William schrieb am Montag, 3. Januar 2022 um 16:43:11 UTC+1:
> On Monday, January 3, 2022 at 4:56:33 AM UTC-4, WM wrote:
> > William schrieb am Sonntag, 2. Januar 2022 um 15:34:04 UTC+1:
> > > On Saturday, January 1, 2022 at 3:34:40 PM UTC-4, WM wrote:
> > > > William schrieb am Freitag, 31. Dezember 2021 um 23:11:00 UTC+1:
> > > > > On Friday, December 31, 2021 at 1:44:41 PM UTC-4, WM wrote:
> > > >
> > > > > > > > {1}
> > > > > > > > {2, 1}
> > > > > > > > {3, 2, 1}
> > > > > > > > {4, 3, 2, 1}
> > > > > > > > {5, 4, 3, 2, 1}
> > > > > > > > ...
> > > > > Correct.. "All lines" is the set which is infinite. Note the set can be larger than each line, i.e. each element.
> > > > >

Simple commercial addition shows the contrary.

> > The collection ℕ_def
> of natural numbers that have the property they can be written down, is a subset of N_P the set of natural numbers that are the largest element of a FISON.

What subset is infinite but less than the smallest infinite set?

> N_P is infinite.

ℕ_def is smaller.

> One of the FISONs is the collection.
> Nope, the colliction is the set N_P. Each FISON has a last element . The set N_P has no last element. N_P is not a FISON. However, N_P is complete.

False. Only ℕ is complete. But
∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo .

Regards, WM

Jim Burns schrieb am Montag, 3. Januar 2022 um 18:00:16 UTC+1:
> On 1/3/2022 4:23 AM, WM wrote:
> > William schrieb
> > am Sonntag, 2. Januar 2022 um 15:22:50 UTC+1:
> >> On Saturday, January 1, 2022 at 3:37:22 PM UTC-4,
> >> WM wrote:
>
> >>> This has been shown:
> >>> Collect all rationals of the interval (0, 1]
> [...]
> >> The bijection exists,
> >
> > If a bijection existed, then every re-ordering of the
> > image set would be a bijection. For instance taking
> > one rational out of every unit interval, for instance
> > simply chosing always n/1.
> You are taking a property which is true for some sets

All sets which can be in bijection have this property, because by reordering the number of elements does not change.

> -- sets which can be totally-ordered with two ends
> in which each _cut_ is a _step_ --
> and you're looking for it in sets without that property.
> And you find it's not there.

I prove that there is no bijection.
>
> You haven't found a contradiction.
> You've found things which cannot be totally-ordered
> with two ends in which each cut is a step.

That is in contradiction with the emaning of bijection.
>
> We have reasons to examine both kinds of things,
> different reasons, in some cases.

There is no reason but only lack of logic.

Collect all rationals of the interval (0, 1] into a reservoir, map one of them by 1 and remove it from the reservoir. Then collect all rationals of the interval(1, 2] into the reservoir, map one of them by 2 and remove it from the reservoir. Then collect all rationals of the interval (2, 3] into the reservoir, map one of mthem by 3 and remove it from the reservoir. Continue until all intervals will have been worked through. Then the reservoir will be empty according to set theory.

The reservoir will be empty only when the audience is hypnotized.

> >> it is not the result of a process.
> >
> > That is the usual lie.
> > Of course every sequence is a process.
> ...and every sheep-number is a sheep.

Don't confuse matheolgians with these intelligent animals.

Never a sheep would believe that the reservoir could be empty.

