on the number line.

Post by Moebius
What IS "between [0, 1] and (0, 1]". (Which real numbers are "between
[0, 1] and (0, 1]"? 0? So why not just talk about 0 in this case?)
Thank you.
(Actually, it sounds like meaningless mumbo-jumbo to me. Something a
psychotic asshole full of shit might utter.)

[Moebius]

Post by olcott

Post by Moebius

[...] between [0, 1] and (0, 1].

Could you please explain to me the meaning of the phrase "between [0,
1] and (0, 1]"?

One can construe them as line segments that differ in length
by a single geometric point on the number line. I came up
with that a few years ago with my infinitesimal number system
that <bla>

Hint: IR does not comtain any "infinitesimal numbers" (except 0 that is).

[olcott]

Post by Moebius

Post by olcott

Post by Moebius

[...] between [0, 1] and (0, 1].

Could you please explain to me the meaning of the phrase "between [0,
1] and (0, 1]"?

One can construe them as line segments that differ in length
by a single geometric point on the number line. I came up
with that a few years ago with my infinitesimal number system
that <bla>

Hint: IR does not comtain any "infinitesimal numbers" (except 0 that is).

It is the case that the infinite set of points specified
by [0, 1] has exactly one less point than the infinite
set of points specified by (0, 1].

[Chris M. Thomasson]

Post by olcott

Post by Moebius

Post by olcott

Post by Moebius

[...] between [0, 1] and (0, 1].

Could you please explain to me the meaning of the phrase "between
[0, 1] and (0, 1]"?

One can construe them as line segments that differ in length
by a single geometric point on the number line. I came up
with that a few years ago with my infinitesimal number system
that <bla>

Hint: IR does not comtain any "infinitesimal numbers" (except 0 that is).

It is the case that the infinite set of points specified
by [0, 1] has exactly one less point than the infinite
set of points specified by (0, 1].

You mean going from:

p0 = {-1, 0};
p1 = {1, 0};
dif = p1 - p0;
unsigned long n = 10;

float normal_base = 1.f / n;

vs:


float normal_base = 1.f / (n - 1);

as in:

for (unsigned long i = 0; i < n; ++i)
{
float normal = normal_base * i;

glm::vec2 k0 = p0 + dif * normal;
}

?

[Moebius]

Post by olcott

Post by Moebius

Post by olcott

Post by Moebius

[...] between [0, 1] and (0, 1].

Could you please explain to me the meaning of the phrase "between
[0, 1] and (0, 1]"?

One can construe them as line segments that differ in length
by a single geometric point on the number line. I came up
with that a few years ago with my infinitesimal number system
that <bla>

Hint: IR does not comtain any "infinitesimal numbers" (except 0 that is).

It is the case that the infinite set of points
(0, 1] has exactly one point less than the infinite
set of points [0, 1].

Indeed! The point 0.

[0, 1] \ (0, 1] = {0}.

[olcott]

Post by Moebius

Post by Moebius

Post by olcott

Post by Moebius

[...] between [0, 1] and (0, 1].

Could you please explain to me the meaning of the phrase "between
[0, 1] and (0, 1]"?

One can construe them as line segments that differ in length
by a single geometric point on the number line. I came up
with that a few years ago with my infinitesimal number system
that <bla>

Hint: IR does not comtain any "infinitesimal numbers" (except 0 that is).

It is the case that the infinite set of points (0, 1] has exactly one
point less than the infinite
set of points [0, 1].

Indeed! The point 0.
[0, 1] \ (0, 1] = {0}.

Yes. That is the point.

[Moebius]

Post by olcott

Post by Moebius

Post by Moebius

Post by olcott

Post by Moebius

[...] between [0, 1] and (0, 1].

Could you please explain to me the meaning of the phrase "between
[0, 1] and (0, 1]"? [...]

One can construe them as line segments that differ in length
by a single geometric point on the number line. I came up
with that a few years ago with my infinitesimal number system
that <bla>

Hint: IR does not comtain any "infinitesimal numbers" (except 0 that is).

It is the case that the infinite set of points (0, 1] has exactly one
point less than the infinite
set of points [0, 1].

Indeed! The point 0.
[0, 1] \ (0, 1] = {0}.

Yes. That is the point.

Remember that I originally wrote:

"Could you please explain to me the meaning of the phrase "between [0,
1] and (0, 1]"? What IS "between [0, 1] and (0, 1]". (Which real numbers
are "between [0, 1] and (0, 1]"? 0? So why not just talk about 0 in this
case?)"

Thank you for your contribution. EOD.

[FromTheRafters]

Post by olcott

Post by Moebius

[...] between [0, 1] and (0, 1].

Could you please explain to me the meaning of the phrase "between [0, 1]
and (0, 1]"?

One can construe them as line segments that differ in length
by a single geometric point on the number line.

An accumulation point. How big is it? How big must it be to affect
length?

[olcott]

Post by olcott

Post by Moebius

[...] between [0, 1] and (0, 1].

Could you please explain to me the meaning of the phrase "between [0,
1] and (0, 1]"?

One can construe them as line segments that differ in length
by a single geometric point on the number line.

An accumulation point. How big is it? How big must it be to affect length?

The reason that I named it the infinitesimal number
system is that its units of measure are infinitesimal: 1/∞

[Ross Finlayson]

Post by olcott

Post by Moebius

[...] between [0, 1] and (0, 1].

Could you please explain to me the meaning of the phrase "between [0,
1] and (0, 1]"?

One can construe them as line segments that differ in length
by a single geometric point on the number line.

An accumulation point. How big is it? How big must it be to affect length?

"Non-standard analysis: continuum infinitesimal analysis, and sweep"

[Jim Burns]

Post by olcott

Post by Moebius

[...] between [0, 1] and (0, 1].

Could you please explain to me
the meaning of the phrase "between [0, 1] and (0, 1]"?

The meaning of that phrase is {}.

In WM's mind, {} proves darkᵂᴹ numbers exist by
the application of (at least)
shifting quantifiers,
assuming what he intends to prove, and
thinking that 'finite' and 'infinite' mean something they don't.

Post by olcott
One can construe them as line segments that
differ in length by a single geometric point
on the number line.
I came up with that a few years ago with my
infinitesimal number system that
seems to map integers to a contiguous set of
immediately adjacent geometric points
on the number line.

Consider a number line in which
each split is situated.

⎛ In a situated split F′ ᵉᵃᶜʰ<ᵉᵃᶜʰ H′
⎜ a number.line.point x′ exists
⎝ either last.in.F′ or first.in.H′

Consider the unit.fractions ⅟ℕ = {⅟j: finite j ∈ ℕ₁}

The set F′ of lower.bounds of ⅟ℕ and
the set H′ of not.lower.bounds of ⅟ℕ
is a split of the number.line

x′ situates F′ ᵉᵃᶜʰ<ᵉᵃᶜʰ H′

0 is a lower.bound of ⅟ℕ
Therefore, x′ isn't < 0

x′ isn't > 0
⎛ Assume otherwise.
⎜ Assume x′ > 0
⎜ 2⋅x′ > x′ > ½⋅x′ > 0
⎜
⎜ ½⋅x′ < x′ is a lower.bound of ⅟ℕ
⎜
⎜ 2⋅x′ > x′ isn't a lower.bound of ⅟ℕ
⎜ finite unit.fraction ⅟k < 2⋅x′
⎜ finite unit.fraction ¼⋅⅟k < ¼⋅2⋅x′ = ½⋅x′
⎜ ½⋅x′ isn't a lower.bound of ⅟ℕ
⎜
⎝ Contradiction.

