If something holds for many natural numbers except almost all natural numbers, then it does not hold for all natural numbers. A property like
the visibility of a natural number holds only for a finite initial
segment {1, 2, 3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
Its existence is the precondition for visibility. Almost all natural
numbers remain invisible. Every natural number that can be chosen as an individual belongs to a finite initial segment of ℕ because for every choice, ℵo natural numbers are not chosen.
All natural numbers together can only be manipulated collectively:
|ℕ \ {1, 2, 3, ...}| = 0 .
Here all natural numbers have been manipulated (subtracted). Note: "All natural numbers" or ℕ denotes the complete set, i.e., a state such that
no remainder exists.
Regards, WM
If something holds for many natural numbers except almost all natural numbers, then it does not hold for all natural numbers. A property like the visibility of a natural number holds only for a finite initial segment {1,
2, 3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
WM <wolfgang.m...@tha.de> writes:
If something holds for many natural numbers except almost all natural numbers, then it does not hold for all natural numbers. A property like the
visibility of a natural number holds only for a finite initial segment {1, 2, 3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .But since ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo the above is true for
every (non-empty) subset of ℕ (including ℕ itself). ℕ_vis is only interesting if there is a n ∈ ℕ such that n ∉ ℕ_vis but since you won't
define the property itself so one but you can say anything about it!
--
Ben.
On Tuesday, 21 November 2023
at 12:37:35 UTC+1, Ben Bacarisse wrote:
[...]
I miss the "Tai" spammer...
*Plonk*
On 11/21/2023 6:44 AM, Julio Di Egidio wrote:
On Tuesday, 21 November 2023
at 12:37:35 UTC+1, Ben Bacarisse wrote:
[...]
I miss the "Tai" spammer...
You read spammers.
Why am I not surprised?
If something holds for
many natural numbers except
almost all natural numbers,
then it does not hold for
all natural numbers.
A property like
the visibility of a natural number
holds only for a finite initial
segment {1, 2, 3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
All natural numbers together
can only be manipulated collectively:
On 11/21/2023 5:33 AM, WM wrote:
A property like
the visibility of a natural number
holds only for a finite initial segment {1, 2, 3, ..., n}:
Because each natural number is in
some {1,2,3,...,n}
∀n ∈ ℕ_vis:
|ℕ_vis\{1,2,3,...,n}| = |ℕ_vis| := ℵ₀
All natural numbers together
can only be manipulated collectively:
When we say numbers are being manipulated,
it is nearly always understood to be a metaphor
for describing a manipulation of them.
WM <wolfgang.mueckenheim@tha.de> writes:
If something holds for many natural numbers except almost all natural
numbers, then it does not hold for all natural numbers. A property like the >> visibility of a natural number holds only for a finite initial segment {1, >> 2, 3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
But since ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
the above is true for
every (non-empty) subset of ℕ (including ℕ itself).
ℕ_vis is only
interesting if there is a n ∈ ℕ such that n ∉ ℕ_vis
but since you won't
define the property itself so one but you can say anything about it!
On 21.11.2023 21:45, Jim Burns wrote:
On 11/21/2023 5:33 AM, WM wrote:
A property like
the visibility of a natural number
holds only for a finite initial segment {1, 2, 3, ..., n}:
Because each natural number is in
some {1,2,3,...,n}
No, that is wrong. Each natural number in a FISON has ℵo successors. The collection of all natural numbers in FISONs cannot have card ℵo because all of them have ℵo successors. Two consecutive ℵo-sets in ℕ are impossible.
On 21.11.2023 21:45, Jim Burns wrote:
On 11/21/2023 5:33 AM, WM wrote:
A property like
the visibility of a natural number
holds only for
a finite initial segment {1, 2, 3, ..., n}:
Because each natural number is in
some {1,2,3,...,n}
No, that is wrong.
Each natural number in a FISON has
ℵo successors.
Each natural number in a FISON has
ℵo successors.
The collection of
all natural numbers in FISONs
cannot have card ℵo because
all of them have ℵo successors.
Two consecutive ℵo-sets in ℕ
are impossible.
Two consecutive ℵo-sets in ℕ
are impossible.
On 21.11.2023 12:37, Ben Bacarisse wrote:
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
That is wrong!
On 21.11.2023 21:45, Jim Burns wrote:
each [and every] natural number is in some {1, 2, 3, ..., n}.
No, that is wrong.
Each natural number in a FISON has ℵo successors.
The collection of all natural numbers in FISONs cannot have card ℵo
because all of them have ℵo successors.
Two <bla>
On 21.11.2023 12:37, Ben Bacarisse wrote:
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
That is wrong!
On 11/21/2023 4:54 PM, WM wrote:
No, that is wrong.
Each natural number in a FISON has
ℵo successors.
Each natural number in a FISON has
ℵ₀ successors in a FISON
Two consecutive ℵ₀-sets in ℕ
are also unnecessary.
If something holds for many natural numbers except almost all natural numbers, then it does not hold for all natural numbers. A property like the visibility of a natural number holds only for a finite initial segment {1, 2, 3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
Every natural number that can be chosen as an individual belongs to a finite initial segment of ℕ because for every choice, ℵo natural numbers are not chosen.
I miss the "Tai" spammer...
On 2023-11-21 10:33:17 +0000, WM said:
If something holds for many natural numbers except almost all natural
numbers, then it does not hold for all natural numbers. A property
like the
visibility of a natural number holds only for a finite initial segment
{1, 2,
3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
There is no "visibility" property of natural numbers.
However, it is
true that
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
Every natural number that can be chosen as an individual belongs to a
finite
initial segment of ℕ because for every choice, ℵo natural numbers are not
chosen.
There is no "can be chosen" property of natural numbers. Every natural
number can be chosen.
Why don't you choose a natural number that has less than ℵo successors?
On 23.11.2023 11:47, Mikko wrote:
On 2023-11-21 10:33:17 +0000, WM said:
If something holds for many natural numbers except almost all natural
numbers, then it does not hold for all natural numbers. A property like the >>> visibility of a natural number holds only for a finite initial segment {1, 2,
3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
There is no "visibility" property of natural numbers.
You can see and choose many, not all though.
However, it is
true that
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
That cannot be true for all natural numbers, because then almost all
(ℵo) natural numbers would follow upon all natural numbers.
Every natural number that can be chosen as an individual belongs to a finite
initial segment of ℕ because for every choice, ℵo natural numbers are not
chosen.
There is no "can be chosen" property of natural numbers. Every natural
number can be chosen.
Why don't you choose a natural number that has less than ℵo successors?
The collection ℕ_vis of all natural numbers that can be chosen has ℵo successors.
The collection of all natural numbers has no successors.
Collecting all natural numbers with the property
|ℕ \ {1, 2, 3, ..., n}| = ℵo
yields ℕ_vis
|ℕ \ ℕ_vis| = ℵo.
On 2023-11-23 15:44:48 +0000, WM said:
On 23.11.2023 11:47, Mikko wrote:
On 2023-11-21 10:33:17 +0000, WM said:
If something holds for many natural numbers except almost all natural
numbers, then it does not hold for all natural numbers. A property
like the
visibility of a natural number holds only for a finite initial
segment {1, 2,
3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
There is no "visibility" property of natural numbers.
You can see and choose many, not all though.
However, it is
true that
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
That cannot be true for all natural numbers, because then almost all
(ℵo) natural numbers would follow upon all natural numbers.
That "because" is unprovable and invalid.
Every natural number that can be chosen as an individual belongs to
a finite
initial segment of ℕ because for every choice, ℵo natural numbers
are not
chosen.
There is no "can be chosen" property of natural numbers. Every natural
number can be chosen.
Why don't you choose a natural number that has less than ℵo successors?
The same reason you can't choose a natural number less than zero:
there is no such natural number.
The collection ℕ_vis of all natural numbers that can be chosen has ℵo
successors.
As every natural number can be chosen, ℕ_vis = ℕ. Every natural number has ℵo successors but ℕ_vis has none.
The collection of all natural numbers has no successors.
Collecting all natural numbers with the property
|ℕ \ {1, 2, 3, ..., n}| = ℵo
yields ℕ_vis
|ℕ \ ℕ_vis| = ℵo.
No, it yields |ℕ \ ℕ_vis| = 0 because ℕ = ℕ_vis.
On 22.11.2023 16:40, Jim Burns wrote:
On 11/21/2023 4:54 PM, WM wrote:
No, that is wrong.
Each natural number in a FISON has
ℵo successors.
Each natural number in a FISON has
ℵ₀ successors in a FISON
Impossible because
two consecutive ℵo-sets in ℕ
are impossible.
On 24.11.2023 09:29, Mikko wrote:
As every natural number can be chosen,
ℕ_vis = ℕ.
Every natural number has ℵo successors
but ℕ_vis has none.
What you claim is a contradiction.
If ℕ_vis has not ℵo successors,
then the reason can only be, that
there are visible numbers with
less [successor].
What else should be the reason?
Without the existence of
visible numbers with less [successor],
all elements of ℕ_vis would have ℵo successors.
That means ℕ_vis would have
ℵo successors.
No, it yields |ℕ\ℕ_vis| = 0
because ℕ = ℕ_vis.
Then
you should be able to remove/subtract,
choosing one by one,
all natural numbers from ℕ.
On 21.11.2023 12:37, Ben Bacarisse wrote:
WM <wolfgang.mueckenheim@tha.de> writes:
If something holds for many natural numbers except almost all naturalBut since ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
numbers, then it does not hold for all natural numbers. A property like the >>> visibility of a natural number holds only for a finite initial segment {1, >>> 2, 3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
That is wrong!
|ℕ \ ℕ_vis| = ℵo .
but since you won't
define the property itself so one but you can say anything about it!
On 11/24/2023 5:16 AM, WM wrote:
Each number n in ℕ_vis
is visible and
determines a split of ℕ_vis into
1x1 2.ended visible predecessors and
1x1 1.ended visible successors,
ℵ₀-many visible successors.
Without the existence of
visible numbers with less [successor],
all elements of ℕ_vis would have ℵo successors.
Yes, they do.
That means ℕ_vis would have ℵo successors.
No.
No element of ℕ_vis succeeds
all elements of ℕ_vis.
No, it yields |ℕ\ℕ_vis| = 0
because ℕ = ℕ_vis.
Then
you should be able to remove/subtract,
choosing one by one,
all natural numbers from ℕ.
They (or you, or I) should not be able to
perform a supertask, such as
removing ℵ₀-many numbers one by one.
However,
reasoning about
removing ℵ₀-many numbers one by one
might not be supertask,
WM <wolfgang.mueckenheim@tha.de> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des Unendlichen" at Hochschule Augsburg.)
On 21.11.2023 12:37, Ben Bacarisse wrote:
WM <wolfgang.mueckenheim@tha.de> writes:
If something holds for many natural numbers except almost all naturalBut since ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
numbers, then it does not hold for all natural numbers. A property like the
visibility of a natural number holds only for a finite initial segment {1, >>>> 2, 3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
That is wrong!
No it's provably true, and I suspect you know that it is.
|ℕ \ ℕ_vis| = ℵo .
Since you can't define ℕ_vis
no one can prove you wrong. All you've
said is that ∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo and there are countless sets with that property.
A visible number can be named as an individual.
In fact, ℕ_vis is potentially infinite.
On 25.11.2023 01:59, Jim Burns wrote:
Each number n in ℕ_vis
is visible and
determines a split of ℕ_vis into
1x1 2.ended visible predecessors and
1x1 1.ended visible successors,
ℵ₀-many visible successors.
A visible number can be named
as an individual.
That is the definition of
the property "visible".
Name ℵ₀ visible successors.
Fail.
On 11/25/2023 12:13 PM, WM wrote:
On 25.11.2023 01:59, Jim Burns wrote:
Each number n in ℕ_vis
is visible and
determines a split of ℕ_vis into
1x1 2.ended visible predecessors and
1x1 1.ended visible successors,
ℵ₀-many visible successors.
A visible number can be named
as an individual.
That is the definition of
the property "visible".
The visibleᵂᴹ is nameableᵂᴹ
The nameableᵂᴹ is definableᵂᴹ
The definableᵂᴹ is accessibleᵂᴹ
The accessibleᵂᴹ is tangibleᵂᴹ
So. That's all cleared up.
From some comments you've made and some
things you have or have not objected to,
I have gathered the impression that visibleᵂᴹ-nameableᵂᴹ-definableᵂᴹ-...-edibleᵂᴹ
means not-darkᵂᴹ
and
darkᵂᴹ means ridiculously-large,
too large to be counted to by
universe-sized computers, et cetera.