Regards, WM

Jim Burns schrieb am Dienstag, 4. Januar 2022 um 03:09:20 UTC+1:
> On 1/3/2022 4:23 AM, WM wrote:
> > William schrieb
> > am Sonntag, 2. Januar 2022 um 15:22:50 UTC+1:
> >> This is not a contradition but a result you do not like.
> >
> > It is a result that I do not like because it is provably
> > wrong. I never met a student who accepted this mess.
> You never met a student who...
>
> You never met a student enrolled in your class who...
>
> You never met a student enrolled in your class
> and required to repeat back whatever you feel like
> telling them, who...
>
> ...who accepted this mess.
>
> ...who was willing to publicly disagree with the old man
> who would give them an easy A if they managed to keep
> their mouth shut.
>
> This version of peer review, your version, certainly
> has advantages over the more common version,
> advantages to you, anyway.
>
> One advantage, to you, at least:
> You could announce to your students that 2 + 2 = 5
> and they would dutifully nod and write that down.
> Your classroom is a safe space for you, in which
> you face zero chance of have any errors pointed out.
> I'm sure you love that.

A nice tirade. I'll give you a tip how to get out of the trap. Ask some person of your surrounding (wife, children, friends, cab driver) which has not yet been infected by matheology what they think of this procedure: Collect all rationals of the interval (0, 1] into a reservoir, map one of them by 1 and remove it from the reservoir. Then collect all rationals of the interval(1, 2] into the reservoir, map one of them by 2 and remove it from the reservoir. Then collect all rationals of the interval (2, 3] into the reservoir, map one of mthem by 3 and remove it from the reservoir. Continue until all intervals will have been worked through. Can the reservoir be empty then?

If necessary explain first what an interval and what a natural number and what a rational numbers is. Would you dare to do so?

Regards, WM

On Tuesday, January 4, 2022 at 12:38:04 PM UTC-4, WM wrote:
> William schrieb am Montag, 3. Januar 2022 um 16:43:11 UTC+1:

> > > One of the FISONs is the collection.
> > Nope, the collrction is the set N_P. Each FISON has a last element . The set N_P has no last element. N_P is not a FISON. However, N_P is complete..
> False. Only ℕ is complete. But
> ∀n ∈ ℕ_P: |ℕ[_P] \ {1, 2, 3, ..., n}| = ℵo .

A correct statement about each element of the set N_P, Has no bearing on the fact that the set N_P is complete.

--
William Hughes

On Tuesday, January 4, 2022 at 12:38:04 PM UTC-4, WM wrote:
> William schrieb am Montag, 3. Januar 2022 um 16:43:11 UTC+1:
> > On Monday, January 3, 2022 at 4:56:33 AM UTC-4, WM wrote:

> > > One of the FISONs is the collection.
> > Nope, the colliction is the set N_P. Each FISON has a last element . The set N_P has no last element. N_P is not a FISON. However, N_P is complete..
> False. Only ℕ is complete. But
> ∀n ∈ ℕ_P: |ℕ[_P] \ {1, 2, 3, ..., n}| = ℵo .

A correct statement about the elements of the set N_P., it is equivalent to saying that N_P does not have a largest element and is thus not a FISON. Each element that is in N_P is followed by an infinite number of elements that are in N_P.

--
William Hughes

Modern problems are rather:

"There are many different sized infinities."
Mathematicians: fantastic.
"There are many geometries, including ones where parallel lines meet."
Mathematicians: awesome.
"There are more than two genders."
Mathematicians: Um, ah, uh, hrmph.
https://twitter.com/Anthony_Bonato/status/1478388698261729283

LoL

WM schrieb am Sonntag, 7. November 2021 um 12:56:59 UTC+1:
> William schrieb am Samstag, 6. November 2021 um 22:42:36 UTC+1:
> > The following statements of about the set |N_F are all true
> >
> > (i) Every element of |N_F is finite
> >
> > (ii) |N_F, like any Peano set, has cardinality aleph_0
> >
> > (iii) Between every element of |N_F and omega there are infinitely many natural numbers.
> >
> > (iv) There is no natural number between the set |N_F and omega.
> What is left-hand side next to omega on the ordinal line?
> >
> > (iv) is true, not because there is an element of |N_F which is not separated from omega by an infinite set of natural numbers, but because there is no largest natural number.
> >
> Please explain with that model what elements are in all infinitely many infinite endsegments, the intersection of which is empty.
>
> Regards, WM

On 1/4/2022 4:18 PM, Mostowski Collapse wrote:
> Modern problems are rather:
>
> "There are many different sized infinities."
> Mathematicians: fantastic.
> "There are many geometries, including ones where parallel lines meet."
> Mathematicians: awesome.
> "There are more than two genders."
> Mathematicians: Um, ah, uh, hrmph.
> https://twitter.com/Anthony_Bonato/status/1478388698261729283
>
> LoL[...]