Therefore, x′ isn't > 0
x′ isn't < 0
x′ = 0
0 situates the split between
lower.bounds and not.lower.bounds of ⅟ℕ

No point δ > 0 is a lower.bound of ⅟ℕ
Each point δ > 0 has a finite unit.fraction
between δ and 0
No point δ > 0 is an infinitesimal.

[WM]

Post by Moebius

[...] between [0, 1] and (0, 1].

Could you please explain to me the meaning of the phrase "between [0, 1]
and (0, 1]"?
What IS "between [0, 1] and (0, 1]". (Which real numbers are "between
[0, 1] and (0, 1]"? 0?

There is nothing. But ℵ₀ unit fractions require ℵ₀*2ℵ₀ smaller
positive points.
Therefore ∀x > 0: NUF(x) = ℵ₀ is wrong.

Regards, WM

[WM]

Post by Jim Burns
NUF(x) = |⅟ℕ∩(0,x]| = ℵ₀
does not imply any unit fraction outside (0,x]

But it implies that you can use only x having ℵ₀ smaller positive
points, in fact even ℵ₀*2ℵ₀.

∀ x > 0: NUF(x) = ℵ₀ would be wrong.

Regards, WM

[Jim Burns]

Post by WM

Post by Jim Burns
NUF(x) = |⅟ℕ∩(0,x]| = ℵ₀
does not imply any unit fraction outside (0,x]

But it implies that you can use only
x having ℵ₀ smaller positive points,
in fact even ℵ₀*2ℵ₀.
∀ x > 0: NUF(x) = ℵ₀ would be wrong.

Each non.{}.subset of ⅟ℕ∩(0,1] is maximummed.
∀S ⊆ ⅟ℕ∩(0,1]: S ≠ {} ⇒ ∃u = max.S

Each unit.fraction in ⅟ℕ∩(0,1] is down.stepped.
∀u ∈ ⅟ℕ∩(0,1] ∃v = ⅟(1+⅟u) = max.⅟ℕ∩(0,u)

Each unit.fraction in ⅟ℕ∩(0,1] is non.max.up.stepped.
∀u ∈ ⅟ℕ∩(0,1]: u ≠ max.⅟ℕ∩(0,1] ⇒
∃v = ⅟(-1+⅟u) = min.⅟ℕ∩(u,1]

⎛ Whatever else is
⎜ maximummed and down.stepped and non.max.up.stepped
⎜ has as many as
⎜ maximummed and down.stepped and non.max.up.stepped
⎝ ⅟ℕ∩(0,1]

Each non.{}.subset of ⅟ℕ∩(0,x] is
a non.{}.subset of ⅟ℕ∩(0,1]
and is maximummed.
∀S ⊆ ⅟ℕ∩(0,x]: S ≠ {} ⇒ ∃u = max.S

Each unit.fraction in ⅟ℕ∩(0,x] is
a unit.fraction in ⅟ℕ∩(0,1]
and is down.stepped.
∀u ∈ ⅟ℕ∩(0,x] ∃v = ⅟(1+⅟u) = max.⅟ℕ∩(0,u)

Each unit.fraction in ⅟ℕ∩(0,x] is
a unit.fraction in ⅟ℕ∩(0,1]
and is non.max.up.stepped.
∀u ∈ ⅟ℕ∩(0,x]: u ≠ max.⅟ℕ∩(0,x] ⇒
∃v = ⅟(-1+⅟u) = min.⅟ℕ∩(u,x]

⅟ℕ∩(0,x] is
maximummed and down.stepped and non.max.up.stepped
⅟ℕ∩(0,x] has as many as ⅟ℕ∩(0,1]
|⅟ℕ∩(0,x]| = |⅟ℕ∩(0,1]| = ℵ₀


∀ᴿx > 0: NUF(x) = ℵ₀
because of
maximumming and down.stepping and non.max.up.stepping.

[WM]

Post by Jim Burns

Post by WM

Post by Jim Burns
NUF(x) = |⅟ℕ∩(0,x]| = ℵ₀
does not imply any unit fraction outside (0,x]

But it implies that you can use only
x having ℵ₀ smaller positive points,
in fact even ℵ₀*2ℵ₀.
∀ x > 0: NUF(x) = ℵ₀ would be wrong.

Each non.{}.subset of ⅟ℕ∩(0,1] is maximummed.
∀S ⊆ ⅟ℕ∩(0,1]: S ≠ {} ⇒ ∃u = max.S

In the dark domain the maximum cannot be discerned.

Post by Jim Burns
Each unit.fraction in ⅟ℕ∩(0,1] is down.stepped.
∀u ∈ ⅟ℕ∩(0,1] ∃v = ⅟(1+⅟u) = max.⅟ℕ∩(0,u)

Assuming all this would lead to potential infinity.
I assume actual infinity however. There are dark numbers. There is a
greates natnumber. There is no chance to lose Bob.

Regards, WM

[Jim Burns]

Post by WM

Post by Jim Burns

Post by WM

Post by Jim Burns
NUF(x) = |⅟ℕ∩(0,x]| = ℵ₀
does not imply any unit fraction outside (0,x]

But it implies that you can use only
x having ℵ₀ smaller positive points,
in fact even ℵ₀*2ℵ₀.
∀ x > 0: NUF(x) = ℵ₀ would be wrong.

Each non.{}.subset of ⅟ℕ∩(0,1] is maximummed.
  ∀S ⊆ ⅟ℕ∩(0,1]: S ≠ {}  ⇒  ∃u = max.S

In the dark domain the maximum cannot be discerned.

⅟ℕᶠⁱⁿ∩(0,1] is the set of finite.unit.fractions.

Each non.{}.subset of ⅟ℕᶠⁱⁿ∩(0,1] is maximummed.
∀S ⊆ ⅟ℕᶠⁱⁿ∩(0,1]: S ≠ {} ⇒ ∃u = max.S

Each finite.unit.fraction in ⅟ℕᶠⁱⁿ∩(0,1] is down.stepped.
∀u ∈ ⅟ℕᶠⁱⁿ∩(0,1] ∃v = ⅟(1+⅟u) = max.⅟ℕᶠⁱⁿ∩(0,u)

Each finite.unit.fraction in ⅟ℕᶠⁱⁿ∩(0,1] is non.max.up.stepped.
∀v ∈ ⅟ℕᶠⁱⁿ∩(0,1]: v ≠ max.⅟ℕᶠⁱⁿ∩(0,1] ⇒
∃u = ⅟(-1+⅟v) = min.⅟ℕᶠⁱⁿ∩(v,1]

Each non.{}.subset of ⅟ℕᶠⁱⁿ∩(0,1] is maximummed.
Each finite.unit.fraction in ⅟ℕᶠⁱⁿ∩(0,1] is down.stepped.
Each finite.unit.fraction in ⅟ℕᶠⁱⁿ∩(0,1] is non.max.up.stepped.