_But finite_
A finite set has an order such that
first and last exists and,
for each split Fᣔ<ᣔH
last in F and first in H exist.
I suspect that you intend darkᵂᴹ and visibleᵂᴹ
to stand in place of infinite and finite
One important distinction between
darkᵂᴹ:visibleᵂᴹ and infinite:finite is that
ridiculously-large can be finite,
with first, last, last-befores and first-afters.
You might remember my mentioning:
infinite is not ridiculously-large finite.
Name ℵ₀ visible successors.
Fail.
Describe the natural numbers and
augment not-first-falsely with the claim
that each natural number has
ℵ₀ natural numbers after that number,
with none being last, none being after all.
On 2023-11-25 17:29:23 +0000, WM said:
On 25.11.2023 01:26, Ben Bacarisse wrote:
ℕ_vis is the collection of all natural numbers n that satisfy
|ℕ \ {1, 2, 3, ..., n}| = ℵo and that can be collected into a collection X
such that |ℕ \ X| = ℵo.
That is true about every finite subset of ℕ
it does not identify any particular subset.
ℕ_vis = X.
Still undefined.
no one can prove you wrong. All you've
said is that ∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo and there are
countless sets with that property.
In fact, ℕ_vis is potentially infinite. But all these countless sets
have a common property, namely |ℕ \ ℕ_vis| = ℵo.
You mean, because ℕ_vis is still undefined, we cannot know whether it
is infinite?
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo (*)
is wrong because collecting all n with the property (*) into ℕ_vis and
assuming |ℕ \ ℕ_vis| = 0, then the property (*) must have vanished by
magic spell. That is not mathematics.
It seems that nothing that mentions ℕ_vis is mathematics.
But in order to exclude also that spell, I define (see above): ℕ_vis
is the collection of all natural numbers n that satisfy
|ℕ \ {1, 2, 3, ..., n}| = ℵo and that can be collected such that the
collection keeps this property: |ℕ \ ℕ_vis| = ℵo. As soon as this
result is violated, we stop to collect.
With this procedure you never stop collecting and therefore never get ℕ_vis.
On 25.11.2023 01:26, Ben Bacarisse wrote:
WM <wolfgang.mueckenheim@tha.de> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)
On 21.11.2023 12:37, Ben Bacarisse wrote:
WM <wolfgang.mueckenheim@tha.de> writes:
If something holds for many natural numbers except almost all natural >>>>> numbers, then it does not hold for all natural numbers. A property like theBut since ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
visibility of a natural number holds only for a finite initial segment {1,
2, 3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
That is wrong!
No it's provably true, and I suspect you know that it is.
It is provable in an inconsistent theory.
|ℕ \ ℕ_vis| = ℵo .
Since you can't define ℕ_vis
ℕ_vis is the collection of all natural numbers n that satisfy
ℕ \ {1, 2, 3, ..., n}| = ℵo and that can be collected into a collection X
such that |ℕ \ X| = ℵo.
ℕ_vis = X.
no one can prove you wrong. All you've
said is that ∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo and there are
countless sets with that property.
In fact, ℕ_vis is potentially infinite. But all these countless sets
have a common property, namely |ℕ \ ℕ_vis| = ℵo.
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo (*)
is wrong because collecting all n with the property (*) into ℕ_vis and assuming |ℕ \ ℕ_vis| = 0, then the property (*) must have vanished by magic spell. That is not mathematics.
But in order to exclude also that spell, I define (see above): ℕ_vis is
the collection of all natural numbers n that satisfy
|ℕ \ {1, 2, 3, ..., n}| = ℵo and that can be collected such that the collection keeps this property: ℕ \ ℕ_vis| = ℵo. As soon as this result is violated, we stop to collect.
On 11/26/2023 5:19 AM, WM wrote:
All visible natnumbers belong toI think you're thinking "ended" and
a finite set which however
is not bounded.
writing "bounded"
All visible natnumbers belong to
an _infinite_ set which
is not _ended_
That there must be a first disagreement
follows from
the predecessors of n being finitely-many.
We cannot determine the order of dark numbers.Per you elsewhere, we also cannot say
that a darkᵂᴹ number equals itself.
For example,
∀n ∈ ℕ_vis: |ℕ_vis\⟨0,…,n⟩| = ℵ₀
On 26.11.2023 05:02, Jim Burns wrote:
A finite set has an order such that
first and last exists and,
for each split Fᣔ<ᣔH
last in F and first in H exist.
I suspect that you intend darkᵂᴹ and visibleᵂᴹ
to stand in place of infinite and finite
All visible natnumbers belong to
a finite set which however
is not bounded.
It is potentially infinite.
It is described by the Peano axioms.
One important distinction between
darkᵂᴹ:visibleᵂᴹ and infinite:finite is that
ridiculously-large can be finite,
with first, last, last-befores and first-afters.
WE cannot determine the order of dark numbers.
On Sunday, November 26, 2023 at 3:11:56 PM UTC-5, Transfinity wrote:that is 2 bit numbers may be called light.
Jim Burns schrieb am Sonntag, 26. November 2023 um 20:35:09 UTC+1:
On 11/26/2023 5:19 AM, WM wrote:
The set is finite at every instance, but the instances can grow without bound.All visible natnumbers belong toI think you're thinking "ended" and
a finite set which however
is not bounded.
writing "bounded"
All visible natnumbers belong toThe set is finite at every instance, but the instances can grow without bound.
an _infinite_ set which
is not _ended_
That there must be a first disagreementThat is fact because n is finite.
follows from
the predecessors of n being finitely-many.
We cannot prove it, but we must assume it. Otherwise dark numbers would be nonsense.We cannot determine the order of dark numbers.Per you elsewhere, we also cannot say
that a darkᵂᴹ number equals itself.
For example,Since ℕ_vis is finite and therefore smaller than ℵ₀, this is wrong.
∀n ∈ ℕ_vis: |ℕ_vis\⟨0,…,n⟩| = ℵ₀
Regards, WMi have logic source code that uses the set 0123 for lowlevel reason on truth trees.
i call set 0123 "light numbers".
from the light maybe you can get dark?
avoid negation and prosper in truth
daniel2380+++
Jim Burns schrieb am Sonntag, 26. November 2023 um 20:35:09 UTC+1:i have logic source code that uses the set 0123 for lowlevel reason on truth trees.
On 11/26/2023 5:19 AM, WM wrote:
The set is finite at every instance, but the instances can grow without bound.All visible natnumbers belong toI think you're thinking "ended" and
a finite set which however
is not bounded.
writing "bounded"
All visible natnumbers belong toThe set is finite at every instance, but the instances can grow without bound.
an _infinite_ set which
is not _ended_
That there must be a first disagreementThat is fact because n is finite.
follows from
the predecessors of n being finitely-many.
We cannot prove it, but we must assume it. Otherwise dark numbers would be nonsense.We cannot determine the order of dark numbers.Per you elsewhere, we also cannot say
that a darkᵂᴹ number equals itself.
For example,Since ℕ_vis is finite and therefore smaller than ℵ₀, this is wrong.
∀n ∈ ℕ_vis: |ℕ_vis\⟨0,…,n⟩| = ℵ₀
Regards, WM
On Sunday, November 26, 2023 at 3:38:02 PM UTC-5, Daniel Pehoushek wrote:with bignums you get 2^32 bit numbers that could be lighter than bigger numbers.
On Sunday, November 26, 2023 at 3:11:56 PM UTC-5, Transfinity wrote:
Jim Burns schrieb am Sonntag, 26. November 2023 um 20:35:09 UTC+1:
On 11/26/2023 5:19 AM, WM wrote:
The set is finite at every instance, but the instances can grow without bound.All visible natnumbers belong toI think you're thinking "ended" and
a finite set which however
is not bounded.
writing "bounded"
All visible natnumbers belong toThe set is finite at every instance, but the instances can grow without bound.
an _infinite_ set which
is not _ended_
That there must be a first disagreementThat is fact because n is finite.
follows from
the predecessors of n being finitely-many.
We cannot prove it, but we must assume it. Otherwise dark numbers would be nonsense.We cannot determine the order of dark numbers.Per you elsewhere, we also cannot say
that a darkᵂᴹ number equals itself.
For example,Since ℕ_vis is finite and therefore smaller than ℵ₀, this is wrong.
∀n ∈ ℕ_vis: |ℕ_vis\⟨0,…,n⟩| = ℵ₀
Regards, WMi have logic source code that uses the set 0123 for lowlevel reason on truth trees.
i call set 0123 "light numbers".
from the light maybe you can get dark?
avoid negation and prosper in truththat is 2 bit numbers may be called light.
daniel2380+++
you could call 32 bit numbers light.
with bignums you get 2^32 numbers that could be lighter than bigger numbers.
so we have degrees of lightness related to size.
with the comparison operator we have to handle the size relation.
does the set of natnumbers require or imply comparison with lessthan? daniel2380+++
On 24.11.2023 09:29, Mikko wrote:
On 2023-11-23 15:44:48 +0000, WM said:
On 23.11.2023 11:47, Mikko wrote:
On 2023-11-21 10:33:17 +0000, WM said:
If something holds for many natural numbers except almost all natural >>>>> numbers, then it does not hold for all natural numbers. A property like the
visibility of a natural number holds only for a finite initial segment {1, 2,
3, ..., n}:
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
There is no "visibility" property of natural numbers.
You can see and choose many, not all though.
However, it is
true that
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
That cannot be true for all natural numbers, because then almost all
(ℵo) natural numbers would follow upon all natural numbers.
That "because" is unprovable and invalid.
The same reason you can't choose a natural number less than zero:Every natural number that can be chosen as an individual belongs to a finite
initial segment of ℕ because for every choice, ℵo natural numbers are not
chosen.
There is no "can be chosen" property of natural numbers. Every natural >>>> number can be chosen.
Why don't you choose a natural number that has less than ℵo successors? >>
there is no such natural number.
Choose all natural number which you can choose as individuals. Collect
them into a set X. Remove that set from ℕ. What will remain?
On 26.11.2023 11:56, Mikko wrote:
On 2023-11-25 17:29:23 +0000, WM said:
On 25.11.2023 01:26, Ben Bacarisse wrote:
ℕ_vis is the collection of all natural numbers n that satisfy
|ℕ \ {1, 2, 3, ..., n}| = ℵo and that can be collected into a collection X
such that |ℕ \ X| = ℵo.
That is true about every finite subset of ℕ
No, it is not true for every finite subset, but only for every finite
initial segment {1, 2, 3, ..., n}. Note the "all" in the definition.
On 2023-11-24 10:16:18 +0000, WM said:
Choose all natural numbers which you can choose as individuals. Collect
them into a set X. Remove that set from ℕ. What will remain?
The empty set.
On 2023-11-26 11:13:19 +0000, WM said:
On 26.11.2023 11:56, Mikko wrote:
On 2023-11-25 17:29:23 +0000, WM said:
On 25.11.2023 01:26, Ben Bacarisse wrote:
ℕ_vis is the collection of all natural numbers n that satisfy
|ℕ \ {1, 2, 3, ..., n}| = ℵo and that can be collected into a collection X
such that |ℕ \ X| = ℵo.
That is true about every finite subset of ℕ
No, it is not true for every finite subset, but only for every finite initial segment {1, 2, 3, ..., n}. Note the "all" in the definition.It is also true for for example {3, 6, 9}.
it is not true?
Mikko
Mikko schrieb am Montag, 27. November 2023 um 12:05:47 UTC+1:
On 2023-11-26 11:13:19 +0000, WM said:
On 26.11.2023 11:56, Mikko wrote:It is also true for for example {3, 6, 9}.
On 2023-11-25 17:29:23 +0000, WM said:
On 25.11.2023 01:26, Ben Bacarisse wrote:
ℕ_vis is the collection of all natural numbers n that satisfy |ℕ \ {1,
2, 3, ..., n}| = ℵo and that can be collected into a collection X such >>>>> that |ℕ \ X| = ℵo.
That is true about every finite subset of ℕ
No, it is not true for every finite subset, but only for every finite
initial segment {1, 2, 3, ..., n}. Note the "all" in the definition.
ℕ_vis is the collection of _all_ natural numbers n ... that can be collected into a collection X such that |ℕ \ X| = ℵo.
On 2023-11-27 11:42:17 +0000, WM said:
Mikko schrieb am Montag, 27. November 2023 um 12:05:47 UTC+1:
On 2023-11-26 11:13:19 +0000, WM said:
On 26.11.2023 11:56, Mikko wrote:It is also true for for example {3, 6, 9}.