I think there is such a thing as a real hermaphrodite... Marijuana
growers know this well. This is a third state, in a sense?

On 1/4/2022 11:56 AM, WM wrote:

> If necessary explain first what an interval and
> what a natural number and what a rational numbers is.
> Would you dare to do so?

I can do that.
I can't promise I'll hold their attention.
Maybe that could be a resolution for the New Year.

----
Natural numbers start with the observation that, for some
collections, if they match member-to-member in one order,
then they match member-to-member in all other orders.

So, because 1<2<3 matches a<b<c,
1<2<3 also matches b<a<c b<c<a c<b<a c<a<b a<c<b

This any-order property allows a farmer sending their
sheep out into the fields to graze all day to put one
pebble in a pile for each sheep, and then, at sunset,
to take a pebble out of that pile, one for each sheep,
as they come in. Because their sheep and their pebbles
have this any-order property, when the sheep return in
the evening , the farmer can check that they're all in
by looking to see if there are still pebbles in the pile.
By the any-order property, it doesn't matter in which
order the sheep return.

This any-order property is something some collections
have and other collections do not have.

----
Natural numbers (pebble-names) are an upgrade of the
farmer's pile-of-pebbles technology. For each pebble-name,
one and only one other pebble-name has been designated
its successor. Pebble-names, not physical pebbles. The pile
of pebbles has become an abstraction, much easier to
carry.

Within the designated-successor system, we are able to
use a pebble-name to refer to a specific pile in a
specific order. '5' means '1,2,3,4,5'
but not '137,17,1001,1,5' not even '4,1,3,2,5'

Using the newer technology, a farmer can _tell_ her son to
bring in the _5_ sheep and he will know to match a sheep
to pebble-name 1, to 2, to 3, to 4, and to 5, and he'll
have rounded up all the sheep -- as though she had handed
him a bag with the sheep-matching pebbles in it.

----
Some collections have the any-order property. If one
matches member-to-member to another in one order,
then they match member-to-member in all the other orders.
Other collections do not have the any-order property.

A herd of sheep in a pasture has the any-order property.
A bag of pebbles has the any-order property.
The pebble-names up to a particular pebble-name has the
any-order property.

With pebble-names, though, it pays us to be a little
careful. The pebble-names 1,2,3,4,5 match to Dasher, Dancer,
Prancer, Vixen, and Rudolph, and, by the any-order property,
those pebble-names match those flying reindeer in the order
4,1,3,2,5 as well. But there is only one order, the one using
designated successors, by which we can refer to the pile
1,2,3,4,5 of pebble-names by its last pebble-name 5.

The reindeer can also be matched to the pile as 4,1,3,5,2
but there aren't only 2 reindeer. It's only the order with
the designated successors which we want, 1,2,3,4,5

----
Definitions and theorems.

For a collection L with a total-order,
a _cut_ D is a non-empty proper subset of L
which is _downward-closed_
( j > i ) -> ( j ∈ D -> i ∈ D )

Its complement U = L\D is a non-empty proper subset
of C which is _upward-closed_
( i < j ) -> ( i ∉ D -> j ∉ D )

All the elements of L are either "before the cut" in D
or "after the cut" not-in D, in U.

(Theorem.)
For a collection F with the any-order property,
F has a total-order such that
F has a minimum and a maximum, and
each cut of F is a _step_

A cut D,U is a _step_ iff
D has a maximum and U has a minimum.

In order to proceed from D to U, it is necessary to step
over the gap between max(D) and min(U) somewhere.

