On Sunday, 9 October 2022 at 06:02:36 UTC-3, WM wrote:
[...]
> Yes, there is no gap. And if ω exists, then there is no gap between ω and ℕ. But every leap from ω into ℕ covers infinitely many natural numbers.

You're really flipped your wig this time! WTF is a "leap from ω into ℕ"? What even do *YOU* mean by a "gap between ω and ℕ"? More basically, is ω =/= ℕ for *YOU*?

On 10/9/2022 5:39 AM, WM wrote:
> Jim Burns schrieb am Samstag,
> 8. Oktober 2022 um 20:39:13 UTC+2:
>> On 10/7/2022 3:42 PM, WM wrote:

>>> We agree that the leap can end at every
>>> desired FISON end.
>>> But aleph_0 natural numbers are covered by
>>> every leap.
>>
>> For simplicity, consider only 0 and ω
>> and whatever is between them.
>>
>> Split them between
>> what can be reached one-by-one from 0
>> and what can't be reached that way.
>>
>> What's needed in order to reach one-by-one?
>>
>> One-by-one, whichever way they're split
>> into before and after,
>> just-one-more must be possible.
>> For each split,
>> a from-before-place and a to-after-place,
>> with none between, must exist.
>>
>> Each place on the 0-side of the split
>> can be reached one-by-one from 0
>
> Every FISON-end can be reached one-by-one.

Everything on the 0-side of the 0-side,ω-side
split can be reached one-by-one.

Nothing on the 0-side of the 0-side,ω-side
split cannot be reached one-by-one.

> But every FISON contains less than half of
> the infinite initial segment,
> i.e., of all natural numbers {1, 2, 3, ...}.
> Proof:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> or
> ∀n ∈ ℕ_def: 2n < ℵo ==> n < ℵo/2.

The 0-side of the 0-side,ω-side split
contains _all_ of the 0-side of the
0-side,ω-side split, without exception.

No FISON is the 0-side of the 0-side,ω-side
split.

> Nevertheless it is impossible to distinguish
> different infinite initial segments.

A unique _least_ (⊆) infinite initial
segment 𝕃𝕀𝕀𝕊 exists
such that
𝕃𝕀𝕀𝕊 is an infinite initial segment
and,
if 𝔸 is any infinite initial segment,
then 𝕃𝕀𝕀𝕊 ⊆ 𝔸
[1]

Suppose
"another" least infinite initial
segment 𝕃𝕀𝕀𝕊₂ exists
such that
𝕃𝕀𝕀𝕊₂ is an infinite initial segment
and,
if 𝔸 is any initial infinite segment,
then 𝕃𝕀𝕀𝕊₂ ⊆ 𝔸

Then
𝕃𝕀𝕀𝕊₂ ⊆ 𝕃𝕀𝕀𝕊 and 𝕃𝕀𝕀𝕊 ⊆ 𝕃𝕀𝕀𝕊₂
and
𝕃𝕀𝕀𝕊₂ = 𝕃𝕀𝕀𝕊

𝕃𝕀𝕀𝕊 is unique.

𝕃𝕀𝕀𝕊 is the 0-side of the 0-side,ω-side
split.

if
P(0) ∧ ∀j ∈ 𝕃𝕀𝕀𝕊 : P(j) ⟹ P(j+1)
then
∀k ∈ 𝕃𝕀𝕀𝕊 : P(k)

We define ℕ to be 𝕃𝕀𝕀𝕊

> Why? Because the additional numbers are dark.

The additional numbers, if dark, if existing,
are on the ω-side of the 0-side,ω-side split.

The 0-side,ω-side split is not just-one-more.
The reason is that
the 0-side does not contain a last-before.
(To be just-one-more, there must be a last-before,
and a first-after, and none between.)

Whatever the ω-side of the 0-side,ω-side split
contains or does not contain, the 0-side,ω-side
split is not just-one-more, and leaps which
cross the 0-side,ω-side split cannot be finite.

Ultimately, the reason is that
𝐹ₖ⊕⟨k+1⟩ is a FISON too.

----
[1]
𝔸 is an initial infinite segment
⟺
0 ∈ 𝔸 ∧ ∀k ∈ 𝔸 : k+1 ∈ 𝔸

----
A unique least (⊆) infinite initial
segment 𝕃𝕀𝕀𝕊 exists.

| Assume that infinite initial segments 𝔸 𝔹
| exist.
| | 𝓘𝓘𝓢(𝔸) =
| { 𝕊 ⊆ 𝔸 : 𝕊 is an inf. init. seg. }
| the set of all infinite initial segments ⊆ 𝔸
| 𝓘𝓘𝓢(𝔸) contains at least 𝔸
| | ⋂𝓘𝓘𝓢(𝔸) =
| the intersection of all infinite initial
| segments ⊆ 𝔸
| == the set of all k in all 𝕊 in 𝓘𝓘𝓢(𝔸)
| | 𝕃𝕀𝕀𝕊(𝔸) = ⋂𝓘𝓘𝓢(𝔸)
| The least infinite initial segment ⊆ 𝔸
| is the intersection of all infinite initial
| segments ⊆ 𝔸
| || ⋂𝓘𝓘𝓢(𝔸) is an infinite initial segment,
|| by lemma 1
||
|| For each 𝕊 ∈ 𝓘𝓘𝓢(𝔸), ⋂𝓘𝓘𝓢(𝔸) ⊆ 𝕊
|| from the definition of 'intersection'
||
|| ⋂𝓘𝓘𝓢(𝔸) = 𝕃𝕀𝕀𝕊(𝔸)
| || Lemma 1.
|| The intersection of infinite initial
|| segments is an infinite initial segment.
|| [2]
| | Similarly, for infinite initial segment 𝔹
| 𝕃𝕀𝕀𝕊(𝔹) = ⋂𝓘𝓘𝓢(𝔹)
| |
| 𝔸∩𝔹 is an infinite initial segment,
| by lemma 1
| | 𝔸∩𝔹 is an infinite initial segment ⊆ 𝔸
| so 𝕃𝕀𝕀𝕊(𝔸) ⊆ 𝔸∩𝔹
| | 𝕃𝕀𝕀𝕊(𝔸) ⊆ 𝔸∩𝔹 ⊆ 𝔹
| 𝕃𝕀𝕀𝕊(𝔸) is an infinite initial segment ⊆ 𝔹
| so 𝕃𝕀𝕀𝕊(𝔹) ⊆ 𝕃𝕀𝕀𝕊(𝔸)
| | Similarly, 𝕃𝕀𝕀𝕊(𝔸) ⊆ 𝕃𝕀𝕀𝕊(𝔹)
| | 𝕃𝕀𝕀𝕊(𝔸) ⊆ 𝕃𝕀𝕀𝕊(𝔹)
| and
| 𝕃𝕀𝕀𝕊(𝔹) ⊆ 𝕃𝕀𝕀𝕊(𝔸)
| so
| 𝕃𝕀𝕀𝕊(𝔸) = 𝕃𝕀𝕀𝕊(𝔹)

Define
𝕃𝕀𝕀𝕊 = 𝕃𝕀𝕀𝕊(𝔸) = 𝕃𝕀𝕀𝕊(𝔹)

𝕃𝕀𝕀𝕊 is an infinite initial segment
and,
if 𝔸 is any infinite initial segment,
then 𝕃𝕀𝕀𝕊 ⊆ 𝔸

----
[2]
Lemma 1.
The intersection of infinite initial
segments is an infinite initial segment.