On 2023-11-25 17:29:23 +0000, WM said:
On 25.11.2023 01:26, Ben Bacarisse wrote:
ℕ_vis is the collection of all natural numbers n that satisfy |ℕ \ >>>>>> {1, 2, 3, ..., n}| = ℵo and that can be collected into a
collection X such that |ℕ \ X| = ℵo.
That is true about every finite subset of ℕ
No, it is not true for every finite subset, but only for every
finite initial segment {1, 2, 3, ..., n}. Note the "all" in the
definition.
ℕ_vis is the collection of _all_ natural numbers n ... that can be
collected into a collection X such that |ℕ \ X| = ℵo.
The collection {3, 6, 9} is a collection such that |ℕ \ {3, 6, 9}| = ℵo.
It contains all natural numbers that can be collected into {3, 6, 9}. Therefore it is a possible X.
On 2023-11-27 11:42:17 +0000, WM said:
Mikko schrieb am Montag, 27. November 2023 um 12:05:47 UTC+1:
On 2023-11-26 11:13:19 +0000, WM said:
On 26.11.2023 11:56, Mikko wrote:It is also true for for example {3, 6, 9}.
On 2023-11-25 17:29:23 +0000, WM said:
On 25.11.2023 01:26, Ben Bacarisse wrote:
ℕ_vis is the collection of all natural numbers n that satisfy |ℕ \ >>>>>> {1, 2, 3, ..., n}| = ℵo and that can be collected into a
collection X such that |ℕ \ X| = ℵo.
That is true about every finite subset of ℕ
No, it is not true for every finite subset, but only for every
finite initial segment {1, 2, 3, ..., n}. Note the "all" in the
definition.
ℕ_vis is the collection of _all_ natural numbers n ... that can be
collected into a collection X such that |ℕ \ X| = ℵo.
The collection {3, 6, 9} is a collection such that |ℕ \ {3, 6, 9}| = ℵo. It contains all natural numbers that can be collected into {3, 6, 9}. Therefore it is a possible X.
On 28.11.2023 10:47, Mikko wrote:
On 2023-11-27 11:42:17 +0000, WM said:
Mikko schrieb am Montag, 27. November 2023 um 12:05:47 UTC+1:
On 2023-11-26 11:13:19 +0000, WM said:
On 26.11.2023 11:56, Mikko wrote:It is also true for for example {3, 6, 9}.
On 2023-11-25 17:29:23 +0000, WM said:
On 25.11.2023 01:26, Ben Bacarisse wrote:
ℕ_vis is the collection of all natural numbers n that satisfy |ℕ \ {1,
2, 3, ..., n}| = ℵo and that can be collected into a collection X such
that |ℕ \ X| = ℵo.
That is true about every finite subset of ℕ
No, it is not true for every finite subset, but only for every finite >>>>> initial segment {1, 2, 3, ..., n}. Note the "all" in the definition.
ℕ_vis is the collection of _all_ natural numbers n ... that can be
collected into a collection X such that |ℕ \ X| = ℵo.
The collection {3, 6, 9} is a collection such that |ℕ \ {3, 6, 9}| = ℵo.
But it does not contain all smaller numbers.
ℕ_vis is the collection of ***all*** natural numbers n that satisfy |ℕ \ {1, 2, 3, ..., n}| = ℵo.
On 28.11.2023 10:47, Mikko wrote:
On 2023-11-27 11:42:17 +0000, WM said:
Mikko schrieb am Montag, 27. November 2023 um 12:05:47 UTC+1:
On 2023-11-26 11:13:19 +0000, WM said:
On 26.11.2023 11:56, Mikko wrote:It is also true for for example {3, 6, 9}.
On 2023-11-25 17:29:23 +0000, WM said:
On 25.11.2023 01:26, Ben Bacarisse wrote:
ℕ_vis is the collection of all natural numbers n that satisfy |ℕ \ {1,
2, 3, ..., n}| = ℵo and that can be collected into a collection X such
that |ℕ \ X| = ℵo.
That is true about every finite subset of ℕ
No, it is not true for every finite subset, but only for every finite >>>>> initial segment {1, 2, 3, ..., n}. Note the "all" in the definition.
ℕ_vis is the collection of _all_ natural numbers n ... that can be
collected into a collection X such that |ℕ \ X| = ℵo.
The collection {3, 6, 9} is a collection such that |ℕ \ {3, 6, 9}| = ℵo.
But it does not contain all smaller numbers.
On Wednesday, November 29, 2023 at 11:30:32 AM UTC+1, WM wrote:
ℕ_vis is the collection of ***all*** natural numbers n that satisfy |ℕ \ {1, 2, 3, ..., n}| = ℵo.In other words, ℕ_vis := {n e ℕ : |ℕ \ {1, 2, 3, ..., n}| = ℵo}.
Then ℕ_vis = ℕ.
Fritz Feldhase schrieb am Mittwoch, 29. November 2023 um 11:42:32 UTC+1:
On Wednesday, November 29, 2023 at 11:30:32 AM UTC+1, WM wrote:
ℕ_vis is the collection of ***all*** natural numbers n that satisfy |ℕ \ {1, 2, 3, ..., n}| = ℵo.
In other words, ℕ_vis := {n e ℕ : |ℕ \ {1, 2, 3, ..., n}| = ℵo}.
Then ℕ_vis = ℕ.
For <bla>
On 28.11.2023 10:47, Mikko wrote:
On 2023-11-27 11:42:17 +0000, WM said:ℕ_vis is the collection of ***all*** natural numbers n that satisfy |ℕ
Mikko schrieb am Montag, 27. November 2023 um 12:05:47 UTC+1:
On 2023-11-26 11:13:19 +0000, WM said:
On 26.11.2023 11:56, Mikko wrote:It is also true for for example {3, 6, 9}.
On 2023-11-25 17:29:23 +0000, WM said:
On 25.11.2023 01:26, Ben Bacarisse wrote:
ℕ_vis is the collection of all natural numbers n that satisfy |ℕ \ {1,
2, 3, ..., n}| = ℵo and that can be collected into a collection X such
that |ℕ \ X| = ℵo.
That is true about every finite subset of ℕ
No, it is not true for every finite subset, but only for every finite >>>>> initial segment {1, 2, 3, ..., n}. Note the "all" in the definition.
ℕ_vis is the collection of _all_ natural numbers n ... that can be
collected into a collection X such that |ℕ \ X| = ℵo.
The collection {3, 6, 9} is a collection such that |ℕ \ {3, 6, 9}| = ℵo. >> It contains all natural numbers that can be collected into {3, 6, 9}.
Therefore it is a possible X.
\ {1, 2, 3, ..., n}| = ℵo. See above.
The collection of ***all*** natural numbers n that satisfy |IN \ {1, 2, 3, ..., n}| = ℵo is IN.
On 2023-11-29 10:26:05 +0000, WM said:
ℕ_vis is the collection of ***all*** natural numbers n that satisfy |ℕ >> \ {1, 2, 3, ..., n}| = ℵo. See above.
Seeing above, it is not, just those that can be collected into a collection
X such that |ℕ \ X| = ℵo.
The collection of ***all*** natural numbers n that satisfy
|ℕ \ {1, 2, 3, ..., n}| = ℵo is N.
Es ist nicht so schwer: Die collection (Menge, Klasse) aller natürliche Zahlen n, für die ℕ \ {1, 2, 3, ..., n}| = ℵo gilt, ist gleich IN, weil für jede natürliche Zahl n |IN \ {1, 2, 3, ..., n}| = ℵo gilt.
On 29.11.2023 18:07, Mikko wrote:
On 2023-11-29 10:26:05 +0000, WM said:
ℕ_vis is the collection of ***all*** natural numbers n that satisfy |ℕ >>> \ {1, 2, 3, ..., n}| = ℵo. See above.
Seeing above, it is not, just those that can be collected into a collection >> X such that |ℕ \ X| = ℵo.
That is the definition.
Simply try to collect definable numbers such that |ℕ \ X| < ℵo. Fail.
Therefore, there is no X that could be used for the definition of ℕ_vis,
so ℕ_vis is still undefined.
On Thursday, November 30, 2023 at 4:41:45 PM UTC+1, WM wrote:
ℕ_vis contains numbers which have ℵo successors.
ℕ contains all numbers, also these successors.
In der Tat. Allerdings ist auch ℕ als ℕ_vis MÖGLICH, denn alle Zahlen in ℕ haben ℵo successors.
On 2023-11-30 15:41:40 +0000, WM said:
On 30.11.2023 11:53, Mikko wrote:
Therefore, there is no X that could be used for the definition of ℕ_vis, >>> so ℕ_vis is still undefined.
ℕ_vis contains numbers which have ℵo successors. ℕ contains all
numbers, also these successors. Therefore there is definitely a
difference.
Which of those numbers which have ℵo successors are contained in ℕ_vis?
Does ℕ_vis contain anything else?
On 30.11.2023 11:53, Mikko wrote:
Therefore, there is no X that could be used for the definition of ℕ_vis, >> so ℕ_vis is still undefined.
ℕ_vis contains numbers which have ℵo successors. ℕ contains all numbers, also these successors. Therefore there is definitely a
difference.
On 03.12.2023 12:00, Mikko wrote:
On 2023-11-30 15:41:40 +0000, WM said:
On 30.11.2023 11:53, Mikko wrote:
Therefore, there is no X that could be used for the definition of ℕ_vis, >>>> so ℕ_vis is still undefined.
ℕ_vis contains numbers which have ℵo successors. ℕ contains all
numbers, also these successors. Therefore there is definitely a
difference.
Which of those numbers which have ℵo successors are contained in ℕ_vis?
What we can say is: not all.
Mikko schrieb am Sonntag, 3. Dezember 2023 um 17:10:55 UTC+1:described by the Peano axioms. Very useful for mathematics.
On 2023-12-03 11:13:24 +0000, WM said:It is the collection (a set is invariable, but ℕ_vis is variable like the collection of known prime numbers or the collection of past days or the collection of numbers you can count to (those you can count to, not those you claim you could count to))
On 03.12.2023 12:00, Mikko wrote:So ℕ_vis is some unknown subset of ℕ. Not useful.
On 2023-11-30 15:41:40 +0000, WM said:
On 30.11.2023 11:53, Mikko wrote:
Therefore, there is no X that could be used for the definition of ℕ_vis,
so ℕ_vis is still undefined.
ℕ_vis contains numbers which have ℵo successors. ℕ contains all >>>>> numbers, also these successors. Therefore there is definitely a
difference.
Which of those numbers which have ℵo successors are contained in ℕ_vis?
What we can say is: not all.
Regards, WM
EVERY Natural number could be used individually, so all are potentially visible, and none are essentially dark.
Your "dark numbers" are not numbers that CAN'T be used individually, but numbers that currently HAVEN'T been used individually (YET).
There ARE infinitely many prime numbers, though almost all of themmay be "dark" (forever).
There's a certain dark number. Now - in a moment - I will tell youits /numeral/ (in decimal form).
In other words, even "dark" numbers are ordered,
On 04.12.2023 13:27, Richard Damon wrote:
EVERY Natural number could be used individually, so all are
potentially visible, and none are essentially dark.
The difference should be easily understandable here: You can empty ℕ by subtracting all natural numbers collectively
|ℕ \ {1, 2, 3, ...}| = 0.
You cannot empty ℕ by subtracting as many numbers as you can define individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Your "dark numbers" are not numbers that CAN'T be used individually,
but numbers that currently HAVEN'T been used individually (YET).
You are wrong. See above. Or consider the prime numbers. Never all prime numbers will be known. Or consider the passed days. Never ll days will
have been passed. There are always almost all remaining not used,
whatever you try.
Regards, WM
On 12/4/23 10:07 AM, WM wrote:
On 04.12.2023 13:27, Richard Damon wrote:
EVERY Natural number could be used individually, so all are
potentially visible, and none are essentially dark.
The difference should be easily understandable here: You can empty ℕ
by subtracting all natural numbers collectively
|ℕ \ {1, 2, 3, ...}| = 0.
You cannot empty ℕ by subtracting as many numbers as you can define
individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Being "Known" doesn't change the properties of the number itself, as it doesn't care if you know about it.
All Natural Numbers are "Visible",
On 05.12.2023 02:50, Richard Damon wrote:
On 12/4/23 10:07 AM, WM wrote:
On 04.12.2023 13:27, Richard Damon wrote:
EVERY Natural number could be used individually, so all are
potentially visible, and none are essentially dark.