For a collection S with the designated-successor order,
S has a total-order such that
each cut of S is a step, and
each step of S is a _count_

What I have not required here is that S have two ends,
or even one end. That's for later.

For a step D,U with max(D) = i and min(U) = j
D,U is a _count_ iff j = i+1

Define
{0,...,k} to be a collection with the designated-
-successor order with min{0,...,k} = 0
and max{0,...,k} = k.

k is a pebble-name.
{0,...,k} is the collection of pebble-names up to k.

(Theorem.)
Let S and S' be collections with the designated-successor
order, min(S) = max(S') = 0 and max(S) = max(S'),
which exists in S and S'.
Then S = S'

The theorem says, in other words,
if {0,...,k} exists at all, there isn't a second one.

(Theorem.)
Let {0,...,j} and {0,...,k} exist.
Then either j in {0,...,k} or k in {0,...,j}

Individual collections-with-the-designated-successor-order
are totally ordered by definition.
It is provable that there is a total-order for
all elements of all collections-with-designated-
-successor-order, such as {0,...,j} and {0,...,k}

Define
k in ℕ iff
a collection {0,...,k} exists
with the designated-successor order,
min{0,...,k} = 0 and
max{0,...,k} = k

ℕ is the set of natural numbers (pebble-names).

William schrieb am Mittwoch, 5. Januar 2022 um 01:04:56 UTC+1:
> On Tuesday, January 4, 2022 at 12:38:04 PM UTC-4, WM wrote:
> > William schrieb am Montag, 3. Januar 2022 um 16:43:11 UTC+1:
> > > On Monday, January 3, 2022 at 4:56:33 AM UTC-4, WM wrote:
>
> > > > One of the FISONs is the collection.
> > > Nope, the colliction is the set N_P. Each FISON has a last element . The set N_P has no last element. N_P is not a FISON. However, N_P is complete.
> > False. Only ℕ is complete. But
> > ∀n ∈ ℕ_P: |ℕ[_P] \ {1, 2, 3, ..., n}| = ℵo .
>
> A correct statement about the elements of the set N_P., it is equivalent to saying that N_P does not have a largest element and is thus not a FISON.

No, it is equivalent to saying that there is a set of ℵo elements which cannot be reached by FISONs.

> Each element that is in N_P is followed by an infinite number of elements that are in N_P.

No, ℵo elements are always outside of the set N_P. If you take the union over all FISONs then ℵo elements remain outside. Compare the endsegments of all elements of N_P. They are *all* infinite. No element comes closer to the end omega. This distance however can be removed by collectively subtracting ℕ:
{0, 1, 2, 3, ..., ω} \ ℕ = {0, ω}.
This proves the existence of numbers in ℕ which can only be handled collectively.

Regards, WM

Jim Burns schrieb am Mittwoch, 5. Januar 2022 um 21:31:05 UTC+1:

> Define
> k in ℕ iff
> a collection {0,...,k} exists
> with the designated-successor order,
> min{0,...,k} = 0 and
> max{0,...,k} = k
>
> ℕ is the set of natural numbers (pebble-names).

Your ℕ is the set ℕ_def of all definable numbers. Between every definable number and omega, there is an infinite endsegment. This cannot be removed by subtracting individual numbers, but it can be removed by collectively subtracting like in
ℕ \ {1, 2, 3, ...} = { }
or
{0, 1, 2, 3, ..., ω} \ ℕ = {0, ω}.

If you disagree then show me how you can accomplish the same by individually subtracting FISONs or natnumbers. Don't "prove" that you can do it. Do it!

Regards, WM

On Thursday, January 6, 2022 at 6:51:00 AM UTC-4, WM wrote:
> William schrieb am Mittwoch, 5. Januar 2022 um 01:04:56 UTC+1:

> > Each element that is in N_P is followed by an infinite number of elements that are in N_P.
> If you take the union over [the set of] FISONs then ℵo elements remain outside.