Therefore,
the finite.unit.fractions in ⅟ℕᶠⁱⁿ∩(0,1] are ℵ₀.many.

∀ᴿx > 0: ⅟ℕᶠⁱⁿ∩(0,x] ≠ {}

⎛ Assume otherwise.
⎜ Assume ⅟ℕᶠⁱⁿ∩(0,x] = {}
⎜
⎜ β ≥ x > 0 situates the split between
⎜ lower.bounds of ⅟ℕᶠⁱⁿ∩(0,1] and
⎜ not.lower.bounds of ⅟ℕᶠⁱⁿ∩(0,1]
⎜ 2⋅β > β > ½⋅β > 0
⎜
⎜ ½⋅β < β is a lower.bound of ⅟ℕᶠⁱⁿ∩(0,1]
⎜
⎜ 2⋅β > β is a not.lower.bound of ⅟ℕᶠⁱⁿ∩(0,1]
⎜ finite.unit.fraction ⅟k < 2⋅β exists
⎜ finite.unit.fraction ¼⋅⅟k < ¼⋅2⋅β = ½⋅β exists
⎜ ½⋅β is a not.lower.bound of ⅟ℕᶠⁱⁿ∩(0,1]
⎝ Contradiction.

Therefore,
∀ᴿx > 0: ⅟ℕᶠⁱⁿ∩(0,x] ≠ {}

(0,x] inherits from its superset (0,1] properties by which
each non.{}.subset of ⅟ℕᶠⁱⁿ∩(0,x] is maximummed, and
each finite.unit.fraction in ⅟ℕᶠⁱⁿ∩(0,x] is down.stepped, and
each finite.unit.fraction in ⅟ℕᶠⁱⁿ∩(0,x] is non.max.up.stepped.

Therefore,
the finite.unit.fractions in ⅟ℕᶠⁱⁿ∩(0,x] are ℵ₀.many.

∀ᴿx > 0: NUFᶠⁱⁿ(x) = ℵ₀

NUFᶠⁱⁿ(x) ≥ NUF(x)

∀ᴿx > 0: NUF(x) ≥ ℵ₀

[Moebius]

Post by WM
In the dark domain

I'd recommend

as a cure in this case.

[WM]

Post by Jim Burns
∀ᴿx > 0: NUF(x) ≥ ℵ₀

This nonsense will not become true by repeating it. ℵ₀ unit fractions
need ℵ₀*2ℵ₀ points above zero. For those x > 0 your claim is
wrong.

Regards, WM

[Jim Burns]

Post by WM

∀ᴿx > 0:  NUF(x) ≥ ℵ₀

This nonsense will not become true by repeating it.

It does not become false by deleting it.

(0,x] inherits from its superset (0,1] properties by which,
for ⅟ℕᶠⁱⁿ∩(0,x] finite.unit.fractions in (0,x]
each non.{}.subset is maximummed, and
each finite.unit.fraction is down.stepped, and
each finite.unit.fraction in is non.max.up.stepped.

Therefore,
the finite.unit.fractions in ⅟ℕᶠⁱⁿ∩(0,x] are ℵ₀.many.

∀ᴿx > 0: NUFᶠⁱⁿ(x) = ℵ₀

NUF(x) ≥ NUFᶠⁱⁿ(x)

∀ᴿx > 0: NUF(x) ≥ ℵ₀

Post by WM
ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.

ℵ₀ is the cardinality of all final ordinals.
Final ordinal α is smaller than α∪{α}

For each final ordinal α: α∪{α} is also a final ordinal.

The cardinality ℵ₀ of final ordinals can't be a final ordinal.
Otherwise,
there would be at least ℵ₀∪{ℵ₀} final ordinals,
which would be more final ordinals than final ordinals.

Therefore,
ℵ₀ is NOT smaller than ℵ₀∪{ℵ₀}

Post by WM
For those x > 0 your claim is wrong.

Finite doesn't need to be small.
Infinite is beyond big.

[Moebius]

ℵ₀ unit fractions need ℵ₀ [corrected --Moebius] points above zero.

Right. And where is the problem?

For those x > 0 your claim is wrong.

Nope. For each and every of these points [here referred to with the
variable "x"]: NUF(x) = ℵ₀ .

[WM]

Post by Moebius
For each and every of these points [here referred to with the
variable "x"]: NUF(x) = ℵ₀ .

I recognized lately that you use the wrong definition of NUF.
Here is the correct definition:
There exist NUF(x) unit fractions u, such that for all y >= x: u < y.
Note that the order is ∃ u ∀ y.
NUF(x) = ℵ₀ for all x > 0 is wrong. NUF(x) = 1 for all x > 0 already
is wrong since there is no unit fraction smaller than all unit fractions.

ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.

Regards, WM

[joes]

Post by WM

For each and every of these points [here referred to with the variable
"x"]: NUF(x) = ℵ₀ .

I recognized lately that you use the wrong definition of NUF.

It is the definition you have previously used.

Post by WM
There exist NUF(x) unit fractions u, such that for all y >= x: u < y.
Note that the order is ∃ u ∀ y.

You are specifying an exact number, not only at least one.

Post by WM
NUF(x) = ℵ₀ for all x > 0 is wrong. NUF(x) = 1 for all x > 0 already is
wrong since there is no unit fraction smaller than all unit fractions.

New sig.

Post by WM
ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.

[Chris M. Thomasson]

Post by WM

For each and every of these points [here referred to with the variable
"x"]: NUF(x) = ℵ₀ .

I recognized lately that you use the wrong definition of NUF.
There exist NUF(x) unit fractions u, such that for all y >= x: u < y.
Note that the order is ∃ u ∀ y.
NUF(x) = ℵ₀ for all x > 0 is wrong. NUF(x) = 1 for all x > 0 already is
wrong since there is no unit fraction smaller than all unit fractions.
ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.

0->(...)->(1/1)

Contains infinite unit fractions.

0->(...)->(1/2)->(1/1)

Contains infinite unit fractions.

0->(...)->(1/3)->(1/2)->(1/1)

Contains infinite unit fractions.

However, (1/3)->(1/1) is finite and only has three unit fractions
expanded to:

(1/3)->(1/2)->(1/1)

Just like the following has four of them:

(1/4)->(1/3)->(1/2)->(1/1)


(0/1) is not a unit fraction. There is no smallest unit fraction.
However, the is a largest one at 1/1.