The difference should be easily understandable here: You can empty ℕ
by subtracting all natural numbers collectively
|ℕ \ {1, 2, 3, ...}| = 0.
You cannot empty ℕ by subtracting as many numbers as you can define
individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Being "Known" doesn't change the properties of the number itself, as
it doesn't care if you know about it.
But it changes the facilities to handle them.
All Natural Numbers are "Visible",
Then empty the set ℕ by subtracting the visible individuals {1, 2, 3,
..., n} (all smaller numbers than n are automatically visible):
|ℕ \ {1, 2, 3, ..., n}|
Regards, WM
On 12/5/23 5:07 AM, WM wrote:
On 05.12.2023 02:50, Richard Damon wrote:Your problem seems to be that you are using operations that are bounded,
On 12/4/23 10:07 AM, WM wrote:
On 04.12.2023 13:27, Richard Damon wrote:
EVERY Natural number could be used individually, so all are
potentially visible, and none are essentially dark.
The difference should be easily understandable here: You can empty ℕ >>>> by subtracting all natural numbers collectively
|ℕ \ {1, 2, 3, ...}| = 0.
You cannot empty ℕ by subtracting as many numbers as you can define
individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
but the Natural Number are unbounded (they are each finite, but no upper limit to them) so bounded operations can't handle them.
Your "dark" numbers are just the number that live in the space between
the
bounded subset of the Natural Numbers that you logic can handle, and
the Unbound set that they actually are.
Show me how you create the actual set ℕ with your system?
Your
operations just don't have what is needed. IF they did, you could use
that exact same tool to create ℕ_vis, and thus show that all Natural Numbers were visible.
On 05.12.2023 13:31, Richard Damon wrote:
On 12/5/23 5:07 AM, WM wrote:
On 05.12.2023 02:50, Richard Damon wrote:Your problem seems to be that you are using operations that are
On 12/4/23 10:07 AM, WM wrote:
On 04.12.2023 13:27, Richard Damon wrote:
EVERY Natural number could be used individually, so all are
potentially visible, and none are essentially dark.
The difference should be easily understandable here: You can empty
ℕ by subtracting all natural numbers collectively
|ℕ \ {1, 2, 3, ...}| = 0.
You cannot empty ℕ by subtracting as many numbers as you can define >>>>> individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
bounded, but the Natural Number are unbounded (they are each finite,
but no upper limit to them) so bounded operations can't handle them.
Every handled number belongs to a finite set. The potentially infinite
set is not bounded by a finite number.
Your "dark" numbers are just the number that live in the space between
the
not finitely
bounded subset of the Natural Numbers that you logic can handle, and
the Unbound set that they actually are.
Show me how you create the actual set ℕ with your system?
That set cannot be created. It must be assumed to exist. Perhaps it does
not exist at all.
Only the visible numbers can be created by induction: with n also n+1 is visible. ℵo always remain dark - if they exist at all.
Your operations just don't have what is needed. IF they did, you could
use that exact same tool to create ℕ_vis, and thus show that all
Natural Numbers were visible.
Show a number that has less than ℵo successors. Show a unit fraction,
that has less than ℵo successors. Such unit fractions do exist, but
cannot be shown.
Regards, WM
On 12/6/23 12:34 PM, WM wrote:
Every handled number belongs to a finite set. The potentially infinite
set is not bounded by a finite number.
Right, but since the upper "bound" for that "finite set" isn't bounded,
the set it represent N_vis, becomes unbounded and equal to N.
Show me how you create the actual set ℕ with your system?
That set cannot be created. It must be assumed to exist. Perhaps it
does not exist at all.
So, if your system can't create N, it can't create your dark numbers, so
they don't actually exist.
Only the visible numbers can be created by induction: with n also n+1
is visible. ℵo always remain dark - if they exist at all.
Yes, ℵo is not a "Natural Number" but oly the value for the cardinality
of the Natural Numbers and there limit.
Anything less than it is what you call "visible", so there is no "dark" number.
0 is visible. But every visible unit fraction has ℵo smaller unitfractions.
On 07.12.2023 02:07, Richard Damon wrote:ℕ is not bounded by ω, since ω is not a member of ℕ. It might be a
On 12/6/23 12:34 PM, WM wrote:
Every handled number belongs to a finite set. The potentially
infinite set is not bounded by a finite number.
Right, but since the upper "bound" for that "finite set" isn't
bounded, the set it represent N_vis, becomes unbounded and equal to N.
The last part is wrong. Unbounded is correct. Dark numbers can be made visible without bound.∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
ℕ_vis is unbounded. But ℕ is bounded by ω
|ℕ \ {1, 2, 3, ...}| = 0, like the unit fractions are bounded by zero.
Show me how you create the actual set ℕ with your system?
That set cannot be created. It must be assumed to exist. Perhaps it
does not exist at all.
So, if your system can't create N, it can't create your dark numbers,
so they don't actually exist.
They do only exist if Cantor was right postulating actual infinity.
Only the visible numbers can be created by induction: with n also n+1
is visible. ℵo always remain dark - if they exist at all.
Yes, ℵo is not a "Natural Number" but oly the value for the
cardinality of the Natural Numbers and there limit.
Anything less than it is what you call "visible", so there is no
"dark" number.
That is wrong. It would mean in the parallel case that any unit fraction
0 is visible. But every visible unit fraction has ℵo smaller unitfractions.
Regards, WM
On 07.12.2023 02:07, Richard Damon wrote:
On 12/6/23 12:34 PM, WM wrote:
Every handled number belongs to a finite set. The potentially infinite
set is not bounded by a finite number.
Right, but since the upper "bound" for that "finite set" isn't bounded,
the set it represent N_vis, becomes unbounded and equal to N.
The last part is wrong. Unbounded is correct. Dark numbers can be made visible without bound.∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
On 2023-12-07 15:58:59 +0000, WM said:
Dark numbers can be made
visible without bound.∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
Same way dark numbers in ℕ_vis can be made visible without bound:
Hint: ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
On 12/7/23 10:58 AM, WM wrote:
ℕ_vis is unbounded. But ℕ is bounded by ωℕ is not bounded by ω, since ω is not a member of ℕ. It might be a limit, but not a bound.
|ℕ \ {1, 2, 3, ...}| = 0, like the unit fractions are bounded by zero.
They do only exist if Cantor was right postulating actual infinity.
No, because if your system can create the Naturals, it can describe all
of them, thus nothing is left to be dark.
That is wrong. It would mean in the parallel case that any unit
fraction > 0 is visible. But every visible unit fraction has ℵo
smaller unit fractions.
Right, which are all visible, just like all the high Natural Numbers.
On 08.12.2023 03:40, Richard Damon wrote:
On 12/7/23 10:58 AM, WM wrote:
ℕ_vis is unbounded. But ℕ is bounded by ωℕ is not bounded by ω, since ω is not a member of ℕ. It might be a
|ℕ \ {1, 2, 3, ...}| = 0, like the unit fractions are bounded by zero.
limit, but not a bound.
ℕ is in the same way restricted to less than ω, as the set of unit fractions is restricted to more than zero. Call it as you like.
They do only exist if Cantor was right postulating actual infinity.
No, because if your system can create the Naturals, it can describe
all of them, thus nothing is left to be dark.
You are wrong. Try to describe the smallest unit fraction.
Note: All unit fractions have distances. Therefore the increase of the function Number of Unit Fractions larger than zero, NUF(x) pauses after
each one. And never more than one are added simultaneously.
That is wrong. It would mean in the parallel case that any unit
fraction > 0 is visible. But every visible unit fraction has ℵo
smaller unit fractions.
Right, which are all visible, just like all the high Natural Numbers.
Prove it by naming the smallest one. In fact infinitely many smallest
ones are dark.
Regards, WM
On 12/8/23 10:16 AM, WM wrote:
Right, which are all visible, just like all the high Natural Numbers.
Prove it by naming the smallest one. In fact infinitely many smallest
ones are dark.
You can't name that which doesn't exist.
If your logic says there is a smallest unit fraction, then it is just incorrect.
You just don't seem to understand Bounded vs Unbounded sets.
On 08.12.2023 18:35, Richard Damon wrote:
On 12/8/23 10:16 AM, WM wrote:
Right, which are all visible, just like all the high Natural Numbers.
Prove it by naming the smallest one. In fact infinitely many smallest
ones are dark.
You can't name that which doesn't exist.
But you claim that for every unit fraction ℵ₀ smaller unit fractions exist. Why don't you name them, if all are existing? They are existing between your named one and zero.
If your logic says that there are ℵ₀ smaller unit fractions but they can't be named because they are not existing, then your logic is incorrect.
If your logic says there is a smallest unit fraction, then it is just
incorrect.
If your logic says that there is no smallest unit fraction, then it is
in contradiction with mathematics which even a Pisa-pupil should master:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 . Note the universal quantifier which does not admit ***any*** exception.
You just don't seem to understand Bounded vs Unbounded sets.
The set of unit fractions is bounded by 0 and 1.
Regards, WM
On 12/9/23 5:55 AM, WM wrote:
But you claim that for every unit fraction ℵ₀ smaller unit fractions
exist. Why don't you name them, if all are existing? They are existing
between your named one and zero.
Ok, for the unit fraction 1/n, the unit fractions smaller than it are:
{ 1/(n+1), 1/(n+2), 1/(n+3) ....}
Of course we can't name them ALL individually, as that is an unbounded operation which your "naming" is defined in a way that is bounded.
Note, to show that something isn't the "smallest", I don't neeed to name
ALL the points smaller, but just one, and that can easily be done.
If your logic says that there are ℵ₀ smaller unit fractions but they
can't be named because they are not existing, then your logic is
incorrect.
But YOU are the one that says they are not existing,
The ℵ₀ smaller unit fractions exist, and each one is individually nameable, we just can't name the full set, as "naming" is a bounded
operation while the set is unbounded.
If your logic says that there is no smallest unit fraction, then it is
in contradiction with mathematics which even a Pisa-pupil should master:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 . Note the universal quantifier which >> does not admit ***any*** exception.
That requirement says nothing about being "first"
There is no finite distance x that we can't get a d_n < x
, so we can
always find an increment that will fit into the gap.
On 09.12.2023 13:27, Richard Damon wrote:
On 12/9/23 5:55 AM, WM wrote:
But you claim that for every unit fraction ℵ₀ smaller unit fractions >>> exist. Why don't you name them, if all are existing? They are
existing between your named one and zero.
Ok, for the unit fraction 1/n, the unit fractions smaller than it are:
{ 1/(n+1), 1/(n+2), 1/(n+3) ....}
They all have ℵo smaller unit fractions. Name those which have less.
Don't less than ℵo exist in the set of ℵo?
Of course we can't name them ALL individually, as that is an unbounded
operation which your "naming" is defined in a way that is bounded.
Note, to show that something isn't the "smallest", I don't neeed to
name ALL the points smaller, but just one, and that can easily be done.
You claim that ℵo are smaller and can be found. But you can't find any
of the smaller ones which have less than ℵo smaller. Therefore your
claim is invalid. The ℵo smaller ones are dark and therefore cannpt be reduced by individual choice, only collectively.
If your logic says that there are ℵ₀ smaller unit fractions but they >>> can't be named because they are not existing, then your logic is
incorrect.
But YOU are the one that says they are not existing,
They are existing. They can be removed by using the complete set, but
not individually. They are dark.
The ℵ₀ smaller unit fractions exist, and each one is individually
nameable, we just can't name the full set, as "naming" is a bounded
operation while the set is unbounded.
Then name the last. You can subtract them completely.
If your logic says that there is no smallest unit fraction, then it
is in contradiction with mathematics which even a Pisa-pupil should
master:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 . Note the universal quantifier which >>> does not admit ***any*** exception.
That requirement says nothing about being "first"
It says that each one sits at its own point, followed by other points
than unit fractions.
There is no finite distance x that we can't get a d_n < x
The first distances are dark.
, so we can always find an increment that will fit into the gap.
Not any without ℵo smaller unit fractions. But you can't removes them individually. Only collectively. They are dark.
Regards, WM
On 12/9/23 11:27 AM, WM wrote:
, so we can always find an increment that will fit into the gap.
Not any without ℵo smaller unit fractions. But you can't removes them
individually. Only collectively. They are dark.
Which is just saying you can't remove the Unbounded set of the Natural Number, or their inverses, the Unit Fractions, by doing a Bounded
operation.
Since the Natural Numbers themselves are Unbounded, that is obvious.
That every Natural Number is "describable" is also an obvious fact, as
each one, individual, as a finite value.