Nope. N_P is defined to be the union of the set of FISONS. There is no largest FISON. Thus there is no largest element of N_P

> Compare the endsegments of all elements of N_P. They are *all* infinite. No element comes closer to the end omega. This distance

The "distance" is not a metric. Every element of N_P is the same "distance" from omega.

>however can be removed by collectively

Correct. N_P is an infinite set (it has no last element). To "remove" every element of N_P you have to "remove" an infinite set (remove collectively). "Removing" a finite set (remove individually) always leaves an infinite set.

--
William Hughes

On 1/6/2022 6:00 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 5. Januar 2022 um 21:31:05 UTC+1:

>>> If necessary explain first [...]
>>> what a natural number [...] is.

>> Define
>> k in ℕ iff
>> a collection {0,...,k} exists
>> with the designated-successor order,
>> min{0,...,k} = 0 and
>> max{0,...,k} = k
>>
>> ℕ is the set of natural numbers (pebble-names).
>
> Your ℕ is the set ℕ_def of all definable numbers.

There are collections such that, if they matched
member-to-member with another in the morning, they will
match member-to-member with the same at night.
The {0,...,k} are some but only some of those collections.

We give special attention to the {0,...,k} because,
(i)
for each match-in-the-morning-match-at-night collection,
one of the {0,...,k} matches it (morning and night), and
(ii)
for each {0,...,k}, it is enough to refer to the last
in {0,...,k} in order to uniquely determine that
particular collection.

So, talking about all the last members of the {0,...,k}
is equivalent in important ways to talking about all
the match-in-the-morning-match-at-night collections,
only much, much more conveniently.

All of that remains true if all of the last members of
the {0,...,k} are named ℕ or if they are named ℕ_def
or if they are not collectively named anything at all.

( For an example in which they are not collectively
( named anything at all, consider the collection of
( hereditarily finite sets. All the k are in it,
( but no collection of all k is in it.
( There is no ℕ in it. There is no ℕ_def in it.
( There surely aren't any dark numbers in it.
( ( Arithmetic, though, has no problem in it.
( Anywhere it seems like you should be saying
( "k is in ℕ" say instead "k ends some {0,...,k}"
( ℕ doesn't exist but, if it did, they would be equivalent.
( It works out the same as if it did.

----
All of the match-in-the-morning-match-at-night collections
are not by themselves all of what are simply collections.

Examples which are not match-in-the-morning-match-at-night
include the collection of all the ends of the {0,...,k}.

Each of the {0,...,k} has a total order with two ends
in which each cut is a step.

None of the {0,...,k} contains all of these ends.
because, for each {0,...,k}, k+1 is another end
and k+1 is not in {0,...,k}

That follows from the definition of {0,...,k}.
ℕ or ℕ_def or nothing-at-all doesn't enter the
conversation. Call the collection what you like,
it doesn't matter.

> Between every definable number and omega,
> there is an infinite endsegment.

Look! ω is not a sheep-number.

I'm deleting your proof that ω is not a sheep-number.
It doesn't matter.

onsdag 22 december 2021 kl. 09:54:15 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 22. Dezember 2021 um 06:27:54 UTC+1:
> > tisdag 21 december 2021 kl. 22:27:57 UTC+1 skrev WM:
>
> > >If infinitely many natnumbers are in all endsegments their intersection is infinite.
> > False, that is a complete non-sequitor. The cardinality of the sets have no effect on the intersection.
> Your level has dropped to Sergio's. EOD.
>
> Regards, WM

