All definable endsegments E(n) of natural numbers
E(n) = {n, n+1, n+2, ...}
contain ℵo natural numbers. But each endsegment has one number less than its predecessor.

A bijection with the initial segments
I(n) = {1, 2, 3, ..., n}
produces the pairs

({1}, {1, 2, 3, ...,})
({1, 2}, {2, 3, 4, ...,})
({1, 2, 3}, {3, 4, 5, ...,})
....
({1, 2, 3, ..., n}, {n, n+1, n+2, ...})
....
({1, 2, 3, ...}, { })

Stop! If infinitely many infinite endsegments are existing, then the empty set cannot be reached in an infinite bijection, can it?

Regards, WM

On Tuesday, May 9, 2023 at 7:07:30 PM UTC+2, WM wrote:

> All [...] endsegments E(n) of natural numbers
> E(n) = {n, n+1, n+2, ...}
> contain ℵo natural numbers. [...]
>
> A [certain] bijection with the initial segments
> I(n) = {1, 2, 3, ..., n}
> produces the pairs
>
> ({1}, {1, 2, 3, ...,})
> ({1, 2}, {2, 3, 4, ...,})
> ({1, 2, 3}, {3, 4, 5, ...,})
> ...
> ({1, 2, 3, ..., n}, {n, n+1, n+2, ...})
> ...

Right. Actually, it's the bijection f: {I(n) : n e IN} --> {E(n) : n e IN} defined with f(x) = E(max(x)) for all x in {I(n) : n e IN}.

> ({1, 2, 3, ...}, { })
>
> Stop!

Indeed!

Hint: ({1, 2, 3, ...}, { }) is nonsense, since {1, 2, 3, ...} is NOT an element in the domain of f, nor is { } an element in the codomain of f.

You just pulled that out of your arse.

Actually, f = {({1}, {1, 2, 3, ...,}), ({1, 2}, {2, 3, 4, ...,}), ({1, 2, 3}, {3, 4, 5, ...,}), ..., ({1, 2, 3, ..., n}, {n, n+1, n+2, ...}), ...}

and [by definition of f]: ({1, 2, 3, ...}, { }) !e f.

Learn some logic, learn some math, Mückenheim!

Fritz Feldhase schrieb am Dienstag, 9. Mai 2023 um 19:45:48 UTC+2:
> On Tuesday, May 9, 2023 at 7:07:30 PM UTC+2, WM wrote:
>
> > All [...] endsegments E(n) of natural numbers
> > E(n) = {n, n+1, n+2, ...}
> > contain ℵo natural numbers. [...]
> >
> > A [certain] bijection with the initial segments
> > I(n) = {1, 2, 3, ..., n}
> > produces the pairs
> >
> > ({1}, {1, 2, 3, ...,})
> > ({1, 2}, {2, 3, 4, ...,})
> > ({1, 2, 3}, {3, 4, 5, ...,})
> > ...
> > ({1, 2, 3, ..., n}, {n, n+1, n+2, ...})
> > ...
> Right. Actually, it's the bijection f: {I(n) : n e IN} --> {E(n) : n e IN} defined with f(x) = E(max(x)) for all x in {I(n) : n e IN}.

Up to every n it is finite.

> > ({1, 2, 3, ...}, { })
> >
> > Stop!
> Indeed!
>
> Hint: ({1, 2, 3, ...}, { }) is nonsense, since {1, 2, 3, ...} is NOT an element in the domain of f, nor is { } an element in the codomain of f.

If we count endsegments by removing n, then the count is finite up to every endsegment that starts with a natural number. It is infinite only when all infinitely many natural numbers have been removed.

Regards, WM

WM brought next idea :
> All definable endsegments E(n) of natural numbers
> E(n) = {n, n+1, n+2, ...}
> contain ℵo natural numbers. But each endsegment has one number less than its
> predecessor.

One fewer symbols, name number of elements.

On 5/9/2023 1:07 PM, WM wrote:

> All definable endsegments E(n)
> of natural numbers
> E(n) = {n, n+1, n+2, ...}
> contain ℵo natural numbers.

Yes.
Each 1×1 1-ended has the same cardinality.
We name that single cardinality ℵ₀

> But each endsegment has
> one number less than its predecessor.

Yes.
n ∈ ⟨n,...⟩ and n ∉ ⟨n+1,...⟩

However,
each end segment is 1×1 1-ended.

1×1
For each split ⟨1,...,i⟩ ⟨i+1,...⟩
some i is last-before
and i+1 is first-after.