A interesting part that breaks the ordering is say well:

(1/4)->(1/2)

has two unit fractions. Then we can make it more fine grain:

(1/4)->(1/2) = ((1/8)+(1/8))->(1/4+1/4)

;^)

[WM]

Post by Jim Burns
(0,x] inherits from its superset (0,1] properties by which,
for ⅟ℕᶠⁱⁿ∩(0,x] finite.unit.fractions in (0,x]
each non.{}.subset is maximummed, and
each finite.unit.fraction is down.stepped, and
each finite.unit.fraction in is non.max.up.stepped.
Therefore,
the finite.unit.fractions in ⅟ℕᶠⁱⁿ∩(0,x] are ℵ₀.many.
∀ᴿx > 0: NUFᶠⁱⁿ(x) = ℵ₀

I recognized lately that you use the wrong definition of NUF.
Here is the correct definition:
There exist NUF(x) unit fractions u, such that for all y >= x: u < y.
Note that the order is ∃ u ∀ y.
NUF(x) = ℵ₀ for all x > 0 is wrong. NUF(x) = 1 for all x > 0 already
is wrong since there is no unit fraction smaller than all unit fractions.

ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.

Regards, WM

[Jim Burns]

Post by WM

Post by Jim Burns
(0,x] inherits from its superset (0,1] properties by which,
for ⅟ℕᶠⁱⁿ∩(0,x] finite.unit.fractions in (0,x]
each non.{}.subset is maximummed,  and
each finite.unit.fraction is down.stepped,  and
each finite.unit.fraction in is non.max.up.stepped.
Therefore,
the finite.unit.fractions in ⅟ℕᶠⁱⁿ∩(0,x] are ℵ₀.many.
∀ᴿx > 0:  NUFᶠⁱⁿ(x) = ℵ₀

I recognized lately that you use
the wrong definition of NUF.
There exist NUF(x) unit fractions u, such that
for all y >= x: u < y.

∀ᴿy ≥ x: y > u ⟺ x > u
⎛ Assume otherwise.
⎜ Assume y ≥ x ∧ ¬(y > u) ∧ x > u
⎜
⎜ However, '>' is transitive.
⎜ y ≥ x ∧ x > u ⇒ y > u
⎝ Contradiction.

Here is an equivalent definition:
There exist NUF(x) unit fractions u, such that
u < x

Post by WM
Note that the order is ∃ u ∀ y.

The order is ∀x ∃u ∀y
∃u ∀x ∀y is an unreliable quantifier shift.

Post by WM
NUF(x) = ℵ₀ for all x > 0 is wrong.
NUF(x) = 1 for all x > 0 already is wrong since
there is no unit fraction smaller than all unit fractions.

NUF(x) > 1 for all x > 0 is correct since
each unit.fraction is larger than at least two unit.fractions.
and
a positive lower.bound of finite unit.fractions
implies
finite unit.fractions below a lower.bound,
a contradiction.

Post by WM
ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.

The finite unit.fractions in (0,x] are
maximummed and down.stepped and non.max.up.stepped.
The finite unit.fractions in (0,x] are
ℵ₀.many.

[Moebius]

Post by Jim Burns

Post by WM
ω-1 ∈ ℕ

Holy shit!

What does "ω-1" even mean?

A "reasonable" interpretation might be:

ω-1 is the natural number k such that k+1 = ω.

The only problem with this interpretation is that there is no natural
number k such that k+1 = ω.

So (in a mathematical context) we can't define:

ω-1 =df the natural number k such that k+1 = ω.

Post by Jim Burns
'Defining into existence' that which doesn't exist
makes nonsense of whatever meaning one's words have.

Right.

Post by Jim Burns

Post by WM
ℕ = {1, 2, 3, ..., ω-1}

*sigh*

[Moebius]

Post by Jim Burns
sci.logic and sci.math

_The rule of subset_ proves that [bla bla bla]

Where did Mückenheim get this "rule" from? Any source?

[Jim Burns]

Post by Moebius

_The rule of subset_ proves that [bla bla bla]

Where did Mückenheim get this "rule" from?
Any source?

Mückenheim is wishcasting.
The rule is that no sets are infiniteⁿᵒᵗᐧᵂᴹ.
The counter.argument is the roster of
the usual suspects: ℕⁿᵒᵗᐧᵂᴹ ℤⁿᵒᵗᐧᵂᴹ ℚⁿᵒᵗᐧᵂᴹ ℝⁿᵒᵗᐧᵂᴹ

It's underwhelming.
And yet,
his rule is better exposition than I can recall,
out of years of much.the.same,
to himself of his own thinking.

Perhaps one old dog can teach himself one new trick?

[FromTheRafters]

Post by Moebius

Post by Jim Burns
sci.logic and sci.math

_The rule of subset_ proves that [bla bla bla]

Where did Mückenheim get this "rule" from? Any source?

He made it up, and it is obviously wrong. He'll 'prove' it though using
his delusion monotony.

[WM]

Post by Moebius

Post by Jim Burns
sci.logic and sci.math

_The rule of subset_

Where did Mückenheim get this "rule" from? Any source?

Logic! If A contains all elements of B, but B does not contain all
elements of A, then A has more elements than B.

Regards, WM

[Tom Bola]

Post by WM

Post by Moebius

Post by Jim Burns
sci.logic and sci.math

_The rule of subset_

Where did Mückenheim get this "rule" from? Any source?

Logic! If A contains all elements of B, but B does not contain all
elements of A, then A has more elements than B.

Cardinality within Math is not just a matter of "more", goofy...

[WM]

Post by Tom Bola

Post by WM
Logic! If A contains all elements of B, but B does not contain all
elements of A, then A has more elements than B.

Cardinality within Math is not just a matter of "more",

No, it is simply nonsense.

Regards, WM

[Tom Bola]

Post by WM

Post by Tom Bola

Post by WM
Logic! If A contains all elements of B, but B does not contain all
elements of A, then A has more elements than B.

Cardinality within Math is not just a matter of "more",

No, it is simply nonsense.

Not for 99,9999999 percent of the educated mankind, but just for
your few parts of that sort of fully idiotic very low IQ clowns...

[FromTheRafters]

Post by Moebius

Post by Jim Burns
sci.logic and sci.math

_The rule of subset_

Where did Mückenheim get this "rule" from? Any source?

Logic! If A contains all elements of B, but B does not contain all elements
of A, then A has more elements than B.

Have you reviewed 'Cardinal Arithmetic' lately? I know it has been
pointed out to you several times. The aforementioned does not work for
infinite sets. Addition like this simply doesn't affect the 'size' of
the set if the set is infinite.

[Tom Bola]

Post by FromTheRafters

Post by Moebius

Post by Jim Burns
sci.logic and sci.math

_The rule of subset_

Where did Mückenheim get this "rule" from? Any source?

Logic! If A contains all elements of B, but B does not contain all elements
of A, then A has more elements than B.

Have you reviewed 'Cardinal Arithmetic' lately? I know it has been
pointed out to you several times. The aforementioned does not work for
infinite sets. Addition like this simply doesn't affect the 'size' of
the set if the set is infinite.

A billion times "discussed" under the buzzword "Dedekind-Infinity"...

The point is, that, for 50++ years, WM does not WANT (accept) our Math.

-
https://de.wikipedia.org/wiki/Unendliche_Menge#Dedekind-Unendlichkeit

[joes]

But what number became ω when doubled?

ω/2

And where is that in {1, 2, 3, ... w} ?

In the midst, far beyond all definable numbers, far beyond ω/10^10.