Let 𝓘𝓘𝓢 be some set of infinite initial segments
and ⋂𝓘𝓘𝓢 be their intersection.

𝕊 is an initial infinite segment
⟺
0 ∈ 𝕊 ∧ ∀k ∈ 𝕊 : k+1 ∈ 𝕊

𝓘𝓘𝓢 is some set of infinite initial segments
∀𝕊 ∈ 𝓘𝓘𝓢 : 0 ∈ 𝕊
0 ∈ ⋂𝓘𝓘𝓢

| Assume k ∈ ⋂𝓘𝓘𝓢
| ∀𝕊 ∈ 𝓘𝓘𝓢 : k ∈ 𝕊
| | 𝓘𝓘𝓢 is some set of infinite initial segments
| ∀𝕊 ∈ 𝓘𝓘𝓢 : (∀k ∈ 𝕊 : k+1 ∈ 𝕊)
| ∀𝕊 ∈ 𝓘𝓘𝓢 : k+1 ∈ 𝕊
| k+1 ∈ ⋂𝓘𝓘𝓢

∀k : k ∈ ⋂𝓘𝓘𝓢 ⟹ k+1 ∈ ⋂𝓘𝓘𝓢

0 ∈ ⋂𝓘𝓘𝓢 ∧ ∀k ∈ ⋂𝓘𝓘𝓢 : k+1 ∈ ⋂𝓘𝓘𝓢
⋂𝓘𝓘𝓢 is an initial infinite segment

On Sunday, October 9, 2022 at 11:23:03 AM UTC-7, Jim Burns wrote:
> On 10/9/2022 5:39 AM, WM wrote:
> > Jim Burns schrieb am Samstag,
> > 8. Oktober 2022 um 20:39:13 UTC+2:
> >> On 10/7/2022 3:42 PM, WM wrote:
>
> >>> We agree that the leap can end at every
> >>> desired FISON end.
> >>> But aleph_0 natural numbers are covered by
> >>> every leap.
> >>
> >> For simplicity, consider only 0 and ω
> >> and whatever is between them.
> >>
> >> Split them between
> >> what can be reached one-by-one from 0
> >> and what can't be reached that way.
> >>
> >> What's needed in order to reach one-by-one?
> >>
> >> One-by-one, whichever way they're split
> >> into before and after,
> >> just-one-more must be possible.
> >> For each split,
> >> a from-before-place and a to-after-place,
> >> with none between, must exist.
> >>
> >> Each place on the 0-side of the split
> >> can be reached one-by-one from 0
> >
> > Every FISON-end can be reached one-by-one.
> Everything on the 0-side of the 0-side,ω-side
> split can be reached one-by-one.
>
> Nothing on the 0-side of the 0-side,ω-side
> split cannot be reached one-by-one.
> > But every FISON contains less than half of
> > the infinite initial segment,
> > i.e., of all natural numbers {1, 2, 3, ...}.
> > Proof:
> > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > or
> > ∀n ∈ ℕ_def: 2n < ℵo ==> n < ℵo/2.
> The 0-side of the 0-side,ω-side split
> contains _all_ of the 0-side of the
> 0-side,ω-side split, without exception.
>
> No FISON is the 0-side of the 0-side,ω-side
> split.
> > Nevertheless it is impossible to distinguish
> > different infinite initial segments.
> A unique _least_ (⊆) infinite initial
> segment 𝕃𝕀𝕀𝕊 exists
> such that
> 𝕃𝕀𝕀𝕊 is an infinite initial segment
> and,
> if 𝔸 is any infinite initial segment,
> then 𝕃𝕀𝕀𝕊 ⊆ 𝔸
> [1]
>
> Suppose
> "another" least infinite initial
> segment 𝕃𝕀𝕀𝕊₂ exists
> such that
> 𝕃𝕀𝕀𝕊₂ is an infinite initial segment
> and,
> if 𝔸 is any initial infinite segment,
> then 𝕃𝕀𝕀𝕊₂ ⊆ 𝔸
>
> Then
> 𝕃𝕀𝕀𝕊₂ ⊆ 𝕃𝕀𝕀𝕊 and 𝕃𝕀𝕀𝕊 ⊆ 𝕃𝕀𝕀𝕊₂
> and
> 𝕃𝕀𝕀𝕊₂ = 𝕃𝕀𝕀𝕊
>
> 𝕃𝕀𝕀𝕊 is unique.
>
>
> 𝕃𝕀𝕀𝕊 is the 0-side of the 0-side,ω-side
> split.
>
> if
> P(0) ∧ ∀j ∈ 𝕃𝕀𝕀𝕊 : P(j) ⟹ P(j+1)
> then
> ∀k ∈ 𝕃𝕀𝕀𝕊 : P(k)
>
> We define ℕ to be 𝕃𝕀𝕀𝕊
> > Why? Because the additional numbers are dark.