Use the rule
| if last-befores of
| ⟨n,...,j⟩ and ⟨n+1,...,k⟩ match
| j↔k
| then first-afters of
| ⟨j+1,...⟩ and ⟨k+1,...⟩ match
| j+1↔k+1

If any j ∈ ⟨n,...⟩ is not-matched
then
because ⟨n,...⟩ is 1×1
a first not-matched j₁ ∈ ⟨n,...⟩ exists
with j₁-1 matched.

But
| j₁-1 matching k₁-1 and
| j₁ not-matching k₁
contradicts the rule.

Therefore,
for each j ∈ ⟨n,...⟩
k ∈ ⟨n+1,...⟩ exists such that
j↔k

Likewise,
for each k ∈ ⟨n+1,...⟩
j ∈ ⟨n,...⟩ exists such that
j↔k

⟨n,...⟩ can match ⟨n+1,...⟩

> A bijection with the initial segments
> I(n) =
⟨1,...,n⟩
> produces the pairs

⟨⟨1⟩ ⟨1,...⟩⟩
⟨⟨1,2⟩ ⟨2,...⟩⟩
⟨⟨1,...,3⟩ ⟨3,...⟩⟩
....
⟨⟨1,...,n⟩ ⟨n,...⟩⟩
....
⟨⟨1,...⟩ ⟨⟩⟩

> Stop!
> If infinitely many infinite endsegments
> are existing,
> then the empty set cannot be reached
> in an infinite bijection, can it?

Your example shows an infinite bijection
which maps ⟨1,...⟩ to ⟨⟩
The pair ⟨⟨1,...⟩ ⟨⟩⟩ exists
More than that isn't needed.

For each n ∈ ⟨1,...⟩
n _can be reached_ from 1 because
for each split S of ⟨1,...,n⟩
S ⟨1,...,n⟩\S =
⟨1,...,j⟩ ⟨j+1,...,n⟩
some last-before j and first-after j+1
steps across that split,

A split without stepping-across
blocks the reach.
n can be reached if
that split S and every other split
does not block the reach.

> If infinitely many infinite endsegments
> are existing,
> then the empty set cannot be reached
> in an infinite bijection, can it?

This sequence of pairs is _almost_ 1×1

⟨⟨1⟩ ⟨1,...⟩⟩
⟨⟨1,2⟩ ⟨2,...⟩⟩
⟨⟨1,...,3⟩ ⟨3,...⟩⟩
....
⟨⟨1,...,n⟩ ⟨n,...⟩⟩
....
⟨⟨1,...⟩ ⟨⟩⟩

For _almost_ each split
some ⟨⟨1,...,i⟩ ⟨i,...⟩⟩ is last-before
and ⟨⟨1,...,i+1⟩ ⟨i+1,...⟩⟩ is first-after

Almost.

Assign those-which-can-be-reached to
one side of a split and
assign the rest to the split's other side.

If
last-before and first-after exist for
that split,
then
last-before and first-after exist for
each split between first and first-after,
and first-after can be reached,
and first-after is on the wrong side.

Contradiction.
So, last-before or first-after
for that split not-exists,
and
after that split cannot be reached.

Thus,
⟨⟨1,...⟩ ⟨⟩⟩ cannot be reached.

However,
⟨⟩ can be bijected to.

Last-befores and first-afters
aren't needed for that, just a pair.
⟨⟨1,...⟩ ⟨⟩⟩ for example.

On Tuesday, May 9, 2023 at 8:35:55 PM UTC+2, FromTheRafters wrote:
> WM brought next idea :
> > All definable endsegments E(n) of natural numbers
> > E(n) = {n, n+1, n+2, ...}
> > contain ℵo natural numbers. But each endsegment has one number less than its
> > predecessor.
> One fewer symbols, name number of elements.

...., same number of elements.

On Tuesday, May 9, 2023 at 8:20:16 PM UTC+2, WM wrote:

> If we count endsegments [...] then the count is finite up to every endsegment that starts with a natural number. It [reaches] infinit[y] only when

Chuck Norris does the counting.

Remember: He counted to infinity - twice.

On Tuesday, May 9, 2023 at 9:30:39 PM UTC+2, Jim Burns wrote:
> On 5/9/2023 1:07 PM, WM wrote:
> >
> > A bijection with the [finite] initial segments
> > I(n) = ⟨1,...,n⟩
> > produces the pairs
> >
> ⟨⟨1⟩ ⟨1,...⟩⟩
> ⟨⟨1,2⟩ ⟨2,...⟩⟩
> ⟨⟨1,...,3⟩ ⟨3,...⟩⟩
> ...
> ⟨⟨1,...,n⟩ ⟨n,...⟩⟩
> ...
> ⟨⟨1,...⟩ ⟨⟩⟩
>
> Your example shows an infinite bijection
> which maps ⟨1,...⟩ to ⟨⟩
And ⟨1,...⟩ is a finite initial segment in your book (i.e. identical with ⟨1,...,n⟩ for some n in IN)?
Fascinating.