Your "dark" numbers don't exist, but are only a "figment" of trying to
use Bounded logic on an Unbounded set.
Every Natural Number is describable, there are no undescribable Natural Numbers, but to try to describe them all at once, is an unbounded
operation.
On 08.12.2023 08:57, Mikko wrote:
On 2023-12-07 15:58:59 +0000, WM said:
Dark numbers can be made visible without bound.∀n ∈ ℕ_vis: |ℕ \ {1, 2,
3, ..., n}| = ℵo .
Same way dark numbers in ℕ_vis can be made visible without bound:
There are no visible dark numbers.
On 09.12.2023 17:43, Richard Damon wrote:
On 12/9/23 11:27 AM, WM wrote:
, so we can always find an increment that will fit into the gap.
Not any without ℵo smaller unit fractions. But you can't removes them
individually. Only collectively. They are dark.
Which is just saying you can't remove the Unbounded set of the Natural
Number, or their inverses, the Unit Fractions, by doing a Bounded
operation.
This means the same as "dark numbers are existing and cannot be removed individually".
Since the Natural Numbers themselves are Unbounded, that is obvious.
It is, but most people don't understand it.
That every Natural Number is "describable" is also an obvious fact, as
each one, individual, as a finite value.
Then you could describe and remove until nothing remains.
Your "dark" numbers don't exist, but are only a "figment" of trying to
use Bounded logic on an Unbounded set.
Every Natural Number is describable, there are no undescribable
Natural Numbers, but to try to describe them all at once, is an
unbounded operation.
It is impossible to perform. That proves the existence of dark numbers.
Regards, WM
On 2023-12-08 15:08:45 +0000, WM said:
On 08.12.2023 08:57, Mikko wrote:
On 2023-12-07 15:58:59 +0000, WM said:
Dark numbers can be made visible without bound.∀n ∈ ℕ_vis: |ℕ \ {1,
2, 3, ..., n}| = ℵo .
Same way dark numbers in ℕ_vis can be made visible without bound:
There are no visible dark numbers.
Without a definition of "dark number" that cannot be proven.
On 12/10/23 3:58 AM, WM wrote:
On 09.12.2023 17:43, Richard Damon wrote:
On 12/9/23 11:27 AM, WM wrote:
, so we can always find an increment that will fit into the gap.
Not any without ℵo smaller unit fractions. But you can't removes
them individually. Only collectively. They are dark.
Which is just saying you can't remove the Unbounded set of the
Natural Number, or their inverses, the Unit Fractions, by doing a
Bounded operation.
This means the same as "dark numbers are existing and cannot be
removed individually".
No, because no Natural Number is itself unbounded, so no Natural Number
is dark.
Boundedness is a property of SETS, not NUMBERS.
That every Natural Number is "describable" is also an obvious fact,
as each one, individual, as a finite value.
Then you could describe and remove until nothing remains.
But that is a SET operation. The SET of Natural Numbers is unbounded, so
can not be
Nope. Again, the SET of Natural Numbers is Unbounded, but each
individual Natural Number is finite (and thus part of a Bounded set)
We can individual describe only finite things (since we are finite) so
we CAN describe ANY Natural Number, so no Natural Number is "Dark".
You are just confusing the SET of Natural Numbers for the individual
elements of the set.
I guess, your concept of "Set Theory" is incapable of handling unbounded sets.
On 10.12.2023 13:28, Richard Damon wrote:
On 12/10/23 3:58 AM, WM wrote:
On 09.12.2023 17:43, Richard Damon wrote:
On 12/9/23 11:27 AM, WM wrote:
, so we can always find an increment that will fit into the gap.
Not any without ℵo smaller unit fractions. But you can't removes
them individually. Only collectively. They are dark.
Which is just saying you can't remove the Unbounded set of the
Natural Number, or their inverses, the Unit Fractions, by doing a
Bounded operation.
This means the same as "dark numbers are existing and cannot be
removed individually".
No, because no Natural Number is itself unbounded, so no Natural
Number is dark.
Every natural number is finite. Every set of individually defined
natnumbers is finite.
Boundedness is a property of SETS, not NUMBERS.
Of course.
That every Natural Number is "describable" is also an obvious fact,
as each one, individual, as a finite value.
Then you could describe and remove until nothing remains.
But that is a SET operation. The SET of Natural Numbers is unbounded,
so can not be
That is just what I said.
The set of numbers smaller than 1000 can be emptied by individual
operations. An actually infinite set cannot be emptied by individual operations. The remainder is called dark.
Nope. Again, the SET of Natural Numbers is Unbounded, but each
individual Natural Number is finite (and thus part of a Bounded set)
The set consists of only natnumbers. Thoese which can be used as
individuals leave almost all natnumbers as remainder.
We can individual describe only finite things (since we are finite) so
we CAN describe ANY Natural Number, so no Natural Number is "Dark".
Wrong. The unit fractions have a first element because they all have
finite distances. But you cannot describe it.
You are just confusing the SET of Natural Numbers for the individual
elements of the set.
The set is nothing but its members.
I guess, your concept of "Set Theory" is incapable of handling
unbounded sets.
It is better than all others because they fail to distinguish potential
and actual infinity.
Regards, WM
On 10.12.2023 11:35, Mikko wrote:
On 2023-12-08 15:08:45 +0000, WM said:
On 08.12.2023 08:57, Mikko wrote:
On 2023-12-07 15:58:59 +0000, WM said:
Dark numbers can be made visible without bound.∀n ∈ ℕ_vis: |ℕ \ {1,
2, 3, ..., n}| = ℵo .
Same way dark numbers in ℕ_vis can be made visible without bound:
There are no visible dark numbers.
Without a definition of "dark number" that cannot be proven.
Dark numbers are well defined.
Definition: A natural number is "identified" or (individually) "defined"
or "instantiated" if it can be communicated such that sender and
receiver understand the same and can link it by a finite initial segment
to the origin 0. All other natural numbers are called dark natural numbers.
Since it is impossible to reduce the undefined numbers to less than ℵo, almost are natural numbers are dark and will remain dark.
Regards, WM
On 10.12.2023 11:35, Mikko wrote:
On 2023-12-08 15:08:45 +0000, WM said:
On 08.12.2023 08:57, Mikko wrote:
On 2023-12-07 15:58:59 +0000, WM said:
Dark numbers can be made visible without bound.∀n ∈ ℕ_vis: |ℕ \ {1, 2,
3, ..., n}| = ℵo .
Same way dark numbers in ℕ_vis can be made visible without bound:
There are no visible dark numbers.
Without a definition of "dark number" that cannot be proven.
Dark numbers are well defined.
Definition: A natural number is "identified" or (individually)
"defined" or "instantiated" if it can be communicated such that sender
and receiver understand the same and can link it by a finite initial
segment to the origin 0. All other natural numbers are called dark
natural numbers.
On 12/11/23 6:20 AM, WM wrote:
Definition: A natural number is "identified" or (individually)
"defined" or "instantiated" if it can be communicated such that sender
and receiver understand the same and can link it by a finite initial
segment to the origin 0. All other natural numbers are called dark
natural numbers.
Since it is impossible to reduce the undefined numbers to less than
ℵo, almost are natural numbers are dark and will remain dark.
If you are implying that they are numbrs that can't be communicated,
then that has nothing to do with being able to list ALL the numbers in
finite time.
Your logic confuses a property of the Set, it being unbounded in size,
and thus not individually listable, with the properties of the elements, which are all finite, and thus defined/instantiateable/communicable.
There are an unbounded number of "Finite Initial Segments" that could be transmitted, so we can cover all of the Natural Numbers in that set, and "reduce" the "undefined" numbers to 0.
So, your Darkness is just a category error of applying a Set Property to
an Element of it.
On 2023-12-11 11:20:21 +0000, WM said:
Dark numbers are well defined.
Definition: A natural number is "identified" or (individually)
"defined" or "instantiated" if it can be communicated such that sender
and receiver understand the same and can link it by a finite initial
segment to the origin 0. All other natural numbers are called dark
natural numbers.
That definition has no significance as long as it has not been included
in a proof.
On 11.12.2023 23:56, Richard Damon wrote:
On 12/11/23 6:20 AM, WM wrote:
Definition: A natural number is "identified" or (individually)
"defined" or "instantiated" if it can be communicated such that
sender and receiver understand the same and can link it by a finite
initial segment to the origin 0. All other natural numbers are called
dark natural numbers.
Since it is impossible to reduce the undefined numbers to less than
ℵo, almost are natural numbers are dark and will remain dark.
If you are implying that they are numbrs that can't be communicated,
then that has nothing to do with being able to list ALL the numbers in
finite time.
But it has to do with not being able to list the last one.
Your logic confuses a property of the Set, it being unbounded in size,
and thus not individually listable, with the properties of the
elements, which are all finite, and thus
defined/instantiateable/communicable.
If all elements were listable, then all could be listed.
There are an unbounded number of "Finite Initial Segments" that could
be transmitted, so we can cover all of the Natural Numbers in that
set, and "reduce" the "undefined" numbers to 0.
No you can't. If you could transmit every one such that none would
remain, the you could list the last one. That is impossible. Therefore
there remain always amost all natnumbers unlisted.
So, your Darkness is just a category error of applying a Set Property
to an Element of it.
We talk about the numbers.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
The set is used here:
|ℕ \ {1, 2, 3, ...}| = 0
There is in fact a difference as can see.
Regards, WM
On 12/12/2023 3:14 AM, Mikko wrote:
On 2023-12-11 11:20:21 +0000, WM said:Why? There have been definitions of interesting mathematical objects
On 10.12.2023 11:35, Mikko wrote:
On 2023-12-08 15:08:45 +0000, WM said:
On 08.12.2023 08:57, Mikko wrote:
On 2023-12-07 15:58:59 +0000, WM said:
Dark numbers can be made visible without bound.∀n ∈ ℕ_vis: |ℕ \ >>>>>>>
{1, 2, 3, ..., n}| = ℵo .
Same way dark numbers in ℕ_vis can be made visible without bound: >>>>>>
There are no visible dark numbers.
Without a definition of "dark number" that cannot be proven.
Dark numbers are well defined.
Definition: A natural number is "identified" or (individually)
"defined" or "instantiated" if it can be communicated such that sender
and receiver understand the same and can link it by a finite initial
segment to the origin 0. All other natural numbers are called dark
natural numbers.
That definition has no significance as long as it has not been included
in a proof.
around for decades before the proof that there was or was not an element
that fit the definition.
In fact you could "prove" that every dark number has infinite
predecessors. That isn't very interesting if there are no dark numbers
though.
On 12.12.2023 11:14, Mikko wrote:
On 2023-12-11 11:20:21 +0000, WM said:
Dark numbers are well defined.
Definition: A natural number is "identified" or (individually)
"defined" or "instantiated" if it can be communicated such that sender
and receiver understand the same and can link it by a finite initial
segment to the origin 0. All other natural numbers are called dark
natural numbers.
That definition has no significance as long as it has not been included
in a proof.
It is the result of many proofs.
Definition: A natural number is "identified" or (individually)
"defined" or "instantiated" if it can be communicated such that
sender and receiver understand the same and can link it by a finite
initial segment to the origin 0. All other natural numbers are called
dark natural numbers.
In fact you could "prove" that every dark number has infinite
predecessors.
If you build a system on things that are not known to exist you don't
know whether the system is consistent.
On 12/12/23 5:16 AM, WM wrote:
We talk about the numbers.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
The set is used here:
|ℕ \ {1, 2, 3, ...}| = 0
There is in fact a difference as we can see.
And, if you want to claim that the remainder is "Dark Numbers", show me a
set that in non-empty, and contains only "Dark Numbers" (and not ANY of N_def)
Note, ℕ \ {1, 2, 3, ..., n} is not such a set, as it contains the
definite number (n+1)
Show me one which has NO definite numbers, but does have some Natural
Numbers (some dark ones)
You can't try to weasel out by saying they are only useable as a
collection, because I am asking for such a collection.
Your problem is going to be that your "Darkness" is actually a property
of Unbounded Sets,
On 2023-12-12 11:55:16 +0000, WM said:
That definition has no significance as long as it has not been included
in a proof.
It is the result of many proofs.
A definition is never the result of a proof.
On 13.12.2023 10:47, Mikko wrote:
On 2023-12-12 11:55:16 +0000, WM said:
That definition has no significance as long as it has not been included >>>> in a proof.
It is the result of many proofs.
A definition is never the result of a proof.
Inaccessible, dark, numbers are the result of many proofs. The
definition has been given to describe them.