> The additional numbers, if dark, if existing,
> are on the ω-side of the 0-side,ω-side split.
>
> The 0-side,ω-side split is not just-one-more.
> The reason is that
> the 0-side does not contain a last-before.
> (To be just-one-more, there must be a last-before,
> and a first-after, and none between.)
>
> Whatever the ω-side of the 0-side,ω-side split
> contains or does not contain, the 0-side,ω-side
> split is not just-one-more, and leaps which
> cross the 0-side,ω-side split cannot be finite.
>
> Ultimately, the reason is that
> 𝐹ₖ⊕⟨k+1⟩ is a FISON too.
>
> ----
> [1]
> 𝔸 is an initial infinite segment
> ⟺
> 0 ∈ 𝔸 ∧ ∀k ∈ 𝔸 : k+1 ∈ 𝔸
>
> ----
> A unique least (⊆) infinite initial
> segment 𝕃𝕀𝕀𝕊 exists.
>
> | Assume that infinite initial segments 𝔸 𝔹
> | exist.
> |
> | 𝓘𝓘𝓢(𝔸) =
> | { 𝕊 ⊆ 𝔸 : 𝕊 is an inf. init. seg. }
> | the set of all infinite initial segments ⊆ 𝔸
> | 𝓘𝓘𝓢(𝔸) contains at least 𝔸
> |
> | ⋂𝓘𝓘𝓢(𝔸) =
> | the intersection of all infinite initial
> | segments ⊆ 𝔸
> | == the set of all k in all 𝕊 in 𝓘𝓘𝓢(𝔸)
> |
> | 𝕃𝕀𝕀𝕊(𝔸) = ⋂𝓘𝓘𝓢(𝔸)
> | The least infinite initial segment ⊆ 𝔸
> | is the intersection of all infinite initial
> | segments ⊆ 𝔸
> |
> || ⋂𝓘𝓘𝓢(𝔸) is an infinite initial segment,
> || by lemma 1
> ||
> || For each 𝕊 ∈ 𝓘𝓘𝓢(𝔸), ⋂𝓘𝓘𝓢(𝔸) ⊆ 𝕊
> || from the definition of 'intersection'
> ||
> || ⋂𝓘𝓘𝓢(𝔸) = 𝕃𝕀𝕀𝕊(𝔸)
> |
> || Lemma 1.
> || The intersection of infinite initial
> || segments is an infinite initial segment.
> || [2]
> |
> | Similarly, for infinite initial segment 𝔹
> | 𝕃𝕀𝕀𝕊(𝔹) = ⋂𝓘𝓘𝓢(𝔹)
> |
> |
> | 𝔸∩𝔹 is an infinite initial segment,
> | by lemma 1
> |
> | 𝔸∩𝔹 is an infinite initial segment ⊆ 𝔸
> | so 𝕃𝕀𝕀𝕊(𝔸) ⊆ 𝔸∩𝔹
> |
> | 𝕃𝕀𝕀𝕊(𝔸) ⊆ 𝔸∩𝔹 ⊆ 𝔹
> | 𝕃𝕀𝕀𝕊(𝔸) is an infinite initial segment ⊆ 𝔹
> | so 𝕃𝕀𝕀𝕊(𝔹) ⊆ 𝕃𝕀𝕀𝕊(𝔸)
> |
> | Similarly, 𝕃𝕀𝕀𝕊(𝔸) ⊆ 𝕃𝕀𝕀𝕊(𝔹)
> |
> | 𝕃𝕀𝕀𝕊(𝔸) ⊆ 𝕃𝕀𝕀𝕊(𝔹)
> | and
> | 𝕃𝕀𝕀𝕊(𝔹) ⊆ 𝕃𝕀𝕀𝕊(𝔸)
> | so
> | 𝕃𝕀𝕀𝕊(𝔸) = 𝕃𝕀𝕀𝕊(𝔹)
>
> Define
> 𝕃𝕀𝕀𝕊 = 𝕃𝕀𝕀𝕊(𝔸) = 𝕃𝕀𝕀𝕊(𝔹)
>
> 𝕃𝕀𝕀𝕊 is an infinite initial segment
> and,
> if 𝔸 is any infinite initial segment,
> then 𝕃𝕀𝕀𝕊 ⊆ 𝔸
>
> ----
> [2]
> Lemma 1.
> The intersection of infinite initial
> segments is an infinite initial segment.
>
>
> Let 𝓘𝓘𝓢 be some set of infinite initial segments
> and ⋂𝓘𝓘𝓢 be their intersection.
>
> 𝕊 is an initial infinite segment
> ⟺
> 0 ∈ 𝕊 ∧ ∀k ∈ 𝕊 : k+1 ∈ 𝕊
>
> 𝓘𝓘𝓢 is some set of infinite initial segments
> ∀𝕊 ∈ 𝓘𝓘𝓢 : 0 ∈ 𝕊
> 0 ∈ ⋂𝓘𝓘𝓢
>
> | Assume k ∈ ⋂𝓘𝓘𝓢
> | ∀𝕊 ∈ 𝓘𝓘𝓢 : k ∈ 𝕊
> |
> | 𝓘𝓘𝓢 is some set of infinite initial segments
> | ∀𝕊 ∈ 𝓘𝓘𝓢 : (∀k ∈ 𝕊 : k+1 ∈ 𝕊)
> | ∀𝕊 ∈ 𝓘𝓘𝓢 : k+1 ∈ 𝕊
> | k+1 ∈ ⋂𝓘𝓘𝓢
>
> ∀k : k ∈ ⋂𝓘𝓘𝓢 ⟹ k+1 ∈ ⋂𝓘𝓘𝓢
>
> 0 ∈ ⋂𝓘𝓘𝓢 ∧ ∀k ∈ ⋂𝓘𝓘𝓢 : k+1 ∈ ⋂𝓘𝓘𝓢
> ⋂𝓘𝓘𝓢 is an initial infinite segment
Zero is not a quantity dark math.
Mitchell Raemsch