We are both smarter than you so I take it as a compliment :)

torsdag 6 januari 2022 kl. 11:51:00 UTC+1 skrev WM:
> William schrieb am Mittwoch, 5. Januar 2022 um 01:04:56 UTC+1:
> > On Tuesday, January 4, 2022 at 12:38:04 PM UTC-4, WM wrote:
> > > William schrieb am Montag, 3. Januar 2022 um 16:43:11 UTC+1:
> > > > On Monday, January 3, 2022 at 4:56:33 AM UTC-4, WM wrote:
> >
> > > > > One of the FISONs is the collection.
> > > > Nope, the colliction is the set N_P. Each FISON has a last element .. The set N_P has no last element. N_P is not a FISON. However, N_P is complete.
> > > False. Only ℕ is complete. But
> > > ∀n ∈ ℕ_P: |ℕ[_P] \ {1, 2, 3, ..., n}| = ℵo .
> >
> > A correct statement about the elements of the set N_P., it is equivalent to saying that N_P does not have a largest element and is thus not a FISON.
> No, it is equivalent to saying that there is a set of ℵo elements which cannot be reached by FISONs.
> > Each element that is in N_P is followed by an infinite number of elements that are in N_P.
> No, ℵo elements are always outside of the set N_P. If you take the union over all FISONs then ℵo elements remain outside. Compare the endsegments of all elements of N_P. They are *all* infinite. No element comes closer to the end omega. This distance however can be removed by collectively subtracting ℕ:
> {0, 1, 2, 3, ..., ω} \ ℕ = {0, ω}.
> This proves the existence of numbers in ℕ which can only be handled collectively.
>
> Regards, WM
it proves nothing of the sort you imbecile

William schrieb am Donnerstag, 6. Januar 2022 um 16:12:43 UTC+1:
> On Thursday, January 6, 2022 at 6:51:00 AM UTC-4, WM wrote:
> > William schrieb am Mittwoch, 5. Januar 2022 um 01:04:56 UTC+1:
>
> > > Each element that is in N_P is followed by an infinite number of elements that are in N_P.
> > If you take the union over [the set of] FISONs then ℵo elements remain outside.
>
> Nope. N_P is defined to be the union of the set of FISONS. There is no largest FISON. Thus there is no largest element of N_P
> > Compare the endsegments of all elements of N_P. They are *all* infinite.. No element comes closer to the end omega. This distance
> The "distance" is not a metric.

Yes. The dark natnumbers have no discernible order.

> Every element of N_P is the same "distance" from omega.

Not quite, 4 is ten steps farther from omega than 14. But the distance is ℵo where ℵo + 10 = ℵo.

> >however can be removed by collectively
> Correct. N_P is an infinite set (it has no last element).

Nevertheless every element has distance ℵo from omega.

> To "remove" every element of N_P you have to "remove" an infinite set (remove collectively). "Removing" a finite set (remove individually) always leaves an infinite set.

So it is. Same however holds for mappings which are necessarily between individuals.

Regards, WM

On Friday, January 7, 2022 at 12:08:45 PM UTC+1, WM wrote:

> The dark natnumbers have no discernible order.

This proves that there are no dark numbers in IN, since IN is a totally orderd set [by the usual <=].

See: https://en.wikipedia.org/wiki/Total_order

You know, for EACH and EVERY element n in IN and EACH and EVERY element m in IN: Either n < m, m < n, or n = m.

Hence there are no "dark numbers" in IN.

> > Every element of N_P is the same "distance" from omega.
> >
> Not quite, 4 is ten steps farther from omega than 14. But the distance is ℵo where ℵo + 10 = ℵo.

Huh?!

That just implies what William said, moron.

> every element has distance ℵo from omega.

That just implies what William said, moron!

So what's your point, man?!

On Friday, January 7, 2022 at 7:08:45 AM UTC-4, WM wrote:
> William schrieb am Donnerstag, 6. Januar 2022 um 16:12:43 UTC+1:
> > On Thursday, January 6, 2022 at 6:51:00 AM UTC-4, WM wrote:
> > > William schrieb am Mittwoch, 5. Januar 2022 um 01:04:56 UTC+1:
> >
> > > > Each element that is in N_P is followed by an infinite number of elements that are in N_P.
> > > If you take the union over [the set of] FISONs then ℵo elements remain outside.
> >
> > Nope. N_P is defined to be the union of the set of FISONS. There is no largest FISON. Thus there is no largest element of N_P

N_P is the union of the set of FISONs. Every natural number is an element of the set N_P. Every element of the set N_P is finite. N_P, like any Peano set has cardinality aleph_0.