Fritz Feldhase was thinking very hard :
> On Tuesday, May 9, 2023 at 8:35:55 PM UTC+2, FromTheRafters wrote:
>> WM brought next idea :
>>> All definable endsegments E(n) of natural numbers
>>> E(n) = {n, n+1, n+2, ...}
>>> contain ℵo natural numbers. But each endsegment has one number less than
>>> its predecessor.
>> One fewer symbols, name number of elements.
>
> ..., same number of elements.

Yes, good catch. :)

Jim Burns schrieb am Dienstag, 9. Mai 2023 um 21:30:39 UTC+2:
> On 5/9/2023 1:07 PM, WM wrote:
>
> > All definable endsegments E(n)
> > of natural numbers
> > E(n) = {n, n+1, n+2, ...}
> > contain ℵo natural numbers.
> Yes.
> Each 1×1 1-ended has the same cardinality.

Cardinality is meaningful only if lossless swaps suffer losses.
Not of interest to me.

> Your example shows an infinite bijection
> which maps ⟨1,...⟩ to ⟨⟩

In fact it swaps natnumbers from endsegments to initial segments.
> ⟨1,...,j⟩ ⟨j+1,...,n⟩
Only when all natnumbers have left the endsegements, the initial segments reach |N.

As long as definable natnumbers are swapped, the sequence is finite.
Endsegments sitting at a definable index of the sequence are infinite in a finite sequence.

> Thus,
> ⟨⟨1,...⟩ ⟨⟩⟩ cannot be reached.

It is impossible to remove all natnumbers? That proves dark numbers.

All definable natnumbers can be swapped from endsegments to initial segments.

Regards, WM

On Wednesday, May 10, 2023 at 3:18:19 PM UTC+2, WM wrote:
> Jim Burns schrieb am Dienstag, 9. Mai 2023 um 21:30:39 UTC+2:

> > Thus,
> > ⟨⟨1,...⟩ ⟨⟩⟩ cannot be reached.
> >
> It is impossible to remove all natnumbers [one by one]?

Only Chuck Norris can do this. (Since he's able to perform supertasks.)

Fritz Feldhase schrieb am Mittwoch, 10. Mai 2023 um 16:45:13 UTC+2:
> On Wednesday, May 10, 2023 at 3:18:19 PM UTC+2, WM wrote:
> > Jim Burns schrieb am Dienstag, 9. Mai 2023 um 21:30:39 UTC+2:
>
> > > Thus,
> > > ⟨⟨1,...⟩ ⟨⟩⟩ cannot be reached.
> > >
> > It is impossible to remove all natnumbers [one by one]?
>
> Only Chuck Norris can do this. (Since he's able to perform supertasks.)

Cantor can it too: "every number p/q comes at an absolutely fixed position of a simple infinite sequence". And you claim to be able too, because you believe in infinite bijections with |N.

Regards, WM

On 5/10/2023 9:18 AM, WM wrote:
> Jim Burns schrieb am Dienstag,
> 9. Mai 2023 um 21:30:39 UTC+2:

>> Thus,
>> ⟨⟨1,...⟩ ⟨⟩⟩ cannot be reached.
>
> It is impossible to remove all
> natnumbers?

It is impossible for each split
to have its last-before and first-after.

A split without both last-before and
first-after cannot be crossed.
Whatever is in the after-part
cannot be reached.

Anything one step from being finite
is itself finite.

A split between finitely-many steps
and after-all-finitely-many steps
cannot be crossed,
because
otherwise,
the infinite first-after would be
one step from the finite last-before.
and
both infinite and finite.

Imagine whatever you want to be
after-all-finitely-many steps.

They can't be reached,
NOT because
the after-part is "wrong" and
it needs to be completed,
but because
each one of the before-part
is not the last-before of that split,
and,
for the after-part to be reachable,
each split needs last-before and
first-after.

> It is impossible to remove all natnumbers?
> That proves dark numbers.

Whatever you imagine dark numbers to be,
for ℕ such that
∀M ⊆ ℕ:
M ∋ 0 ∧ ∀m ∈ M: M ∋ m⁺⁺ ⟺ M = ℕ

a second end not-exists.

Whatever you append after ℕ
that new set will have
a split without a last-before, and
the appendix cannot be reached.

----
Consider the set {reachable} of reachables.
Nothing outside of {reachable} is
one step from anything inside {reachable}
because
anything one step from
something reachable inside {reachable}
is itself reachable,
and, so, it would not be outside {reachable}.