That is a bit imprecise. Even though you keep on talking about
consecutive infinities, you can't compare natural and "dark" numbers.

The input set was the Natural Numbers and w,

ω/10^10 and ω/10 are dark natural numbers.

If all natural numbers exist, then ω-1 exists.

Why?

Because otherwise there was a gap below ω.

But you combined two different sets, so why can't there be a gap?

I assume completness.

Completeness of N? No number n reaches omega.

∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every unit
fraction 1/n, there exists another unit fraction smaller than itself.

No. My formula says ∀n ∈ ℕ.

That is not a contradiction.

Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists.

Not for all dark numbers.

[WM]

Post by joes

But what number became ω when doubled?

ω/2

And where is that in {1, 2, 3, ... w} ?

In the midst, far beyond all definable numbers, far beyond ω/10^10.

That is a bit imprecise. Even though you keep on talking about
consecutive infinities, you can't compare natural and "dark" numbers.

Dark natnumbers are larger than defined natnumbers. Even dar natnumbers
can be compared by size. ω/10^10 < ω/10 < ω/2, < ω-1.

Post by joes

ω/10^10 and ω/10 are dark natural numbers.

If all natural numbers exist, then ω-1 exists.

Why?

Because otherwise there was a gap below ω.

But you combined two different sets, so why can't there be a gap?

I assume completness.

Completeness of N? No number n reaches omega.

What is immediately before ω? Is it a blasphemy to ask such questions?

Post by joes

∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every unit
fraction 1/n, there exists another unit fraction smaller than itself.

No. My formula says ∀n ∈ ℕ.

That is not a contradiction.

It is not a contradiction to my formula if some n has no n+1.

Regards, WM

[joes]

Post by WM

Post by joes

But what number became ω when doubled?

ω/2

And where is that in {1, 2, 3, ... w} ?

In the midst, far beyond all definable numbers, far beyond ω/10^10.

That is a bit imprecise. Even though you keep on talking about
consecutive infinities, you can't compare natural and "dark" numbers.

Dark natnumbers are larger than defined natnumbers. Even dar natnumbers
can be compared by size. ω/10^10 < ω/10 < ω/2, < ω-1.

How is this different from two consecutive infinities?

Post by WM

Post by joes

ω/10^10 and ω/10 are dark natural numbers.

If all natural numbers exist, then ω-1 exists.

Why?

Because otherwise there was a gap below ω.

But you combined two different sets, so why can't there be a gap?

I assume completness.

Completeness of N? No number n reaches omega.

What is immediately before ω? Is it a blasphemy to ask such questions?

There is no "before".

Post by WM

Post by joes

∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every
unit fraction 1/n, there exists another unit fraction smaller than
itself.

No. My formula says ∀n ∈ ℕ.

That is not a contradiction.

... to infinitely many unit fractions.

[Moebius]

Post by WM
ω/10^10 < ω/10 < ω/2, < ω-1.

<facepalm>

Keiner dieser Ausdrücke ist definiert. Diese "Zahlen" existieren nur in
Mückenheims Wahnsystem (->delusion).

How is this [...]

It's pure nonsense.

Post by WM
ω/10^10 and ω/10 are dark [...] numbers.

Ja, in seinem Wahnsystem wird es wohl so sein.

Es könnte aber auch brauner Scheißdreck sein.

Post by WM
If all natural numbers exist, then ω-1 exists.

Ah? (->Wahnsystem)

Post by WM
What is immediately before ω? Is it <bla>

NOTHING is "immediately before ω".

Genauer: Keine Ordinalzahl.

In Zeichen: ~Eo e ORD: o + 1 = ω.

@Mückenheim: Sie faseln nur saudummen Scheißdreck daher.

[WM]

Post by Moebius

Post by WM
ω/10^10 < ω/10 < ω/2, < ω-1.

Keiner dieser Ausdrücke ist definiert.

They are not defined by FISONs. Their relative sizes do not need FISONs to
be comparable.

Post by Moebius

Post by WM
What is immediately before ω?

NOTHING is "immediately before ω".
Genauer: Keine Ordinalzahl.

How close do the ordinal numbers 1, 2, 3, ... come to their limit ω?

Regards, WM

[joes]

Post by WM

Post by Moebius

Post by WM
ω/10^10 < ω/10 < ω/2, < ω-1.

Keiner dieser Ausdrücke ist definiert.

They are not defined by FISONs. Their relative sizes do not need FISONs
to be comparable.

Do you imagine N as symmetrically countable from either end, with
something in between?

Post by WM

Post by Moebius

Post by WM
What is immediately before ω?

NOTHING is "immediately before ω".
Genauer: Keine Ordinalzahl.

How close do the ordinal numbers 1, 2, 3, ... come to their limit ω?

Not at all. They are all infinitely far away, or rather omega is
infinitely far away from all of them. The infinity is at infinity.
Being a limit, this distance does not shrink. In the same sense as
cardinality, none of them get any closer. The limit is not a part of
the sequence, i.e. the natural numbers. It lies outside and beyond.
It is not even a proper value: the sequence diverges, it transcends
all bounds.

[WM]

Post by joes
Do you imagine N as symmetrically countable from either end, with
something in between?

Yes. An unbridgeable dark space exists between both ends like in the reals
between two known real numbers.

Post by joes

Post by WM

Post by Moebius

Post by WM
What is immediately before ω?

NOTHING is "immediately before ω".
Genauer: Keine Ordinalzahl.

How close do the ordinal numbers 1, 2, 3, ... come to their limit ω?

Not at all. They are all infinitely far away, or rather omega is
infinitely far away from all of them. The infinity is at infinity.
Being a limit, this distance does not shrink. In the same sense as
cardinality, none of them get any closer.

Cardinality is nonsense. It concerns only definable numbers. 2 is closer
to ω than 1, but it is not easy to see. We cannot count through the dark
numbers. But 1/2 is closerto 0 than 1/1. That is fact.

Post by joes
The limit is not a part of
the sequence, i.e. the natural numbers. It lies outside and beyond.

The limit is the ωth term.

Post by joes
It is not even a proper value: the sequence diverges, it transcends
all bounds.

The reciprocal is easier to understand.

Regards, WM

[FromTheRafters]

Post by Moebius

Post by WM
ω/10^10 < ω/10 < ω/2, < ω-1.

Keiner dieser Ausdrücke ist definiert.

They are not defined by FISONs. Their relative sizes do not need FISONs to be
comparable.

Post by Moebius

Post by WM
What is immediately before ω?

NOTHING is "immediately before ω".
Genauer: Keine Ordinalzahl.

How close do the ordinal numbers 1, 2, 3, ... come to their limit ω?

It is merely the symbol for the countably infinite ordered set of
naturals. It represents all of the ordinals which came before it, and
yet it is the first of its kind, the first transfinite ordinal.

You might as well be asking what natural number comes before the least
natural number.

[WM]

Post by FromTheRafters

Post by WM
How close do the ordinal numbers 1, 2, 3, ... come to their limit ω?