On 13.12.2023 11:00, Mikko wrote:
If you build a system on things that are not known to exist you don't
know whether the system is consistent.
That is impossible to know for all Gödel-systems.
Regards, WM
On 2023-12-13 17:05:37 +0000, WM said:
On 13.12.2023 10:47, Mikko wrote:
On 2023-12-12 11:55:16 +0000, WM said:
That definition has no significance as long as it has not been
included
in a proof.
It is the result of many proofs.
A definition is never the result of a proof.
Inaccessible, dark, numbers are the result of many proofs. The
definition has been given to describe them.
Not of any proofs about natural numbers. That there are real numbers that cannot be described is a well known result.
On 13.12.2023 18:47, Mikko wrote:
On 2023-12-13 17:05:37 +0000, WM said:
On 13.12.2023 10:47, Mikko wrote:
On 2023-12-12 11:55:16 +0000, WM said:
That definition has no significance as long as it has not been
included
in a proof.
It is the result of many proofs.
A definition is never the result of a proof.
Inaccessible, dark, numbers are the result of many proofs. The
definition has been given to describe them.
Not of any proofs about natural numbers. That there are real numbers that
cannot be described is a well known result.
There are natural numbers that cannot be described.
Proof: For every x > 0 there are infinitely many smaller unit fractions.
That means infinitely many unit fractions and their natural numbers
cannot be chosen or described as individuals by choosing or describing
any individual x > 0.
Regards, WM
On 12.12.2023 13:13, Richard Damon wrote:
On 12/12/23 5:16 AM, WM wrote:
We talk about the numbers.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
The set is used here:
|ℕ \ {1, 2, 3, ...}| = 0
There is in fact a difference as we can see.
And, if you want to claim that the remainder is "Dark Numbers", show
me a set that in non-empty, and contains only "Dark Numbers" (and not
ANY of N_def)
Impossible. ℕ_def is potentially infinite. That makes this matter so difficult to understand for you.
Note, ℕ \ {1, 2, 3, ..., n} is not such a set, as it contains the
definite number (n+1)
To be precise, it is not a set at all, because it can grow (and shrink).
We call it a collection.
Show me one which has NO definite numbers, but does have some Natural
Numbers (some dark ones)
Impossible.
You can't try to weasel out by saying they are only useable as a
collection, because I am asking for such a collection.
Your problem is going to be that your "Darkness" is actually a
property of Unbounded Sets,
It is a property of actually infinite sets. The collection ℕ_def is unbounded but without dark numbers.
Regards, WM
On 13.12.2023 11:00, Mikko wrote:
If you build a system on things that are not known to exist you don't
know whether the system is consistent.
That is impossible to know for all Gödel-systems.
On 13.12.2023 18:47, Mikko wrote:
On 2023-12-13 17:05:37 +0000, WM said:
On 13.12.2023 10:47, Mikko wrote:
On 2023-12-12 11:55:16 +0000, WM said:
That definition has no significance as long as it has not been included >>>>>> in a proof.
It is the result of many proofs.
A definition is never the result of a proof.
Inaccessible, dark, numbers are the result of many proofs. The
definition has been given to describe them.
Not of any proofs about natural numbers. That there are real numbers that
cannot be described is a well known result.
There are natural numbers that cannot be described.
Proof: For every x > 0 there are infinitely many smaller unit
fractions. That means infinitely many unit fractions and their natural numbers cannot be chosen or described as individuals by choosing or describing any individual x > 0.
Richard Damon schrieb am Freitag, 15. Dezember 2023 um 01:11:04 UTC+1:
On 12/14/23 9:51 AM, WM wrote:
There are natural numbers that cannot be described.You haven't show that they can't be described/chosen individually, just
Proof: For every x > 0 there are infinitely many smaller unit fractions. >>> That means infinitely many unit fractions and their natural numbers
cannot be chosen or described as individuals by choosing or describing
any individual x > 0.
that you can't describe them all at once.
Then describe a unit fraction that has remained undescribed by me.
Any Natural Number can be chosen or described as an individual. It just
is we can't do that to EVERY Number AT ONCE.
Describe not every one at once but only one of those which
remain forever undescribed by me.
Regards, WM
Mikko schrieb am Freitag, 15. Dezember 2023 um 12:19:12 UTC+1:
On 2023-12-14 14:51:54 +0000, WM said:
On 13.12.2023 18:47, Mikko wrote:
On 2023-12-13 17:05:37 +0000, WM said:
On 13.12.2023 10:47, Mikko wrote:
On 2023-12-12 11:55:16 +0000, WM said:
There are natural numbers that cannot be described.That "this means" above is not bvious and is not proven and is actually false.
Proof: For every x > 0 there are infinitely many smaller unit
fractions. That means infinitely many unit fractions and their natural
numbers cannot be chosen or described as individuals by choosing or
describing any individual x > 0.
Then describe how these numbers can be chosen as individuals.
On 2023-12-15 17:33:33 +0000, WM said:
Mikko schrieb am Freitag, 15. Dezember 2023 um 12:19:12 UTC+1:
On 2023-12-14 14:51:54 +0000, WM said:
There are natural numbers that cannot be described.That "this means" above is not bvious and is not proven and is actually false.
Proof: For every x > 0 there are infinitely many smaller unit
fractions. That means infinitely many unit fractions and their natural >>>> numbers cannot be chosen or described as individuals by choosing or
describing any individual x > 0.
Then describe how these numbers can be chosen as individuals.
Every natural number has a name,
On 12/15/23 12:26 PM, WM wrote:
Richard Damon schrieb am Freitag, 15. Dezember 2023 um 01:11:04 UTC+1:
On 12/14/23 9:51 AM, WM wrote:
There are natural numbers that cannot be described.You haven't show that they can't be described/chosen individually, just
Proof: For every x > 0 there are infinitely many smaller unit fractions. >>>> That means infinitely many unit fractions and their natural numbers
cannot be chosen or described as individuals by choosing or describing >>>> any individual x > 0.
that you can't describe them all at once.
Then describe a unit fraction that has remained undescribed by me.
You have admitted you can't even give a collective set of your dark
numbers for me to even attempt it.
EVERY Natural Number is described
Describe not every one at once but only one of those which
remain forever undescribed by me.
There are no numbers that you haven't described, since you "describe"
EVERY natural number.
Le 16/12/2023 à 09:01, Mikko a écrit :
On 2023-12-15 17:33:33 +0000, WM said:
Mikko schrieb am Freitag, 15. Dezember 2023 um 12:19:12 UTC+1:
On 2023-12-14 14:51:54 +0000, WM said:
There are natural numbers that cannot be described.That "this means" above is not bvious and is not proven and is
Proof: For every x > 0 there are infinitely many smaller unit
fractions. That means infinitely many unit fractions and their natural >>>>> numbers cannot be chosen or described as individuals by choosing or
describing any individual x > 0.
actually false.
Then describe how these numbers can be chosen as individuals.
Every natural number has a name,
Name the first unit fraction.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
Note smallest universal quantifier. There is no exception. Therefore
NUF(x) can only increase one by one.
∀x ∈ (0, 1]: NUF(x) = ℵo
is wrong.
Regards, WM
Le 16/12/2023 à 02:05, Richard Damon a écrit :
On 12/15/23 12:26 PM, WM wrote:
Richard Damon schrieb am Freitag, 15. Dezember 2023 um 01:11:04 UTC+1:
On 12/14/23 9:51 AM, WM wrote:
There are natural numbers that cannot be described.You haven't show that they can't be described/chosen individually, just >>>> that you can't describe them all at once.
Proof: For every x > 0 there are infinitely many smaller unit
fractions.
That means infinitely many unit fractions and their natural numbers
cannot be chosen or described as individuals by choosing or describing >>>>> any individual x > 0.
Then describe a unit fraction that has remained undescribed by me.
You have admitted you can't even give a collective set of your dark
numbers for me to even attempt it.
Collectively I can: All numbers which cannot be described as individuals
and have no finite initial segmente {1, 2, 3, ..., n} are dark.
EVERY Natural Number is described
Describe the smallest unit fraction.
Describe not every one at once but only one of those which
remain forever undescribed by me.
There are no numbers that you haven't described, since you "describe"
EVERY natural number.
Not by their finite initial segment.
Regards, WM
On 12/16/23 7:24 AM, WM wrote:
1/1 is the "first" as the unit fractions are numbered from the 1 end.
Name the first unit fraction.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
Note smallest universal quantifier. There is no exception. Therefore
NUF(x) can only increase one by one.
∀x ∈ (0, 1]: NUF(x) = ℵo
is wrong.
There is no other end, as Natural Numbers are Unbounded.
Any logic that claims otherwise isn't able to handle Unbounded sets.
Thus, NUF isn't actually properly defined, as it can only handle inputs
that are unboundedly small
and thus (in your
words) numbers not individually usable, that must be used individually
but can't be becuase they are "dark".
Le 16/12/2023 à 13:45, Richard Damon a écrit :
On 12/16/23 7:24 AM, WM wrote:
1/1 is the "first" as the unit fractions are numbered from the 1 end.
Name the first unit fraction.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
Note smallest universal quantifier. There is no exception. Therefore
NUF(x) can only increase one by one.
∀x ∈ (0, 1]: NUF(x) = ℵo
is wrong.
There is no other end, as Natural Numbers are Unbounded.
There is NUF(x) which increases one by one with gaps.
Any logic that claims otherwise isn't able to handle Unbounded sets.
My logic handles actually infinite sets.
Thus, NUF isn't actually properly defined, as it can only handle
inputs that are unboundedly small
All natural numbers are xisting according to set theory. Hence all unit fractions are existing. None is missing. And after each point 1/n the is
a constant level of NUF(x).
and thus (in your words) numbers not individually usable, that must be
used individually but can't be becuase they are "dark".
They cannot be used individually. Nevertheless, according to Cantor,
they exist.
Regards, WM
On 12/16/23 9:06 AM, WM wrote:
There is NUF(x) which increases one by one with gaps.
Which you have been unnable to actually show a value of at any defined number.
By your logic, it increases one by one at numbers that are dark,
What value of x does NUF(x) have a finite value?
NUF(x) for all defined numbers is "infinte", which isn't actually a
Natural Number.
So your NUF is a mapping from non-Natural Numbers or to non-Natural
Numbers so isn't actually a function in the domain of Natural Numbers.
Quote the ACTUAL WORDS that Cantor uses to say that there exists numbersand thus (in your words) numbers not individually usable, that must be
used individually but can't be becuase they are "dark".
They cannot be used individually. Nevertheless, according to Cantor,
they exist.
that can not be used as an individual.
Le 16/12/2023 à 18:15, Richard Damon a écrit :
On 12/16/23 9:06 AM, WM wrote:
There is NUF(x) which increases one by one with gaps.
Which you have been unnable to actually show a value of at any defined
number.
At defined numbers the value is infinite. But it is 0 at 0 and all unit fractions have distances from each other.
By your logic, it increases one by one at numbers that are dark,
The one-by-one increase is dictated by mathematics. It cannot be seen.
That implies the existence of dark numbers.
What value of x does NUF(x) have a finite value?
x is smaller than every definable eps > 0.
NUF(x) for all defined numbers is "infinte", which isn't actually a
Natural Number.
Yes.
So your NUF is a mapping from non-Natural Numbers or to non-Natural
Numbers so isn't actually a function in the domain of Natural Numbers.
The domain is the real axis.
Quote the ACTUAL WORDS that Cantor uses to say that there existsand thus (in your words) numbers not individually usable, that must
be used individually but can't be becuase they are "dark".
They cannot be used individually. Nevertheless, according to Cantor,
they exist.
numbers that can not be used as an individual.
The completed infinite, das vollendete Unendliche or Vollendetunendliche
as Cantor called it [letter to Lipschitz (19 Nov. 1883) and E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 391], is a
prerequisite of set theory. Hence no fraction imust be missing.
Regards, WM
On 12/17/23 4:31 PM, WM wrote:
Le 16/12/2023 à 18:15, Richard Damon a écrit :
On 12/16/23 9:06 AM, WM wrote:
There is NUF(x) which increases one by one with gaps.
Which you have been unnable to actually show a value of at any defined
number.
At defined numbers the value is infinite. But it is 0 at 0 and all unit
fractions have distances from each other.
So?
The distance is unboundedly small, and moves from 0 to infinity in an unboundedly small distance.
This is just the affect of being at the unbounded
By your logic, it increases one by one at numbers that are dark,
The one-by-one increase is dictated by mathematics. It cannot be seen.
That implies the existence of dark numbers.
No, it is not.