On Sunday, October 9, 2022 at 11:23:03 AM UTC-7, Jim Burns wrote:
> On 10/9/2022 5:39 AM, WM wrote:
> > Jim Burns schrieb am Samstag,
> > 8. Oktober 2022 um 20:39:13 UTC+2:
> >> On 10/7/2022 3:42 PM, WM wrote:
>
> >>> We agree that the leap can end at every
> >>> desired FISON end.
> >>> But aleph_0 natural numbers are covered by
> >>> every leap.
> >>
> >> For simplicity, consider only 0 and ω
> >> and whatever is between them.
> >>
> >> Split them between
> >> what can be reached one-by-one from 0
> >> and what can't be reached that way.
> >>
> >> What's needed in order to reach one-by-one?
> >>
> >> One-by-one, whichever way they're split
> >> into before and after,
> >> just-one-more must be possible.
> >> For each split,
> >> a from-before-place and a to-after-place,
> >> with none between, must exist.
> >>
> >> Each place on the 0-side of the split
> >> can be reached one-by-one from 0
> >
> > Every FISON-end can be reached one-by-one.
> Everything on the 0-side of the 0-side,ω-side
> split can be reached one-by-one.
>
> Nothing on the 0-side of the 0-side,ω-side
> split cannot be reached one-by-one.
> > But every FISON contains less than half of
> > the infinite initial segment,
> > i.e., of all natural numbers {1, 2, 3, ...}.
> > Proof:
> > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > or
> > ∀n ∈ ℕ_def: 2n < ℵo ==> n < ℵo/2.
> The 0-side of the 0-side,ω-side split
> contains _all_ of the 0-side of the
> 0-side,ω-side split, without exception.
>
> No FISON is the 0-side of the 0-side,ω-side
> split.
> > Nevertheless it is impossible to distinguish
> > different infinite initial segments.
> A unique _least_ (⊆) infinite initial
> segment 𝕃𝕀𝕀𝕊 exists
> such that
> 𝕃𝕀𝕀𝕊 is an infinite initial segment
> and,
> if 𝔸 is any infinite initial segment,
> then 𝕃𝕀𝕀𝕊 ⊆ 𝔸
> [1]
>
> Suppose
> "another" least infinite initial
> segment 𝕃𝕀𝕀𝕊₂ exists
> such that
> 𝕃𝕀𝕀𝕊₂ is an infinite initial segment
> and,
> if 𝔸 is any initial infinite segment,
> then 𝕃𝕀𝕀𝕊₂ ⊆ 𝔸
>
> Then
> 𝕃𝕀𝕀𝕊₂ ⊆ 𝕃𝕀𝕀𝕊 and 𝕃𝕀𝕀𝕊 ⊆ 𝕃𝕀𝕀𝕊₂
> and
> 𝕃𝕀𝕀𝕊₂ = 𝕃𝕀𝕀𝕊
>
> 𝕃𝕀𝕀𝕊 is unique.
>
>
> 𝕃𝕀𝕀𝕊 is the 0-side of the 0-side,ω-side
> split.
>
> if
> P(0) ∧ ∀j ∈ 𝕃𝕀𝕀𝕊 : P(j) ⟹ P(j+1)
> then
> ∀k ∈ 𝕃𝕀𝕀𝕊 : P(k)
>
> We define ℕ to be 𝕃𝕀𝕀𝕊
> > Why? Because the additional numbers are dark.
> The additional numbers, if dark, if existing,
> are on the ω-side of the 0-side,ω-side split.
>
> The 0-side,ω-side split is not just-one-more.
> The reason is that
> the 0-side does not contain a last-before.
> (To be just-one-more, there must be a last-before,
> and a first-after, and none between.)
>
> Whatever the ω-side of the 0-side,ω-side split
> contains or does not contain, the 0-side,ω-side
> split is not just-one-more, and leaps which
> cross the 0-side,ω-side split cannot be finite.
>
> Ultimately, the reason is that
> 𝐹ₖ⊕⟨k+1⟩ is a FISON too.
>
> ----
> [1]
> 𝔸 is an initial infinite segment
> ⟺
> 0 ∈ 𝔸 ∧ ∀k ∈ 𝔸 : k+1 ∈ 𝔸
>
> ----
> A unique least (⊆) infinite initial
> segment 𝕃𝕀𝕀𝕊 exists.
>
> | Assume that infinite initial segments 𝔸 𝔹
> | exist.
> |
> | 𝓘𝓘𝓢(𝔸) =
> | { 𝕊 ⊆ 𝔸 : 𝕊 is an inf. init. seg. }
> | the set of all infinite initial segments ⊆ 𝔸
> | 𝓘𝓘𝓢(𝔸) contains at least 𝔸
> |
> | ⋂𝓘𝓘𝓢(𝔸) =
> | the intersection of all infinite initial
> | segments ⊆ 𝔸
> | == the set of all k in all 𝕊 in 𝓘𝓘𝓢(𝔸)
> |
> | 𝕃𝕀𝕀𝕊(𝔸) = ⋂𝓘𝓘𝓢(𝔸)
> | The least infinite initial segment ⊆ 𝔸
> | is the intersection of all infinite initial
> | segments ⊆ 𝔸
> |
> || ⋂𝓘𝓘𝓢(𝔸) is an infinite initial segment,
> || by lemma 1
> ||
> || For each 𝕊 ∈ 𝓘𝓘𝓢(𝔸), ⋂𝓘𝓘𝓢(𝔸) ⊆ 𝕊
> || from the definition of 'intersection'
> ||
> || ⋂𝓘𝓘𝓢(𝔸) = 𝕃𝕀𝕀𝕊(𝔸)
> |
> || Lemma 1.
> || The intersection of infinite initial
> || segments is an infinite initial segment.
> || [2]
> |
> | Similarly, for infinite initial segment 𝔹
> | 𝕃𝕀𝕀𝕊(𝔹) = ⋂𝓘𝓘𝓢(𝔹)
> |
> |
> | 𝔸∩𝔹 is an infinite initial segment,
> | by lemma 1
> |
> | 𝔸∩𝔹 is an infinite initial segment ⊆ 𝔸
> | so 𝕃𝕀𝕀𝕊(𝔸) ⊆ 𝔸∩𝔹
> |
> | 𝕃𝕀𝕀𝕊(𝔸) ⊆ 𝔸∩𝔹 ⊆ 𝔹
> | 𝕃𝕀𝕀𝕊(𝔸) is an infinite initial segment ⊆ 𝔹
> | so 𝕃𝕀𝕀𝕊(𝔹) ⊆ 𝕃𝕀𝕀𝕊(𝔸)
> |
> | Similarly, 𝕃𝕀𝕀𝕊(𝔸) ⊆ 𝕃𝕀𝕀𝕊(𝔹)
> |
> | 𝕃𝕀𝕀𝕊(𝔸) ⊆ 𝕃𝕀𝕀𝕊(𝔹)
> | and
> | 𝕃𝕀𝕀𝕊(𝔹) ⊆ 𝕃𝕀𝕀𝕊(𝔸)
> | so
> | 𝕃𝕀𝕀𝕊(𝔸) = 𝕃𝕀𝕀𝕊(𝔹)
>
> Define
> 𝕃𝕀𝕀𝕊 = 𝕃𝕀𝕀𝕊(𝔸) = 𝕃𝕀𝕀𝕊(𝔹)
>
> 𝕃𝕀𝕀𝕊 is an infinite initial segment
> and,
> if 𝔸 is any infinite initial segment,
> then 𝕃𝕀𝕀𝕊 ⊆ 𝔸
>
> ----
> [2]
> Lemma 1.
> The intersection of infinite initial
> segments is an infinite initial segment.
>
>
> Let 𝓘𝓘𝓢 be some set of infinite initial segments
> and ⋂𝓘𝓘𝓢 be their intersection.
>
> 𝕊 is an initial infinite segment
> ⟺
> 0 ∈ 𝕊 ∧ ∀k ∈ 𝕊 : k+1 ∈ 𝕊
>
> 𝓘𝓘𝓢 is some set of infinite initial segments
> ∀𝕊 ∈ 𝓘𝓘𝓢 : 0 ∈ 𝕊
> 0 ∈ ⋂𝓘𝓘𝓢
>
> | Assume k ∈ ⋂𝓘𝓘𝓢
> | ∀𝕊 ∈ 𝓘𝓘𝓢 : k ∈ 𝕊
> |
> | 𝓘𝓘𝓢 is some set of infinite initial segments
> | ∀𝕊 ∈ 𝓘𝓘𝓢 : (∀k ∈ 𝕊 : k+1 ∈ 𝕊)
> | ∀𝕊 ∈ 𝓘𝓘𝓢 : k+1 ∈ 𝕊
> | k+1 ∈ ⋂𝓘𝓘𝓢
>
> ∀k : k ∈ ⋂𝓘𝓘𝓢 ⟹ k+1 ∈ ⋂𝓘𝓘𝓢
>
> 0 ∈ ⋂𝓘𝓘𝓢 ∧ ∀k ∈ ⋂𝓘𝓘𝓢 : k+1 ∈ ⋂𝓘𝓘𝓢
> ⋂𝓘𝓘𝓢 is an initial infinite segment