--
William Hughes

Fritz Feldhase schrieb am Freitag, 7. Januar 2022 um 13:39:42 UTC+1:
> On Friday, January 7, 2022 at 12:08:45 PM UTC+1, WM wrote:
>
> > The dark natnumbers have no discernible order.
> This proves that there are no dark numbers in IN, since IN is a totally orderd set [by the usual <=].

What is the point on the real axis next to 0? There is something. There is no gap. What is it? It cannot be put in any order.
>
> You know, for EACH and EVERY element n in IN and EACH and EVERY element m in IN: Either n < m, m < n, or n = m.

For all those elements subtraction leaves a big rest
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
But subtracting all elements
|ℕ \ {1, 2, 3, ...}| = 0
leaves nothing.
Hence there is a difference between ℕ_def and ℕ.
>
> Hence there are no "dark numbers" in IN.

There is a difference between ℕ_def and ℕ. This is dark.

Regards, WM

William schrieb am Freitag, 7. Januar 2022 um 15:57:15 UTC+1:
> On Friday, January 7, 2022 at 7:08:45 AM UTC-4, WM wrote:

> > > > > Each element that is in N_P is followed by an infinite number of elements that are in N_P.
> > > > If you take the union over [the set of] FISONs then ℵo elements remain outside.
> > >
> > > Nope. N_P is defined to be the union of the set of FISONS. There is no largest FISON. Thus there is no largest element of N_P

Fine. I call this ℕ_def. But every element has an ℵ₀-endsegment.

> N_P is the union of the set of FISONs.

Fine. The intersection of ℵ₀-endsegments is an ℵ₀-endsegment by inclusion monotony.

> Every natural number is an element of the set N_P.

No. Every element of ℕ_P has the distance ℵ₀ from omega, i.e., between every element of ℕ_P and omega, there are ℵ₀ naturak numbers, i.e., almost all natural numbers.

Every element of the set N_P is finite. N_P, like any Peano set has cardinality aleph_0.

That is wrong. There are not two consecutive ℵ₀-sets possible in the normal order of ℕ. But every element of ℕ_P has an ℵ₀-endsegment. Therefore all these elements make the set ℕ_P which cannot be an ℵ₀-set.

Note: If every element of a subset of ℕ has distance ℵ₀ from omega, then the minimum distance is omega. Then the whole subset has distance ℵ₀ from omega.

Regards, WM

On 1/8/2022 5:35 AM, WM wrote:
> Fritz Feldhase schrieb am Freitag, 7. Januar 2022 um 13:39:42 UTC+1:
>> On Friday, January 7, 2022 at 12:08:45 PM UTC+1, WM wrote:
>>
>>> The dark natnumbers have no discernible order.
>> This proves that there are no dark numbers in IN, since IN is a totally orderd set [by the usual <=].
>
> What is the point on the real axis next to 0? There is something. There is no gap. What is it? It cannot be put in any order.

there is no "next to"

Proof

pick a number "next to 0", say k
now choose k/2
k/2 is closer to 0
contradiction
therefore there is no number you can pick "next to 0"

this is OLD stuff, discussed back in 1900s

>>
>> You know, for EACH and EVERY element n in IN and EACH and EVERY element m in IN: Either n < m, m < n, or n = m.
>
> For all those elements subtraction leaves a big rest
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

you stopped at n

> But subtracting all elements
> |ℕ \ {1, 2, 3, ...}| = 0
> leaves nothing.
> Hence there is a difference between ℕ_def and ℕ.

you stopped at n.

>>
>> Hence there are no "dark numbers" in IN.
>
> There is a difference between ℕ_def and ℕ. This is dark.

wrong, you stopped at n.