It is merely the symbol for the countably infinite ordered set of
naturals. It represents all of the ordinals which came before it, and
yet it is the first of its kind, the first transfinite ordinal.
You might as well be asking what natural number comes before the least
natural number.

No. Something comes there on the real line. No gaps larger than 1. If we
extend the real line to ω, then my question is legitimate.

Regards, WM

[WM]

Post by joes

Dark natnumbers are larger than defined natnumbers. Even dark natnumbers
can be compared by size. ω/10^10 < ω/10 < ω/2, < ω-1.

How is this different from two consecutive infinities?

All numbers above a finite natural numbers.

Post by joes

What is immediately before ω? Is it a blasphemy to ask such questions?

There is no "before".

There is the sequence 1, 2, 3, ... and ω is its limit. How close to ω do
the terms reach?

Regards, WM

[Jim Burns]

Post by WM

Post by joes

∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Note the universal quantifier.

Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1),
so that for every unit fraction 1/n, there exists
another unit fraction smaller than itself.

No. My formula says ∀n ∈ ℕ.

That is not a contradiction.

It is not a contradiction to my formula
if some n has no n+1.

No, it literally contradicts your formula
for some n e N to not.have n+1

Your formula uses 'n+1'
Along with saying other things,
using 'n+1' asserts that 'n+1' exists, ∀n ∈ ℕ

That assertion is so often uncontroversial
that it's easy to overlook.
Nonetheless, it is there.

[Moebius]

Post by Jim Burns

Post by WM

Post by joes

∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Note the universal quantifier.

Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1),
so that for every unit fraction 1/n, there exists
another unit fraction smaller than itself.

No. My formula says ∀n ∈ ℕ.

That is not a contradiction.

It is not a contradiction to my formula
if some n has no n+1.

No, it literally contradicts your formula
for some n e N to not.have n+1
Your formula uses 'n+1'
Along with saying other things,
using 'n+1' asserts*) that 'n+1' exists [for all] n ∈ ℕ
That assertion is so often uncontroversial
that it's easy to overlook.
Nonetheless, it is there.

Yes, I already told him that in dsm, but to no avail.

________________________________

*) or presupposes

[WM]

Post by Jim Burns

Post by WM
It is not a contradiction to my formula
if some n has no n+1.

No, it literally contradicts your formula
for some n e N to not.have n+1

My formula is explicitly valid only for natural numbers.

Regards, WM

[Jim Burns]

Post by WM

Post by Jim Burns

Post by WM

∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Note the universal quantifier.

It is not a contradiction to my formula
if some n has no n+1.

No, it literally contradicts your formula
for some n e N to not.have n+1

My formula is explicitly valid only for natural numbers.

Your formula means
∀n ∈ ℕ: ∃k ∈ ℕ: 1/n - 1/k > 0 ∧
k=n+1 ∧ ¬∃k₂≠k: k₂=n+1

Operations have unique values.
'+' is an operation.
∀n ∈ ℕ: ∃k ∈ ℕ: k=n+1

[Moebius]

Post by WM

Post by Jim Burns

Post by WM

∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Note the universal quantifier.

It is not a contradiction to my formula
if some n has no n+1.

No, it literally contradicts your formula
for some n e N to not.have n+1

My formula is explicitly valid only for natural numbers.

This is one of Mückenheim's brainfarts.

I guess what he means is:

"1/n - 1/(n+1) > 0" is (especially) valid for (all) natural numbers n ,

or something like that.

To state that a closed formula / sentence / statement "is valid ... for
natural numbers" is just mumbo-jumbo.

[Richard Damon]

Post by WM

Post by joes

But what number became ω when doubled?

ω/2

And where is that in {1, 2, 3, ... w} ?

In the midst, far beyond all definable numbers, far beyond ω/10^10.

That is a bit imprecise. Even though you keep on talking about
consecutive infinities, you can't compare natural and "dark" numbers.

Dark natnumbers are larger than defined natnumbers. Even dar natnumbers
can be compared by size. ω/10^10 < ω/10 < ω/2, < ω-1.

Post by joes

ω/10^10 and ω/10 are dark natural numbers.

If all natural numbers exist, then ω-1 exists.

Why?

Because otherwise there was a gap below ω.

But you combined two different sets, so why can't there be a gap?

I assume completness.

Completeness of N? No number n reaches omega.

What is immediately before ω? Is it a blasphemy to ask such questions?

It has no predicessor, just like in the Natural Numbers 0 has nothing
before it.

You can expand your number system to define number there, which seems to
be what you "dark numbers" are, numbers bigger than all the finite
Natural Numbers, but smaller than w.

Post by WM

Post by joes

∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every unit
fraction 1/n, there exists another unit fraction smaller than itself.

No. My formula says ∀n ∈ ℕ.

That is not a contradiction.

It is not a contradiction to my formula if some n has no n+1.

It is a violation of the DEFINITION of the Natural Numbers.

Now, if you admit that you system doesn't actaully HAVE the actaual
Natura Numbers, but some other set built by some other methd, go ahead
and try to actually define that set and see what you can do with it.

Just don't claim that anyone else needs to use your inferior number system.

Post by WM
Regards, WM

[WM]

Post by Richard Damon

Post by WM
What is immediately before ω? Is it a blasphemy to ask such questions?

It has no predicessor, just like in the Natural Numbers 0 has nothing
before it.

0 has a continuum above it, no gap! Likewise there must be no gap below
ω.

Post by Richard Damon
You can expand your number system to define number there, which seems to
be what you "dark numbers" are, numbers bigger than all the finite
Natural Numbers, but smaller than w.

Thank you.

Post by Richard Damon

Post by WM
It is not a contradiction to my formula if some n has no n+1.

It is a violation of the DEFINITION of the Natural Numbers.

Otherwise there is a contradiction of mathematics: separated unit
fractions.

Regards, WM

[Richard Damon]

Post by Richard Damon

Post by WM
What is immediately before ω? Is it a blasphemy to ask such questions?

It has no predicessor, just like in the Natural Numbers 0 has nothing
before it.

0 has a continuum above it, no gap! Likewise there must be no gap below ω.

Post by Richard Damon
You can expand your number system to define number there, which seems
to be what you "dark numbers" are, numbers bigger than all the finite
Natural Numbers, but smaller than w.

Thank you.

Post by Richard Damon

Post by WM
It is not a contradiction to my formula if some n has no n+1.

It is a violation of the DEFINITION of the Natural Numbers.

Otherwise there is a contradiction of mathematics: separated unit
fractions.
Regards, WM

No, all Unit fractions are separated in distance by a finite amount, it
is just that you can't specify any finite distance that separates all
unit fractions.

The reversal of the order of qualifications causes the problem.

This is the nature of unbounded numbers, something your logic doesn't
seem to be able to handle.

Every number 1/n is separated from the next smaller unit fraction,
1/(n+1) by a distance of 1/(n*(n+1)) which is a value that is greater
than zero, so we always have a finite difference between all unit
fractions, but that distance gets arbitrarily small, so we can't choose
a single finite eps that all unit fractions are seperated by, even
though all unit fractions are seperated by a finite distance.