So, there's a 0 side and a 1-side,
and the elements betwen them are in terms of each other,
and blind their side are statements one for the other, thus a symmetry.
You brought in omega, can you reduce the argument to [0,1]?
Then that there's, "1/2", is some notion, that for [0, 1/2], and [1/2, 1],
are all sorts same properties, until such time as they vanish to a point.
The de-generate interval, or for de-generate point, ..., is here
the generated partitions, are utterly generated, what results is
de-generate, not dis-generate.

It's the "generate" pronounced like "infinite": generated.
I see you at least have the one side drawn out, ....

On 10/9/2022 2:02 AM, WM wrote:
> Chris M. Thomasson schrieb am Freitag, 7. Oktober 2022 um 22:35:00 UTC+2:
>> On 10/7/2022 12:09 AM, WM wrote:
>
>>> below omega there is no gap. No leap can end in a gap. The result of this is the existence of dark nunbers.
>> With regard to artificially restricting ones self to the natural
>> numbers, there is no gap between say, [41...42].
>
> Yes, there is no gap. And if ω exists, then there is no gap between ω and ℕ. But every leap from ω into ℕ covers infinitely many natural numbers.
A leap? You mean from say, 31 ---> 42? jumping from 31 to 42? Humm...
What is the absolute value of the difference between 31 and 42?
abs(31-42) = 11. 11 is not one or zero, so there is a "gap" wrt the
natural numbers. Remember unit 1 is atomic and is not able to be
breached wrt strictly adhering to the properties of the natural numbers.

A gap:

say, 10 ---> 13.

Well, abs(13-10) = 3, therefore, the "in between" wrt the granularity of
the natural numbers is:

10 + 1 = 11
10 + 1 + 1 = 12
10 + 1 + 1 + 1 = 13

Or:

10 + 1 = 11
10 + 2 = 12
10 + 3 = 13

Sound Kosher to you?

On 9/27/2022 11:53 AM, WM wrote:
> Jim Burns schrieb am Dienstag, 27. September 2022 um 01:04:48 UTC+2:
>> On 9/26/2022 9:16 AM, WM wrote:
>>> Jim Burns schrieb am Sonntag,
>>> 25. September 2022 um 23:28:56 UTC+2:
>>>> However,
>>>> _all_ the exchanges are NOT within any FISON,
>>>> ⋃𝓕 is NOT a FISON.
>>>
>>> It is a FISON.
>> ⋃𝓕 is NOT a FISON.
>
> Up to every n UF(n) is a FISON. But other FISONs are not available.
>>
>>>> "After each" and "at the end" are not the same.
>>>
>>> It is synonymous.
>> ∅ is after (⊃) each infinite end segment of ⋃𝓕
>>
>> ∅ is not an infinite end segment of ⋃𝓕
>> so ∅ is not at the end (⊃) of
>> the infinite end segments of ⋃𝓕
>>
>> "After each" and "at the end" are not synonymous.
>
> ∅ is an end segment . But if you deny, we can say: "After each" and "after the end" are synonymous.
[...]

There is no such thing as an "_end_ segment" in an infinite process?

torsdag 6 oktober 2022 kl. 21:25:22 UTC+2 skrev WM:
> Fritz Feldhase schrieb am Mittwoch, 5. Oktober 2022 um 18:36:21 UTC+2:
> > On Wednesday, October 5, 2022 at 11:10:59 AM UTC+2, WM wrote:
> >
> > > We can start [our leap at] omega and [jump to] 17. That is a leap.
> > > why must [there] be infinitely many natnumbers between omega and the end of the leap.
> >
> > Because the end of the leap is a natural number, and there are infinitely many natural numbers (finite ordinals) between any natural number and omega. That's why.
> If all of them were definable then all of them could be subtracted and nothing below omega would remain.
> > > But the leap from omega cannot pass less than aleph_0 natnumbers. Why that?
> > Because for each and every natural number n there are infinitely many natural numbers (finite ordinals) between n and omega. That's why.
> >
> > Hope this helps.
> It helps to understand why they are dark. They cannot be removed. They cannot be the ends of leaps. They are undefinable.
>
> Regards, WM

No, it helps nothign because what you say has no meaning! It literally always apply to either none of the natural numbers or all of the natural numbers, this making it pointless!

On 9/19/2022 10:05 AM, WM wrote:
> Gus Gassmann schrieb am Montag, 19. September 2022 um 17:00:52 UTC+2:
>> On Sunday, 18 September 2022 at 16:18:04 UTC-3, WM wrote:
>> [...]
>>> I understand: Every fraction can be indexed.
>> Of course every fraction can be indexed. That is the whole point of setting up a bijection. There even is a very simple formula for it.
>>> That amounts to removing all O's.
>> It does not. There never is an 'O' to begin with.
>
> By definition an O is a fraction without index. When all indices have been issued to the first column, then all fractions in other columns are without index. That is marked by an O.
>
>> impose a stepwise process in the pretense that this makes things clearer.
>
> This is Cantor's stepwise process: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
>
> Look, he talks about processes, for instance here: ein drittes Prinzip, welches ich das Hemmungs- oder Beschränkungsprinzip nenne, entgegen, wodurch dem durchaus endlosen Bildungsprozeß sukzessive gewisse Schranken auferlegt werden, [Cantor]
> and here: Es wird daher zunächst der Anschein erweckt, als ob wir uns bei dieser Bildungsweise neuer ganzer bestimmt-unendlicher Zahlen ins Grenzenlose hin verlieren müßten, und daß wir außerstande seien, diesem endlosen Prozeß einen gewissen vorläufigen Abschluß zu geben, [Cantor]
>
>> (It does not.)
>
> Everybody can see that you are wrong. Can you see it too?
Puke! Yuck.

On 9/2/2022 2:20 PM, Sergio wrote:
> On 9/2/2022 3:31 PM, WM wrote:
>> Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:42:47 UTC+2:
>>> WM <askas...@gmail.com> writes:
>>>
>>>> Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:00:32 UTC+2:
>>>>> WM <askas...@gmail.com> writes:
>>>>>
>>>>>> That means there is a bijection between visible natnumbers
>>>>>> and visible fractions.
>>>>> Your textbook does not explain what a visible number is
>>>>
>>>> My textbook, like claisscal maths deals only with visible numbers. No
>>>> reason to mention this. Further in 2015 when it was published I did
>>>> not know about dark numbers.
>>> I have always been talking about the maths in your book. Did I not make
>>> that clear enough? Sorry.
>>>
>>> If your book is no longer adequate for this task, please say so, but I
>>> hope it is because you are not good at answering direct questions, but
>>> the book is there for all to see. What I wanted to know is:
>>>
>>> (a) Is k (now cut) a function from NxN to N as the set N and the term
>>> function are defined in your book?
>>
>> In my book there are, as I clearly stated in the introduction, only
>> potentially infinite sets. That are the visible numbers which are
>> sufficient to do classical mathematics.
>
> so your book does not apply to "modern" math, but only "visible" Ants.

Humm... invisible ants on the skin. Shit. Dark? Yikes!