>
> Regards, WM

On 1/8/2022 6:29 AM, WM wrote:
> William schrieb am Freitag, 7. Januar 2022 um 15:57:15 UTC+1:
>> On Friday, January 7, 2022 at 7:08:45 AM UTC-4, WM wrote:
>
>>>>>> Each element that is in N_P is followed by an infinite number of elements that are in N_P.
>>>>> If you take the union over [the set of] FISONs then ℵo elements remain outside.
>>>>
>>>> Nope. N_P is defined to be the union of the set of FISONS. There is no largest FISON. Thus there is no largest element of N_P
>

>
> Note: If every element of a subset of ℕ has distance ℵ₀ from omega, then the minimum distance is omega. Then the whole subset has distance ℵ₀ from omega.

wrong, you again confuse sets and elements.

>
> Regards, WM

WM has brought this to us :
> Fritz Feldhase schrieb am Freitag, 7. Januar 2022 um 13:39:42 UTC+1:
>> On Friday, January 7, 2022 at 12:08:45 PM UTC+1, WM wrote:
>>
>>> The dark natnumbers have no discernible order.
>> This proves that there are no dark numbers in IN, since IN is a totally
>> orderd set [by the usual <=].
>
> What is the point on the real axis next to 0?

There should be two -- say "Howdy neighbor(s)!"

On 1/8/2022 9:14 AM, FromTheRafters wrote:
> WM has brought this to us :
>> Fritz Feldhase schrieb am Freitag, 7. Januar 2022 um 13:39:42 UTC+1:
>>> On Friday, January 7, 2022 at 12:08:45 PM UTC+1, WM wrote:
>>>> The dark natnumbers have no discernible order.
>>> This proves that there are no dark numbers in IN, since IN is a totally orderd set [by the usual <=].
>>
>> What is the point on the real axis next to 0?
>
> There should be two -- say "Howdy neighbor(s)!"

hmm... if numbers were teeth, the "next to" could be a cavity...

Extra Credit Problem;

Is the point "next to" 0 is 0.000...001 ?

Or is it -0.000...001 ?

On Saturday, January 8, 2022 at 8:29:24 AM UTC-4, WM wrote:
> William schrieb am Freitag, 7. Januar 2022 um 15:57:15 UTC+1:
> > On Friday, January 7, 2022 at 7:08:45 AM UTC-4, WM wrote:
>
> > > > > > Each element that is in N_P is followed by an infinite number of elements that are in N_P.
> > > > > If you take the union over [the set of] FISONs then ℵo elements remain outside.
> > > >
> > > > Nope. N_P is defined to be the union of the set of FISONS. There is no largest FISON. Thus there is no largest element of N_P
> Fine. I call this ℕ_def.

Nope. |N_def is the set of natural numbers that have the property they can be written down. |N_def is a proper subset of N_P

--
William Hughes

lørdag 8. januar 2022 kl. 12:35:16 UTC+1 skrev WM:
> Fritz Feldhase schrieb am Freitag, 7. Januar 2022 um 13:39:42 UTC+1:
> > On Friday, January 7, 2022 at 12:08:45 PM UTC+1, WM wrote:
> >
> > > The dark natnumbers have no discernible order.
> > This proves that there are no dark numbers in IN, since IN is a totally orderd set [by the usual <=].
> What is the point on the real axis next to 0? There is something. There is no gap. What is it? It cannot be put in any order.
> >
> > You know, for EACH and EVERY element n in IN and EACH and EVERY element m in IN: Either n < m, m < n, or n = m.
> For all those elements subtraction leaves a big rest
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
> But subtracting all elements
> |ℕ \ {1, 2, 3, ...}| = 0
> leaves nothing.
> Hence there is a difference between ℕ_def and ℕ.
> >
> > Hence there are no "dark numbers" in IN.
> There is a difference between ℕ_def and ℕ. This is dark.
>
> Regards, WM

If |ℕ \ {1, 2, 3, ...}| = 0 leaves nothing, why don't |ℕ \ {1, 2, 3, ..., w}| = 0 leaves nothing?
KON