Sets are not built "Step by Step".
Your "One by One" logic limits you to finite sets,
What value of x does NUF(x) have a finite value?
x is smaller than every definable eps > 0.
What value is that?
You are just admitting your funciton doesn't exist.
NUF(x) for all defined numbers is "infinte", which isn't actually a
Natural Number.
Yes.
So your NUF is a mapping from non-Natural Numbers or to non-Natural
Numbers so isn't actually a function in the domain of Natural Numbers.
The domain is the real axis.
"Axis" isn't a thing for number.
You are just showing you don't know what you are talking about.
You don't seem to understand how you generate the numbers you want to
talk about.
Ok, so you can't actually quote the material you are using.
Quote the ACTUAL WORDS that Cantor uses to say that there existsand thus (in your words) numbers not individually usable, that must
be used individually but can't be becuase they are "dark".
They cannot be used individually. Nevertheless, according to Cantor,
they exist.
numbers that can not be used as an individual.
The completed infinite, das vollendete Unendliche or Vollendetunendliche
as Cantor called it [letter to Lipschitz (19 Nov. 1883) and E. Zermelo:
"Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 391], is a
prerequisite of set theory. Hence no fraction imust be missing.
Note, there IS an indexing that has no fraction missing, so it satisfies
the requirements
Le 18/12/2023 à 01:09, Richard Damon a écrit :
On 12/17/23 4:31 PM, WM wrote:
Le 16/12/2023 à 18:15, Richard Damon a écrit :
On 12/16/23 9:06 AM, WM wrote:
There is NUF(x) which increases one by one with gaps.
Which you have been unnable to actually show a value of at any
defined number.
At defined numbers the value is infinite. But it is 0 at 0 and all
unit fractions have distances from each other.
So?
The distance is unboundedly small, and moves from 0 to infinity in an
unboundedly small distance.
This is just the affect of being at the unbounded
By your logic, it increases one by one at numbers that are dark,
The one-by-one increase is dictated by mathematics. It cannot be
seen. That implies the existence of dark numbers.
No, it is not.
Sets are not built "Step by Step".
Your "One by One" logic limits you to finite sets,
What value of x does NUF(x) have a finite value?
x is smaller than every definable eps > 0.
What value is that?
You are just admitting your funciton doesn't exist.
NUF(x) for all defined numbers is "infinte", which isn't actually a
Natural Number.
Yes.
So your NUF is a mapping from non-Natural Numbers or to non-Natural
Numbers so isn't actually a function in the domain of Natural Numbers.
The domain is the real axis.
"Axis" isn't a thing for number.
You are just showing you don't know what you are talking about.
You don't seem to understand how you generate the numbers you want to
talk about.
Ok, so you can't actually quote the material you are using.
Quote the ACTUAL WORDS that Cantor uses to say that there existsand thus (in your words) numbers not individually usable, that
must be used individually but can't be becuase they are "dark".
They cannot be used individually. Nevertheless, according to
Cantor, they exist.
numbers that can not be used as an individual.
The completed infinite, das vollendete Unendliche or
Vollendetunendliche as Cantor called it [letter to Lipschitz (19 Nov.
1883) and E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen
mathematischen und philosophischen Inhalts", Springer, Berlin (1932)
p. 391], is a prerequisite of set theory. Hence no fraction imust be
missing.
I did it. You claimed: while any distance is finite, you can have an
infinite number of them before any finite distance. How to use them as
an individual if they cannot be separated?
Note, there IS an indexing that has no fraction missing, so it
satisfies the requirements
Is there an indexing of the unit fractions although there is "one of the strange properties of unboundely small values, while any distance is
finite, you can have an infinite number of them before any finite
distance"? How can you index fractions ehich cannot be separated and
used as individuals?
Regards, WM
Le 16/12/2023 à 09:01, Mikko a écrit :
On 2023-12-15 17:33:33 +0000, WM said:
Mikko schrieb am Freitag, 15. Dezember 2023 um 12:19:12 UTC+1:
On 2023-12-14 14:51:54 +0000, WM said:
There are natural numbers that cannot be described.That "this means" above is not bvious and is not proven and is actually false.
Proof: For every x > 0 there are infinitely many smaller unit
fractions. That means infinitely many unit fractions and their natural >>>>> numbers cannot be chosen or described as individuals by choosing or
describing any individual x > 0.
Then describe how these numbers can be chosen as individuals.
Every natural number has a name,
Name the first unit fraction.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
Note smallest universal quantifier. There is no exception. Therefore
NUF(x) can only increase one by one.
∀x ∈ (0, 1]: NUF(x) = ℵo
is wrong.
On 2023-12-16 12:24:03 +0000, WM said:
Le 16/12/2023 à 09:01, Mikko a écrit :
On 2023-12-15 17:33:33 +0000, WM said:
Mikko schrieb am Freitag, 15. Dezember 2023 um 12:19:12 UTC+1:
On 2023-12-14 14:51:54 +0000, WM said:
There are natural numbers that cannot be described.That "this means" above is not bvious and is not proven and is actually false.
Proof: For every x > 0 there are infinitely many smaller unit
fractions. That means infinitely many unit fractions and their natural >>>>>> numbers cannot be chosen or described as individuals by choosing or >>>>>> describing any individual x > 0.
Then describe how these numbers can be chosen as individuals.
Every natural number has a name,
Name the first unit fraction.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
Note smallest universal quantifier. There is no exception. Therefore
NUF(x) can only increase one by one.
∀x ∈ (0, 1]: NUF(x) = ℵo
is wrong.
You can pick any unit fraction and say that it is the "first" in some sense. You cant pick the smallest one because no unit fraction is the smallest one. There are infinitely many unit fractions and everyone has a name.
On 12/18/23 5:45 AM, WM wrote:
Is there an indexing of the unit fractions although there is "one of the
strange properties of unboundely small values, while any distance is
finite, you can have an infinite number of them before any finite
distance"? How can you index fractions which cannot be separated and
used as individuals?
But you CAN index the fractions, and use them individually.
Le 19/12/2023 à 10:54, Mikko a écrit :
On 2023-12-16 12:24:03 +0000, WM said:
Le 16/12/2023 à 09:01, Mikko a écrit :
On 2023-12-15 17:33:33 +0000, WM said:
Mikko schrieb am Freitag, 15. Dezember 2023 um 12:19:12 UTC+1:
On 2023-12-14 14:51:54 +0000, WM said:
There are natural numbers that cannot be described.That "this means" above is not bvious and is not proven and is actually false.
Proof: For every x > 0 there are infinitely many smaller unit
fractions. That means infinitely many unit fractions and their natural >>>>>>> numbers cannot be chosen or described as individuals by choosing or >>>>>>> describing any individual x > 0.
Then describe how these numbers can be chosen as individuals.
Every natural number has a name,
Name the first unit fraction.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
Note smallest universal quantifier. There is no exception. Therefore
NUF(x) can only increase one by one.
∀x ∈ (0, 1]: NUF(x) = ℵo
is wrong.
You can pick any unit fraction and say that it is the "first" in some sense. >> You cant pick the smallest one because no unit fraction is the smallest one. >> There are infinitely many unit fractions and everyone has a name.
ℵo have no names. Every one you name has ℵo smaller ones unnamed.
On 2023-12-19 12:07:45 +0000, WM said:
You can pick any unit fraction and say that it is the "first" in some sense.
You cant pick the smallest one because no unit fraction is the smallest one.
There are infinitely many unit fractions and everyone has a name.
ℵo have no names. Every one you name has ℵo smaller ones unnamed.
How do you know they are unit fractions if they have no names?
Anyway, for every real x > 0 there are ℵo unit fractions between
0 and x that do have a name.
Perhaps there are even more unit
fractions that don't have a name but you can't prove that there
are any.
Le 20/12/2023 à 13:00, Mikko a écrit :
On 2023-12-19 12:07:45 +0000, WM said:
You can pick any unit fraction and say that it is the "first" in
some sense.
You cant pick the smallest one because no unit fraction is the
smallest one.
There are infinitely many unit fractions and everyone has a name.
ℵo have no names. Every one you name has ℵo smaller ones unnamed.
How do you know they are unit fractions if they have no names?
I know it from
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo (*)
|ℕ \ {1, 2, 3, ...}| = 0
when accepting Cantor'a actual infinity.
Anyway, for every real x > 0 there are ℵo unit fractions between
0 and x that do have a name.
No. There cannot be two consecutive sets of card ℵo in ℕ. If there were ℵo named natnumbers, then none would remain unnamed. But (*) is correct.
Perhaps there are even more unit
fractions that don't have a name but you can't prove that there
are any.
They cannot be proved. They must be assumed. But your claim is incorrect because (*) is correct.
Regards, WM
Le 20/12/2023 à 13:00, Mikko a écrit :
On 2023-12-19 12:07:45 +0000, WM said:
You can pick any unit fraction and say that it is the "first" in some sense.
You cant pick the smallest one because no unit fraction is the smallest one.
There are infinitely many unit fractions and everyone has a name.
ℵo have no names. Every one you name has ℵo smaller ones unnamed.
How do you know they are unit fractions if they have no names?
I know it from
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo (*)
|ℕ \ {1, 2, 3, ...}| = 0
when accepting Cantor'a actual infinity.
Anyway, for every real x > 0 there are ℵo unit fractions between
0 and x that do have a name.
No. There cannot be two consecutive sets of card ℵo in ℕ.
If there were ℵo named natnumbers, then none would remain unnamed. But
(*) is correct.
Perhaps there are even more unit
fractions that don't have a name but you can't prove that there
are any.
They cannot be proved. They must be assumed.
But your claim is incorrect because (*) is correct.
On 2023-12-20 16:26:49 +0000, WM said:
ℕ is not a set of sets but a set of numbers.
Le 20/12/2023 à 18:52, Mikko a écrit :
On 2023-12-20 16:26:49 +0000, WM said:
ℕ is not a set of sets but a set of numbers.
But if there were a subset of ℵo nameable numbers first then nothing
could follow.
On 2023-12-21 11:01:50 +0000, WM said:
ℕ is not a set of sets but a set of numbers.
But if there were a subset of ℵo nameable numbers first then nothing
could follow.
Why not?
Anyway, nothing needn't follow so that is not a problem.
Le 21/12/2023 à 12:50, Mikko a écrit :
On 2023-12-21 11:01:50 +0000, WM said:
ℕ is not a set of sets but a set of numbers.
In ZFC everything is a set.
But if there were a subset of ℵo nameable numbers first then nothing
could follow.
Why not?
After ℵo numbers no finite number can follow.
Anyway, nothing needn't follow so that is not a problem.
So you withdraw your claim?
On 2023-12-22 16:30:06 +0000, WM said:
But if there were a subset of ℵo nameable numbers first then nothing >>>> could follow.
Why not?
After ℵo numbers no finite number can follow.
But transfinite and hyperfinite numbers can.
Anyway, nothing needn't follow so that is not a problem.
So you withdraw your claim?
What claim do you mean?
Le 23/12/2023 à 11:01, Mikko a écrit :
On 2023-12-22 16:30:06 +0000, WM said:
But if there were a subset of ℵo nameable numbers first then
nothing could follow.
Why not?
After ℵo numbers no finite number can follow.
But transfinite and hyperfinite numbers can.
Anyway, nothing needn't follow so that is not a problem.
So you withdraw your claim?
What claim do you mean?
Natural numbers following upon a set of ℵo natural numbers.
Regards, WM
On 12/26/23 5:02 AM, WM wrote:
Le 23/12/2023 à 11:01, Mikko a écrit :
On 2023-12-22 16:30:06 +0000, WM said:
But if there were a subset of ℵo nameable numbers first then
nothing could follow.
Why not?
After ℵo numbers no finite number can follow.
But transfinite and hyperfinite numbers can.
Anyway, nothing needn't follow so that is not a problem.
So you withdraw your claim?
What claim do you mean?
Natural numbers following upon a set of ℵo natural numbers.
All of which are Ordinary Natural Numbers, all finitely definable,
usable and nameable, thus none of them are "Dark"
Le 27/12/2023 à 03:29, Richard Damon a écrit :
On 12/26/23 5:02 AM, WM wrote:
Le 23/12/2023 à 11:01, Mikko a écrit :
On 2023-12-22 16:30:06 +0000, WM said:
But if there were a subset of ℵo nameable numbers first then
nothing could follow.
Why not?
After ℵo numbers no finite number can follow.
But transfinite and hyperfinite numbers can.
Anyway, nothing needn't follow so that is not a problem.
So you withdraw your claim?
What claim do you mean?