On 9/6/2022 3:19 AM, WM wrote:
> Ben Bacarisse schrieb am Montag, 5. September 2022 um 17:48:48 UTC+2:
>> WM <askas...@gmail.com> writes:
>>
>>> Fritz Feldhase schrieb am Montag, 5. September 2022 um 10:34:43 UTC+2:
>>>> On Sunday, September 4, 2022 at 11:25:27 PM UTC+2, Ben Bacarisse wrote:
>>>>
>>>>> in WMaths (specifically for some potentially infinite sets)
>>>
>>> only for such!
>>>
>>>> it is possible to have both e ∈ S and S \ {e} = S.
>>>
>>> Dark elements can become visible and vice versa.
>> I've not been following all the interminable threads, but I thought
>> WMaths did not have dark numbers.
>
> It doesn't. Nevertheless potential infinity has a growing character. Therefore larger finite sets are created from smaller sets.
>
> In actual infinity the newly created visible numbers are taken from the dark part of the set, in potential infinity they are created from nothing. In both cases only the potentially infinite sets of visible numbers are part of mathematics.
>
> Regards, WM
>

is the point 0+1/2 dark in the interval [0...1]?

Well, not dark since I named it. Humm... Think of a number, is it dark now?

On 10/9/2022 12:10 PM, mitchr...@gmail.com wrote:
> On Sunday, October 9, 2022 at 11:23:03 AM UTC-7, Jim Burns wrote:
>> On 10/9/2022 5:39 AM, WM wrote:
>>> Jim Burns schrieb am Samstag,
>>> 8. Oktober 2022 um 20:39:13 UTC+2:
>>>> On 10/7/2022 3:42 PM, WM wrote:
>>[...]
> Zero is not a quantity dark math.

Well, the poor old fraction that had a zero, has a story to tell, not
now, basically:

42/0 = oh shit!

Lets define divide by zero for fun! :^)

On 9/27/2022 7:58 PM, Sergi o wrote:
> On 9/27/2022 1:53 PM, WM wrote:
>> Jim Burns schrieb am Dienstag, 27. September 2022 um 01:04:48 UTC+2:
>>> On 9/26/2022 9:16 AM, WM wrote:
>>>> Jim Burns schrieb am Sonntag,
>>>> 25. September 2022 um 23:28:56 UTC+2:
>>>>> However,
>>>>> _all_ the exchanges are NOT within any FISON,
>>>>> ⋃𝓕 is NOT a FISON.
>>>>
>>>> It is a FISON.
>>> ⋃𝓕 is NOT a FISON.
>>
>> Up to every n UF(n) is a FISON. But other FISONs are not available.
>
> did you give them part time jobs ?
>
>>>
>>>>> "After each" and "at the end" are not the same.
>>>>
>>>> It is synonymous.
>>> ∅ is after (⊃) each infinite end segment of ⋃𝓕
>>>
>>> ∅ is not an infinite end segment of ⋃𝓕
>>> so ∅ is not at the end (⊃) of
>>> the infinite end segments of ⋃𝓕
>>>
>>> "After each" and "at the end" are not synonymous.
>>
>> ∅ is an end segment .
>
> Oh No!!  your final cheese has slid off your last cracker !!
>
>          ∅ is never an end segment.
>
>
>
>> But if you deny, we can say: "After each" and "after the end" are
>> synonymous.
>
> you may go now, and mumble or potentially mutter about "potentially
> after each" OR "actual after the end of what ? "
>
>>
>> Regards, WM
>>
>
> sorry to seem that insulting, but your math is so bad, and you should
> expect it. Road Killed Math.
> Need to get some Ants in here...
>

Wrt starting conditions, I have to reply here because I accidentally
deleted your message! Sorry about that.

For some reason I like to dedicate index zero as a "starting condition"
for a recursive definition, say:

i[0] = 0
i[1] = i[0] + 1
i[2] = i[1] + 1
....

Therefore:

i[0] = 0
i[n] = i[n - 1] + 1
....

Simple generator that depends on its predecessor. index zero, i[0] =
zero. So, fine. Another initial starting condition can be i[0] = 41, so:

i[1] = i[1-1] + 1 = 41 + 1 = 42

:^)

On 10/9/2022 3:10 PM, mitchr...@gmail.com wrote:
> On Sunday, October 9, 2022
> at 11:23:03 AM UTC-7, Jim Burns wrote:

>> 0 ∈ ⋂𝓘𝓘𝓢 ∧ ∀k ∈ ⋂𝓘𝓘𝓢 : k+1 ∈ ⋂𝓘𝓘𝓢
>> ⋂𝓘𝓘𝓢 is an initial infinite segment
>
> Zero is not a quantity dark math.

For natural numbers...

Zero is not a successor.
~∃i = 0⁻⁻

Zero has a successor.
1 = 0⁺⁺

Each successor has a successor.
∀j : ∃i = j⁻⁻ ⟹ ∃k = j⁺⁺

No sharing of successors or predecessors.
∀j : ∃k = j⁺⁺ ⟹ (~∃j₂ ≠ j : j₂⁺⁺ = j⁺⁺)
∀j : ∃i = j⁻⁻ ⟹ (~∃j₂ ≠ j : j₂⁻⁻ = j⁻⁻)

IIS(𝔸) == "𝔸 is an Infinite Initial Segment"

Any infinite initial segment 𝔸
contains 0 and contains all its successors.
IIS(𝔸) ⟺
0 ∈ 𝔸 ∧ ∀j ∈ 𝔸 : j⁺⁺ ∈ 𝔸

A Least Infinite Initial Segment 𝕃𝕀𝕀𝕊
exists and is unique.
IIS(𝕃𝕀𝕀𝕊) ∧
∀𝔸 : IIS(𝔸) ⟹ 𝕃𝕀𝕀𝕊 ⊆ 𝔸

The natural numbers are
the least infinite initial segment.
ℕ = 𝕃𝕀𝕀𝕊

On Monday, October 10, 2022 at 1:46:05 PM UTC+2, Jim Burns wrote:

> The [set of] natural numbers [is]
> the least infinite initial segment.
> ℕ = 𝕃𝕀𝕀𝕊

On Sunday, October 9, 2022 at 11:02:36 AM UTC+2, WM wrote:

> there is no gap between ω and ℕ.

I'll interprete this to mean: There is no ordinal o such that for all n in IN: n < o < ω.

Meaning: There are the ordinals in IN (i. e. all elements in IN) and the ordinal ω (which is larger than all ordinals in IN), but there is no ordinal which is largen than all ordinals in IN and smaller than ω.

> But every leap from ω into ℕ covers infinitely many natural numbers.

I'll interprete this to mean: For each and every n in IN: card({k e IN: n < k < ω} = aleph_0.

So what?

Hint: For all n e IN: card({k e IN: n < k} = aleph_0.

Meaning: For each and every n in IN there are countably-infintely many elements in IN which are larger than n.

(And: For all n e IN: n < ω.)

On Sunday, October 9, 2022 at 11:39:23 AM UTC+2, WM wrote:

> The [sequence of all] FISONs [is] an infinite sequence.

Indeed!

> But they never become [...] infinite.

Well, each and every FISON is _finite_ by definition (or at least by a trivial consequence of the definition of FISONs), you know.

And right, there are INFINITELY many FISONs. :-)

Hint: card({{0, ..., n} : n e IN} = aleph_0.