This just shows that this spacing, APPROACHES 0, as a limit, as n
increases. But approaching a limit of 0 is not the same as being 0.

[WM]

Post by Richard Damon

Post by WM

Post by Richard Damon

Post by WM
It is not a contradiction to my formula if some n has no n+1.

It is a violation of the DEFINITION of the Natural Numbers.

Otherwise there is a contradiction of mathematics: separated unit
fractions.

No, all Unit fractions are separated in distance by a finite amount, it
is just that you can't specify any finite distance that separates all
unit fractions.

I don't try. But I know that one unit fraction must be the first.

Post by Richard Damon
The reversal of the order of qualifications causes the problem.

Distances on the real line imply an order, even it it cannot be discerned.

Post by Richard Damon
Every number 1/n is separated from the next smaller unit fraction,
1/(n+1) by a distance of 1/(n*(n+1)) which is a value that is greater
than zero, so we always have a finite difference between all unit
fractions, but that distance gets arbitrarily small,

Can they get smaller than 2^ℵ₀ points?

If not ℵ₀ unit fractions need ℵ₀*2^ℵ₀ points above zero. For
those x > 0 the claim
∀x > 0: NUF(x) ≥ ℵ₀
is wrong.

Regards, WM

[joes]

But what number became ω when doubled?

ω/2

And where is that in {1, 2, 3, ... w} ?

In the midst, far beyond all definable numbers, far beyond ω/10^10.

In other words, outside the Natural Nubmer, all of which are defined
and definable.

That is simply nonsense. Do you know what an accumalation point is?
Every eps interval around 0 contains unit fractions which cannot be
separated from 0 by any eps. Therefore your claim is wrong.

No. There is ALWAYS an epsilon.

The input set was the Natural Numbers and w,

ω/10^10 and ω/10 are dark natural numbers.

They may be "dark" but they are not Natural Numbers.

They are natural numbers.

How are they defined?

Natural numbers, by their definition, are reachable by a finite number
of successor operations from 0.

That is the opinion of Peano and his disciples. It holds only for
potetial infinity, i.e., definable numbers.

I assume completness.

I guess you definition of "completeness" is incorrect.
If I take the set of all cats, and the set of all doges, can there not
be a gap between them?

What is the reason for the gap before omega? How large is it? Are these
questions a blasphemy?

A "gap" implies some sort of space that is not filled. There is no such
space (it would be filled with infinitely many natural numbers).
We just condense the whole of N into one concept and call that omega,
or add it on the next level of infinity.
Your questions are only a display of your unwillingness to understand
infinity, even though you would like to imagine yourself as some sort
of martyr.

∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every
unit fraction 1/n, there exists another unit fraction smaller than
itself.

No. My formula says ∀n ∈ ℕ.

Right, for ALL n in ℕ, there exist another number in ℕ that is n+1,

That does ny formula not say. It says for all n which have successors,
there is distance between 1/n and 1/(n+1).

If k did not have a successor, what would k+1 be?

Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists.

Not for all dark numbers.

Maybe not for dark numbers, but it does for all Natural Numbers, as
that is part of their DEFINITION.

It is the definition of definable numbers. Study the accumulation point.
Define (separate by an eps from 0) all unit fractions. Fail.

[WM]

Post by joes

separated from 0 by any eps. Therefore your claim is wrong.

No. There is ALWAYS an epsilon.

Failing to separate almost all unit fractions.
Don't claim the contrary. Define (separate by an eps from 0) all unit
fractions. Fail.

Post by joes

What is the reason for the gap before omega? How large is it? Are these
questions a blasphemy?

A "gap" implies some sort of space that is not filled. There is no such
space (it would be filled with infinitely many natural numbers).

Hence there is only the sequence of natnumbers.

Post by joes
We just condense the whole of N into one concept and call that omega,

That is nonsense. ω is the first number following upon all natural
numbers.

Post by joes
or add it on the next level of infinity.

Yes.

Post by joes
Your questions are only a display of your unwillingness to understand
infinity,

They prove that I understand the real infinity while are (con)fusing
potential and actual infinity.

Post by joes
If k did not have a successor, what would k+1 be?

ω

Regards, WM

[joes]

Post by WM

Post by joes

Every eps interval around 0 contains unit fractions which cannot be
separated from 0 by any eps. Therefore your claim is wrong.

No. There is ALWAYS an epsilon.

Failing to separate almost all unit fractions.
Don't claim the contrary. Define (separate by an eps from 0) all unit
fractions. Fail.

Well, there's no epsilon that separates all positive numbers from zero.
But every fraction has an epsilon that is smaller.

Post by WM

Post by joes

What is the reason for the gap before omega? How large is it? Are
these questions a blasphemy?

A "gap" implies some sort of space that is not filled. There is no such
space (it would be filled with infinitely many natural numbers).

Hence there is only the sequence of natnumbers.

And no sequence of omega - k.

Post by WM

Post by joes
We just condense the whole of N into one concept and call that omega,

That is nonsense. ω is the first number following upon all natural
numbers.

Following the WHOLE of the natural numbers. The successor of a natural
is also one.

Post by WM

Post by joes
If k did not have a successor, what would k+1 be?

ω

Ah, a natural number.

[WM]

Post by joes

Post by WM
Don't claim the contrary. Define (separate by an eps from 0) all unit
fractions. Fail.

Well, there's no epsilon that separates all positive numbers from zero.
But every fraction has an epsilon that is smaller.

Note that epsilons must be chosen. You cannot choose an eps that separates
more than few unit fractions.

If you object, then simply apply all of them. Then every fraction could be
separated and none would remain. Then you had contradicted the existence
of accumulation points.

Post by joes

Post by WM
Hence there is only the sequence of natnumbers.

And no sequence of omega - k.

Those are natnumbers.

Post by joes

Post by WM

Post by joes
We just condense the whole of N into one concept and call that omega,

That is nonsense. ω is the first number following upon all natural
numbers.

Following the WHOLE of the natural numbers. The successor of a natural
is also one.

Not for all.

Post by joes

Post by WM

Post by joes
If k did not have a successor, what would k+1 be?

ω

Ah, a natural number.

No, ω is the first transfinite number. But we cannot count from 1 to ω.

Regards, WM

[Richard Damon]

Post by WM

Post by joes

separated from 0 by any eps. Therefore your claim is wrong.

No. There is ALWAYS an epsilon.

Failing to separate almost all unit fractions.
Don't claim the contrary. Define (separate by an eps from 0) all unit
fractions. Fail.

Improperly revesing the conditionals.

Post by WM

Post by joes

What is the reason for the gap before omega? How large is it? Are these
questions a blasphemy?

A "gap" implies some sort of space that is not filled. There is no such
space (it would be filled with infinitely many natural numbers).

Hence there is only the sequence of natnumbers.

Nope, just proof of the ignorance of WM.

Post by WM

Post by joes
We just condense the whole of N into one concept and call that omega,

That is nonsense. ω is the first number following upon all natural numbers.

No, it is the first TRANSFINITE number beyond all the natural numbers,
one which has no predicesors.

Post by WM

Post by joes
or add it on the next level of infinity.

Yes.