Natural numbers following upon a set of ℵo natural numbers.
All of which are Ordinary Natural Numbers, all finitely definable,
usable and nameable, thus none of them are "Dark"
ℵo natural numbers follow upon every definable number. Hence they are undefined and remain undefinable.
Regards, WM
On 12/27/23 4:24 AM, WM wrote:
ℵo natural numbers follow upon every definable number. Hence they are
undefined and remain undefinable.
Except that they all are definable, as I have previously shown.
Le 27/12/2023 à 14:37, Richard Damon a écrit :
On 12/27/23 4:24 AM, WM wrote:
ℵo natural numbers follow upon every definable number. Hence they are
undefined and remain undefinable.
Except that they all are definable, as I have previously shown.
Define one natnumber that has not ℵo undefined successors. That is not possible because
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
It is possible however to exhaust all natural numbers collectively, such
that no successors remain,
|ℕ \ {1, 2, 3, ...}| = 0.
Regards, WM
Why does there need to be a Natural Number that has less than ℵo successors?
All the Natural Numbers are also nameable individually.
Le 27/12/2023 à 21:03, Richard Damon a écrit :
Why does there need to be a Natural Number that has less than ℵo
successors?
Collectively you can take all natural numbers with no successors remaining. When you want prove that all natural numbers cannot only be taken collectively but can be named individually, then you must do it with no successors remaining.
All the Natural Numbers are also nameable individually.
Show that none remains.
Regards, WM
Thus any Natural Number n can be names as n Ses then 0, as
in: 0, S0, SS0, SSS0, ...
IF it can't be named that way, it isn't a "Natural Number".
Thus, ALL Natural Numbers have a name.
Le 28/12/2023 à 03:58, Richard Damon a écrit :
Thus any Natural Number n can be names as n Ses then 0, as in: 0, S0,
SS0, SSS0, ...
"..." is naming collectively.
IF it can't be named that way, it isn't a "Natural Number".
Thus, ALL Natural Numbers have a name.
Of course all natural numbers differ from each other by at least one
unit or S. But every name that you choose will leave almost all natural numbers unchosen.
Regards, WM
Le 28/12/2023 à 03:58, Richard Damon a écrit :
Thus any Natural Number n can be names as n Ses then 0, as in: 0, S0,
SS0, SSS0, ...
"..." is naming collectively.
IF it can't be named that way, it isn't a "Natural Number".
Thus, ALL Natural Numbers have a name.
Of course all natural numbers differ from each other by at least one
unit or S. But every name that you choose will leave almost all natural numbers unchosen.
Regards, WM
Where do you see a number that can't be named coming up?
You seem to be having problems using the right qualifier.
Le 28/12/2023 à 15:51, Richard Damon a écrit :
Where do you see a number that can't be named coming up?
They are not visible. But they are there like dark energy. All we can
see are numbers with FISONs
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo having ℵo dark successors.
You seem to be having problems using the right qualifier.
What qualifiers do you know?
Regards, WM
On 12/30/23 7:21 AM, WM wrote:
Le 28/12/2023 à 15:51, Richard Damon a écrit :
Where do you see a number that can't be named coming up?
They are not visible. But they are there like dark energy. All we can
see are numbers with FISONs
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo having ℵo dark successors.
But we can name all the successors, so they are not "dark"
You seem to be having problems using the right qualifier.
What qualifiers do you know?
THere are Universal Qualifiers, say that something is true for ALL or
NONE of the elements of the set, and there are Existential Qualifiers
that state tha there exist with (or without) a given property.
On 30.12.2023 17:43, Richard Damon wrote:
On 12/30/23 7:21 AM, WM wrote:
Le 28/12/2023 à 15:51, Richard Damon a écrit :
Where do you see a number that can't be named coming up?
They are not visible. But they are there like dark energy. All we can
see are numbers with FISONs
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo having ℵo dark successors.
But we can name all the successors, so they are not "dark"
Try it!
You seem to be having problems using the right qualifier.
What qualifiers do you know?
THere are Universal Qualifiers, say that something is true for ALL or
NONE of the elements of the set, and there are Existential Qualifiers
that state tha there exist with (or without) a given property.
I know only quantifiers of that sort.
Regards, WM
On 12/30/23 12:09 PM, WM wrote:
Where do you see a number that can't be named coming up?
They are not visible. But they are there like dark energy. All we can
see are numbers with FISONs
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo having ℵo dark successors.
But we can name all the successors, so they are not "dark"
Try it!
I did.
I showed the naming formula.
Thus, while is has ℵo successors, they are all "Visible" and eligable to have been chosen as the starting n.
Le 30/12/2023 à 18:32, Richard Damon a écrit :
On 12/30/23 12:09 PM, WM wrote:
Where do you see a number that can't be named coming up?
They are not visible. But they are there like dark energy. All we
can see are numbers with FISONs
∀n ∈ ℕ_vis: |ℕ \ {1, 2, 3, ..., n}| = ℵo having ℵo dark successors.
But we can name all the successors, so they are not "dark"
Try it!
I did.
I showed the naming formula.
This formula concerns the collection, not any individual. To name an individual you have to insert a name.
Thus, while is has ℵo successors, they are all "Visible" and eligable
to have been chosen as the starting n.
The successors are not visible. Remember that you cannot distinguish ℵo unit fractions by any choice of an x > 0.
Regards, WM
On 12/30/23 2:21 PM, WM wrote:
The successors are not visible. Remember that you cannot distinguish ℵo
unit fractions by any choice of an x > 0.
Of course I can, and DID.
Le 30/12/2023 à 20:42, Richard Damon a écrit :
On 12/30/23 2:21 PM, WM wrote:
The successors are not visible. Remember that you cannot distinguish
ℵo unit fractions by any choice of an x > 0.
Of course I can, and DID.
If there are ℵo unit fractions before every, any, all x > 0, how will
you distinguish them by x > 0?
Regards, WM
On 12/31/23 4:54 AM, WM wrote:
If there are ℵo unit fractions before every, any, all x > 0, how willGiven x, find n such that 1/(n+1) < x <= 1/n, which can be done for ALL
you distinguish them by x > 0?
x in (0, 1].
Le 31/12/2023 à 13:46, Richard Damon a écrit :
On 12/31/23 4:54 AM, WM wrote:
If there are ℵo unit fractions before every, any, all x > 0, how willGiven x, find n such that 1/(n+1) < x <= 1/n, which can be done for
you distinguish them by x > 0?
ALL x in (0, 1].
But not for all 1/n. Note that ℵo unit fractions are before every given x.
Regards, WM
On 1/1/24 10:29 AM, WM wrote:
...]
"Not All" SPECIFICALLY requires that
there be a member of the set that
doesn't meet the requirement,
and that member can be named
to show its existance.
and that member can be named
to show its existance.
ALL members of the set 1/n have BY DEFINITION
a natural number n they are based on,
and all natural numbers are constructable with
a finite number of steps,
and thus can be named by that procedure.
On 1/1/2024 10:37 AM, Richard Damon wrote:
ALL members of the set 1/n have BY DEFINITION
a natural number n they are based on,
and all natural numbers are constructable with
a finite number of steps,
and thus can be named by that procedure.
Le 01/01/2024 à 20:30, Jim Burns a écrit :
On 1/1/2024 10:37 AM, Richard Damon wrote:
ALL members of the set 1/n have BY DEFINITION
a natural number n they are based on,
and all natural numbers are constructable with
a finite number of steps,
and thus can be named by that procedure.
No.
After all constructed natnumbers
there are almost all natnumbers.
On Tuesday, January 2, 2024
at 12:05:30 PM UTC-8, Jim Burns wrote:
After all
cardinals which.can.change.by.1
there are only
cardinals which.CANNOT.change.by.1.
It is contradictory for
a cardinal which.can.change.by.1
and
a cardinal which.CANNOT.change.by.1.
to be apart by 1.
There are various models of
large numbers and infinite [numbers],
about fragments and parts,
that there are
models of numbers with none infinite and
models of integers with many.
Le 01/01/2024 à 20:30, Jim Burns a écrit :
On 1/1/2024 10:37 AM, Richard Damon wrote:
ALL members of the set 1/n have BY DEFINITION
a natural number n they are based on,
and all natural numbers are constructable with
a finite number of steps,
and thus can be named by that procedure.
No. After all constructed natnumbers there are almost all natnumbers.
Regards, WM
On 1/2/24 2:33 PM, WM wrote:
After all constructed natnumbers there are almost all natnumbers.So, what is the last constructed Natural Number?
By your logic, there must be one,
On Tuesday, January 2, 2024
at 3:59:34 PM UTC-8, Richard Damon wrote:
[...]
Actually
it's by Russell's logic that
a set of all sets that don't contain themselves,
like a model of all finite ordinals,
contains itself.
(Yeah, I know
ordinary set theory sticks its head in the sand
and reckons the grains from there.)
Le 03/01/2024 à 00:59, Richard Damon a écrit :
On 1/2/24 2:33 PM, WM wrote:
After all constructed natnumbers there are almost all natnumbers.So, what is the last constructed Natural Number?
That depends on the system.
By your logic, there must be one,
Yes, but it can be surpassed.
Regards, WM
On 1/3/24 5:24 AM, WM wrote:
Le 03/01/2024 à 00:59, Richard Damon a écrit :
On 1/2/24 2:33 PM, WM wrote:
After all constructed natnumbers there are almost all natnumbers.So, what is the last constructed Natural Number?
That depends on the system.
Really?
Where do you get that from?
Le 03/01/2024 à 13:08, Richard Damon a écrit :
On 1/3/24 5:24 AM, WM wrote:
Le 03/01/2024 à 00:59, Richard Damon a écrit :
On 1/2/24 2:33 PM, WM wrote:
After all constructed natnumbers there are almost all natnumbers.So, what is the last constructed Natural Number?
That depends on the system.
Really?
Where do you get that from?
It is like the known prime numbers. There is a last one, temporarily.
Regards, WM
Le 03/01/2024 à 13:08, Richard Damon a écrit :
On 1/3/24 5:24 AM, WM wrote:
Le 03/01/2024 à 00:59, Richard Damon a écrit :
On 1/2/24 2:33 PM, WM wrote:
After all constructed natnumbers there are almost all natnumbers.So, what is the last constructed Natural Number?
That depends on the system.
Really?
Where do you get that from?
It is like the known prime numbers. There is a last one, temporarily.
Regards, WM
On 1/4/24 5:46 AM, WM wrote:
It is like the known prime numbers. There is a last one, temporarily.
But there is a difference betweeh "known" and "Is".
On 1/4/24 5:46 AM, WM wrote:
Le 03/01/2024 à 13:08, Richard Damon a écrit :
On 1/3/24 5:24 AM, WM wrote:
Le 03/01/2024 à 00:59, Richard Damon a écrit :
On 1/2/24 2:33 PM, WM wrote:
After all constructed natnumbers there are almost all natnumbers.So, what is the last constructed Natural Number?
That depends on the system.
Really?
Where do you get that from?
It is like the known prime numbers. There is a last one, temporarily.
And if "known" is your criteria, then, yes, most (in fact, it would be "almost all") Natural Numbers are "unknown" in the sense of not having
been written down,
but they are all "knowable",
So, perhaps you are just a victim of not properly defining the terms you
are trying to use.
Almost all Natural Numbers are unexpressed, but all are still expressible.
Richard Damon schrieb am Donnerstag, 4. Januar 2024 um 14:00:04 UTC+1:
On 1/4/24 5:46 AM, WM wrote: > Le 03/01/2024 à 13:08, Richard Damon a
écrit : >> On 1/3/24 5:24 AM, WM wrote: >>> Le 03/01/2024 à 00:59,
Richard Damon a écrit : >>>> On 1/2/24 2:33 PM, WM wrote: >>>>> After
all constructed natnumbers there are almost all natnumbers. >>>>> >>>>
So, what is the last constructed Natural Number? >>> >>> That depends
on the system. >> >> Really? >> >> Where do you get that from? > > It
is like the known prime numbers. There is a last one, temporarily. >
And if "known" is your criteria, then, yes, most (in fact, it would be
"almost all") Natural Numbers are "unknown" in the sense of not having
been written down,
Right.
but they are all "knowable",
Not all prime numbers are knowable because the set is infinite but
always only a finite set can be known.
So, perhaps you are just a victim of not properly defining the terms
you are trying to use.
Almost all Natural Numbers are unexpressed, but all are still
expressible.
That is counterfactual credo in absurdum. It is useless to talk to
people who are so anti-mathematical. EOD
Regards, WM
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