> Cantor however claims the existence of |ℕ| = ℵo.

However? Huh?!

> Cantor claims that from 0 all natural nunbers can be reached such that none remains unreached.

Oh, really? Where does *he* claim that. Could you give a quote?

> That is precondition for every bijection with ℕ.

If you say so. Can you PROVE your claim?

Hint: It's not clear what you mean be "reached" here. Can you DEFINE this term/notion, please? (This might help others to check the correctness of the proof you were asked for.)

> Cantor claims that from 0 all |ω| natural nunbers can be reached such that none remains unreached.

Oh, really? Where does *he* claim that. Could you give a quote?

> Everything that can be reached one-by-one is in a FISON.

Hmmm...

It's unclear if your claim _means_:

a) For every n in IN that can be reached one-by-one there is a FISON F, such that n is in F.

or

b) There is a FISION F such that for every n in IN that can be reached one-by-one n is in F.

If all natural number "can be reached from 0" then a) is TRUE, but b) is FALSE. After all, IN is not a FISON, you know.

> The infinite set with |ω| elements can never be reached step-by-step..

Well, maybe IN "can never be reached step-by-step from 0", but it might be possible for each and every element in IN.

It would be helpful, if you could give definition for

x can be be reached step-by-step from 0

=> short: x can be reached

=> shorter: x is reachable

(for any ordinal x).

How about

x is /reachable/ iff card({y e ORD : 0 < y < x}) e IN

?

Meaning: there are FINITELY MANY consecutive ordinals "between" 0 and x.

This way we would immediately get

0 is reachable, 1 is reachable, 2 is reachable, etc.

Or more general:
For all n e IN: n is reachable.
(easy proof by induction).

And, for example,
ω is NOT reachable.

Btw. So if "Cantor" claimed, as you say, that "all natural nunbers can be reached", then he's right.

That's quite a good thing after all, as you say, this "is [a] precondition for every bijection with ℕ".

Jim Burns wrote on 10/10/2022 :
> On 10/9/2022 3:10 PM, mitchr...@gmail.com wrote:
>> On Sunday, October 9, 2022
>> at 11:23:03 AM UTC-7, Jim Burns wrote:
>
>>> 0 ∈ ⋂??? ∧ ∀k ∈ ⋂??? : k+1 ∈ ⋂???
>>> ⋂??? is an initial infinite segment
>>
>> Zero is not a quantity dark math.
>
> For natural numbers...
>
> Zero is not a successor.
> ~∃i = 0⁻⁻
>
> Zero has a successor.
> 1 = 0⁺⁺
>
> Each successor has a successor.
> ∀j : ∃i = j⁻⁻ ⟹ ∃k = j⁺⁺
>
> No sharing of successors or predecessors.
> ∀j : ∃k = j⁺⁺ ⟹ (~∃j₂ ≠ j : j₂⁺⁺ = j⁺⁺)
> ∀j : ∃i = j⁻⁻ ⟹ (~∃j₂ ≠ j : j₂⁻⁻ = j⁻⁻)
>
> IIS(?) == "? is an Infinite Initial Segment"
>
> Any infinite initial segment ?
> contains 0 and contains all its successors.
> IIS(?) ⟺
> 0 ∈ ? ∧ ∀j ∈ ? : j⁺⁺ ∈ ?
>
> A Least Infinite Initial Segment ????
> exists and is unique.
> IIS(????) ∧
> ∀? : IIS(?) ⟹ ???? ⊆ ?
>
> The natural numbers are
> the least infinite initial segment.
> ℕ = ????

I don't think a well ordered set has an infinite initial segment. I
also don't think an initial segment can be the whole set.

Consider the set of n-gons in the order of their number of sides. An
initial segment is defined as all elements strictly preceding some
element n (a natural ordinal number) in the ordering. Since a circle is
not an n-gon (infinity is not a number) and it is a 'strictly
preceding' set of elements in the initial segment, you would need magic
to include *all* elements of |N in your new set.

On Monday, 10 October 2022 at 11:22:48 UTC-3, FromTheRafters wrote:
[...]
> I don't think a well ordered set has an infinite initial segment.

I think that is wrong. Consider the set of natural numbers ordered by << in the following manner

if a is odd and b is even, then a << b
if a and b have the same parity, then A << b iff a < b in the conventional order
if a is even and b is odd, then b << a.

This well-order(!) can also be written as 1 << 3 << 5 << ... << 2 << 4 << 6 << ... .

I would think that for every /even/ natural number the initial segment is infinite.

On Monday, October 10, 2022 at 4:22:48 PM UTC+2, FromTheRafters wrote:

> I don't think a well ordered set has an infinite initial segment.
> I also don't think an initial segment can be the whole set.

No matter what you think, it just depends on the DEFINITION of the notion /initial segment/.

The following supports your view: https://mathworld.wolfram.com/InitialSegment.html

This one too: https://proofwiki.org/wiki/Definition:Initial_Segment

On the other hand: https://math.stackexchange.com/questions/464347/definition-of-initial-segment

Concerning the natural numbers, the notion "FISON" for _finite_ initial segement [...] seems to avoid this (slight) ambiguity.

On Monday, October 10, 2022 at 6:34:47 PM UTC+2, Fritz Feldhase wrote:
> On Monday, October 10, 2022 at 4:22:48 PM UTC+2, FromTheRafters wrote:
>
> > I don't think a well ordered set has an infinite initial segment.
> > I also don't think an initial segment can be the whole set.
> No matter what you think, it just depends on the DEFINITION of the notion /initial segment/.
>
> The following supports your view: https://mathworld.wolfram.com/InitialSegment.html
>
> This one too: https://proofwiki.org/wiki/Definition:Initial_Segment
>
> On the other hand: https://math.stackexchange.com/questions/464347/definition-of-initial-segment
>
> Concerning the natural numbers, the notion "FISON" for _finite_ initial segement [...] seems to avoid this (slight) ambiguity.

No ambiguity <=> no Muckmeatics

On 10/10/2022 10:22 AM, FromTheRafters wrote:
> Jim Burns wrote on 10/10/2022 :

>> IIS(A)  ==  "A is an Infinite Initial Segment"
>>
>> Any infinite initial segment A
>> contains 0 and contains all its successors.
>> IIS(A)  ⟺
>> 0 ∈ A  ∧  ∀j ∈ A : j⁺⁺ ∈ A
>>
>> A Least Infinite Initial Segment LIIS
>> exists and is unique.
>> IIS(LIIS)  ∧
>> ∀A : IIS(A) ⟹ LIIS ⊆ A
>>
>> The natural numbers are
>> the least infinite initial segment.
>> ℕ = LIIS
>
> I don't think a well ordered set has
> an infinite initial segment.
>
> I also don't think an initial segment
> can be the whole set.
>
> Consider the set of n-gons in the order of
> their number of sides. An initial segment is
> defined as all elements strictly preceding
> some element n (a natural ordinal number)
> in the ordering. Since a circle is not
> an n-gon (infinity is not a number) and it is
> a 'strictly preceding' set of elements in
> the initial segment, you would need magic to
> include *all* elements of |N in your new set.