Post by joes
Your questions are only a display of your unwillingness to understand
infinity,

They  prove that I understand the real infinity while are (con)fusing
potential and actual infinity.

Nope, just that you use broken logic,

Post by WM

Post by joes
If k did not have a successor, what would k+1 be?

ω
Regards, WM

[WM]

Post by Richard Damon

Post by WM

Post by joes

separated from 0 by any eps. Therefore your claim is wrong.

No. There is ALWAYS an epsilon.

Failing to separate almost all unit fractions.
Don't claim the contrary. Define (separate by an eps from 0) all unit
fractions. Fail.

Improperly revesing the conditionals.

Not at all! Recognizing that eps must be chosen. You cannot choose a eps
that separates more than few unit fractions. That is why most cranks claim
∀x > 0: NUF(x) ≥ ℵ₀. It holds or all x that can be chosen. How
should there be always an epsilon smaller than every x > 0 which fails? ?
?

Regards, WM

[Richard Damon]

Post by WM

Post by Richard Damon

Post by WM

Post by joes

separated from 0 by any eps. Therefore your claim is wrong.

No. There is ALWAYS an epsilon.

Failing to separate almost all unit fractions.
Don't claim the contrary. Define (separate by an eps from 0) all unit
fractions. Fail.

Improperly revesing the conditionals.

Not at all! Recognizing that eps must be chosen. You cannot choose a eps
that separates more than few unit fractions. That is why most cranks
claim ∀x > 0:  NUF(x) ≥ ℵ₀. It holds or all x that can be chosen. How
should there be always an epsilon smaller than every x > 0 which fails? ? ?
Regards, WM

But you have the condition backwards.

The claim is that there is a finite difference that seperates any two
specific unit fractions,

Not that there is a single finite difference that seperates any two
arbirtry unit fractions.

Not understanding the order of the arguement just blow your logic up.

For your example, for any x, there is an eps = 1/(ceil(1/x)+1) that
provides a unit fraction smaller than x, and thus NUF(x) can not be 1
for any finite x.

The fact that we get a different eps for every x is not a problem, it is
just a property of the unbounded nature of the numbers we are using.

[WM]

Post by Richard Damon
The claim is that there is a finite difference that seperates any two
specific unit fractions,

Of course.

Post by Richard Damon
Not that there is a single finite difference that seperates any two
arbirtry unit fractions.

Between any two arbitrary unit fractions there are ℵ₀*2ℵ₀ points.

Post by Richard Damon
Not understanding the order of the arguement

Between any two arbitrary unit fractions there are ℵ₀*2ℵ₀ points.
No order necessary.

Post by Richard Damon
For your example, for any x, there is an eps = 1/(ceil(1/x)+1) that
provides a unit fraction smaller than x,

No, eps has to be chosen. x has to be chosen. My claim is that most are
dark and cannot be chosen.
Note: For every 1/n there exists a smaller real number. But they cannot be
chosen because most 1/n already cannot be chosen. Proof: For every chosen
eps you fail to separate infinitely many unit fractions.

Post by Richard Damon
and thus NUF(x) can not be 1
for any finite x.

But for x belonging to the first ℵ₀*2ℵ₀ points we have less than
ℵ₀ unit fractions.

Regards, WM

[Richard Damon]

Post by WM

Post by Richard Damon
The claim is that there is a finite difference that seperates any two
specific unit fractions,

Of course.

Post by Richard Damon
Not that there is a single finite difference that seperates any two
arbirtry unit fractions.

Between any two arbitrary unit fractions there are ℵ₀*2ℵ₀ points.

So? that number is just ℵo (you don't understand the mathematics of
transfinite values)

Post by WM

Post by Richard Damon
Not understanding the order of the arguement

Between any two arbitrary unit fractions there are ℵ₀*2ℵ₀ points. No
order necessary.

So? That just shows that between any two unit fractions there IS a
distance, and thus they are seperated, as required.

Post by WM

Post by Richard Damon
For your example, for any x, there is an eps = 1/(ceil(1/x)+1) that
provides a unit fraction smaller than x,

No, eps has to be chosen. x has to be chosen. My claim is that most are
dark and cannot be chosen.

It has to be chosen for a given pair of unit fractions.

Post by WM
Note: For every 1/n there exists a smaller real number. But they cannot
be chosen because most 1/n already cannot be chosen. Proof: For every
chosen eps you fail to separate infinitely many unit fractions.

WHY CAN'T IT BW CHOSEN? That seems to be the flaw in your logic, you
think there exist numbers that are defined mathematically, but can't be
chosen.

If there WAS just a single eps that seperated ALL unit fractions, then
the could be no more than 1/eps unit fractions, and thus that would be
larger than the largest Natural Number.

But EVERY Natural Number has a successor, so there is no largest Natural
Number.

This just shows that your logic is limited and doesn't handle the sets
you are trying to us it one, and it breaks.

Post by WM

Post by Richard Damon
and thus NUF(x) can not be 1 for any finite x.

But for x belonging to the first ℵ₀*2ℵ₀ points we have less than ℵ₀ unit
fractions.

Nope, because ℵo-1 = ℵo, and thus there are ℵo points below (and above)
every point. All values of ℵo are not ordered within themselves.

If you math can't handle that, it can't handle the infinities. Since it
doesn't handle the unbounded numbers, that it doesn't handle infinities
isn't surprizing.

Post by WM
Regards, WM

[WM]

Post by Richard Damon

Post by WM
Note: For every 1/n there exists a smaller real number. But they cannot
be chosen because most 1/n already cannot be chosen. Proof: For every
chosen eps you fail to separate infinitely many unit fractions.

WHY CAN'T IT BW CHOSEN?

Because it is dark.

Post by Richard Damon
If there WAS just a single eps that seperated ALL unit fractions, then
the could be no more than 1/eps unit fractions,

But there is not even an eps that separates half of all unit fractions.

Regards, WM

[Richard Damon]

Post by WM

Post by WM
Note: For every 1/n there exists a smaller real number. But they
For every chosen eps you fail to separate infinitely many unit
fractions.

WHY CAN'T IT BE CHOSEN?

Because it is dark.

Why is it dark, we know a possible value for it, and thus can choose it.

You just incorrectly assume that some numbers just can't be reached
because some 'magic" made them dark.

No, your dark numbers are non-finite numbers that you invent to try to
explain why your logic doesn't work with unbounded numbers.

Post by WM

If there WAS just a single eps that seperated ALL unit fractions, then
the could be no more than 1/eps unit fractions,

But there is not even an eps that separates half of all unit fractions.

Because such a question is meaningles, as there isn't a finite number
that is half of the count of unit fractions.

This is the problem with your logic, it doens't understand that the size
of the unbounded set isn't a finite number that you can do normal math with.

Post by WM
Regards, WM

[WM]

Post by joes
A "gap" implies some sort of space that is not filled. There is no such
space (it would be filled with infinitely many natural numbers).
We just condense the whole of N into one concept and call that omega,
or add it on the next level of infinity.

The limit is not a next level but extends the ordinals just like 0 extends
their reciprocals.

Regards, WM