It's not evident in that, my most recent post, but
I am trying to follow WM's use.

<WM>
> Nevertheless it is impossible to distinguish
> different infinite initial segments.
> Why? Because the additional numbers are dark.

If there are dark numbers (or Robinson
non-standard naturals or infinite ordinals or ...)
then there is an "infinite initial segment"
which can be distinguished from all the others:
the least one, in which no post-finites elements
occur.

Can an "initial segment" lack a last element?
Previous usage suggests it can.
All the end segments of ℕ lack last elements.
WM's (unordered) dark numbers don't change that.

Can an "initial segment" include _all_ the set?
We have been including ℕ among the end segments
of ℕ. It seems as though it can.

To be honest, I might prefer a different term.
But it seems to me that the simplest, most
effective way to say
| I (JB) am not going off somewhere else to
| talk about something else (a thing I do).
is to use WM's terms to answer him.

----
>> For natural numbers...
>>
>> Zero is not a successor.
>> ~∃i = 0⁻⁻
>>
>> Zero has a successor.
>> 1 = 0⁺⁺
>>
>> Each successor has a successor.
>> ∀j : ∃i = j⁻⁻ ⟹ ∃k = j⁺⁺
>>
>> No sharing of successors or predecessors.
>> ∀j : ∃k = j⁺⁺ ⟹ (~∃j₂ ≠ j : j₂⁺⁺ = j⁺⁺)
>> ∀j : ∃i = j⁻⁻ ⟹ (~∃j₂ ≠ j : j₂⁻⁻ = j⁻⁻)

Something bothers me about the way I said that.

Better:
No sharing of successors or predecessors.
~∃j, ∃j₂ ≠ j : j₂⁺⁺ = j⁺⁺
~∃j, ∃j₂ ≠ j : j₂⁻⁻ = j⁻⁻

Hmmm. But still...
Can I abbreviate this
∀j : ∃i = j⁻⁻ ⟹ ∃k = j⁺⁺

as this?
∀j : ∀i ≠ j⁻⁻ : ∃k = j⁺⁺

I don't think I want to say that.
I want to say
Each successor has a successor.

That says, essentially,
Everything has a successor.

The only alternative I see is
∀j : ∃i = j⁻⁻ : (...)

which is worse:
Everything has a predecessor? No thank you.

On 10/10/2022 7:22 AM, FromTheRafters wrote:
> Jim Burns wrote on 10/10/2022 :
>> On 10/9/2022 3:10 PM, mitchr...@gmail.com wrote:
>>> On Sunday, October 9, 2022
>>> at 11:23:03 AM UTC-7, Jim Burns wrote:
>>
>>>> 0 ∈ ⋂??? ∧ ∀k ∈ ⋂??? : k+1 ∈ ⋂???
>>>> ⋂??? is an initial infinite segment
>>>
>>> Zero is not a quantity dark math.
>>
>> For natural numbers...
>>
>> Zero is not a successor.
>> ~∃i = 0⁻⁻
>>
>> Zero has a successor.
>> 1 = 0⁺⁺
>>
>> Each successor has a successor.
>> ∀j : ∃i = j⁻⁻  ⟹  ∃k = j⁺⁺
>>
>> No sharing of successors or predecessors.
>> ∀j : ∃k = j⁺⁺  ⟹  (~∃j₂ ≠ j : j₂⁺⁺ = j⁺⁺)
>> ∀j : ∃i = j⁻⁻  ⟹  (~∃j₂ ≠ j : j₂⁻⁻ = j⁻⁻)
>>
>> IIS(?)  ==  "? is an Infinite Initial Segment"
>>
>> Any infinite initial segment ?
>> contains 0 and contains all its successors.
>> IIS(?)  ⟺
>> 0 ∈ ?  ∧  ∀j ∈ ? : j⁺⁺ ∈ ?
>>
>> A Least Infinite Initial Segment ????
>> exists and is unique.
>> IIS(????)  ∧
>> ∀? : IIS(?) ⟹ ???? ⊆ ?
>>
>> The natural numbers are
>> the least infinite initial segment.
>> ℕ = ????
>
> I don't think a well ordered set has an infinite initial segment. I also
> don't think an initial segment can be the whole set.
>
> Consider the set of n-gons in the order of their number of sides. An
> initial segment is defined as all elements strictly preceding some
> element n (a natural ordinal number) in the ordering. Since a circle is
> not an n-gon (infinity is not a number) and it is a 'strictly preceding'
> set of elements in the initial segment, you would need magic to include
> *all* elements of |N in your new set.

Imvvho, it's a bit fun to think about a circle as an n-gon where n is
infinity.

Gus Gassmann schrieb am Sonntag, 9. Oktober 2022 um 15:58:35 UTC+2:
> On Sunday, 9 October 2022 at 06:02:36 UTC-3, WM wrote:
> [...]
> > Yes, there is no gap. And if ω exists, then there is no gap between ω and ℕ. But every leap from ω into ℕ covers infinitely many natural numbers.
> WTF is a "leap from ω into ℕ"?

It means to enter ℕ from above.

> What even do *YOU* mean by a "gap between ω and ℕ"?

Nothing existing.

> More basically, is ω =/= ℕ?

ω is the first infinite ordinal number.
ℕ is the set of all finite ordinal numbers.
There is an obvious difference.

Regards, WM

Jim Burns schrieb am Sonntag, 9. Oktober 2022 um 20:23:03 UTC+2:
> On 10/9/2022 5:39 AM, WM wrote:

> > Every FISON-end can be reached one-by-one.

> Nothing on the 0-side of the 0-side,ω-side
> split cannot be reached one-by-one.

Not clear what you mean. But irrelevant since:
> > But every FISON contains less than half of
> > the infinite initial segment,
> > i.e., of all natural numbers {1, 2, 3, ...}.
> > Proof:
> > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > or
> > ∀n ∈ ℕ_def: 2n < ℵo ==> n < ℵo/2.
> The 0-side of the 0-side,ω-side split
> contains _all_ of the 0-side of the
> 0-side,ω-side split, without exception.

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

Your discussion deleted since it does not refer to dark numbers.

Regards, WM

zelos...@gmail.com schrieb am Montag, 10. Oktober 2022 um 07:32:27 UTC+2:
> torsdag 6 oktober 2022 kl. 21:25:22 UTC+2 skrev WM:
> > Fritz Feldhase schrieb am Mittwoch, 5. Oktober 2022 um 18:36:21 UTC+2:
> > > Because for each and every natural number n there are infinitely many natural numbers (finite ordinals) between n and omega. That's why.
> > >
> > > Hope this helps.
> > It helps to understand why they are dark. They cannot be removed. They cannot be the ends of leaps. They are undefinable.

> No, it helps nothign because what you say has no meaning!

So it appears to you. Don't worry.

Regards, WM