On Thursday, January 19, 2023 at 6:22:58 AM UTC-4, WM wrote:

> The set of unit fractions 1/n has cardinality ℵo.
> No further unit fraction fits between all unit fractions and zero.
> There are less Peano unit fractions because for every Peano unit fraction ℵo smaller unit fractions fit between it and zero.

Please rephrase without using the word "all".

On 1/19/2023 5:45 AM, WM wrote:
> Jim Burns schrieb am Donnerstag,
> 19. Januar 2023 um 02:59:41 UTC+1:

>> Our meta-reasoning treats claims the way that
>> our reasoning treats sheep and pebbles.
>> Our logic has not seen the Fall.
>
> Can it distinguish this?
> No further unit fraction
> fits between all unit fractions and zero.
> For every Peano unit fraction
> ℵo smaller unit fractions fit between it and zero.

If a sequence of claims has, in order,
the claims
| | P ⇒ For every Peano unit fraction
| ℵo smaller unit fractions fit between it and zero.
| | P
| | For every Peano unit fraction
| ℵo smaller unit fractions fit between it and zero.
| then the claim
| | For every Peano unit fraction
| ℵo smaller unit fractions fit between it and zero.
| is not first possibly-false.

If each claim in a sequence is
not first possibly-false
then each claim in that sequence is
not possibly-false.

> Can it distinguish this?

It works correctly on it.
Since that's probably what you mean, yes.

On Thursday, January 19, 2023 at 11:37:23 AM UTC+1, WM wrote:

> You claim that all endegments are infinite.

Yes, we claim that, because it can be proved from the axioms and definitions in set theory (especially from the definitios of /endsegement/ and /infinite/).

> That is true[.]

Of course.

> But it is not true that their intersection is empty.

Nope. It's equally "true" in the context of set theory (say ZFC), since we can _prove_ it too. Moreover, it's a quite obvious set-theoretic state of affairs, you silly idiot.

> You conclude that only from the fact that you cannot see the dark numbers.

No, we "conclude" that, because it is a theorem in set theory. Moreoer, it's a quite obvious set-theoretic state of affairs. That means that the theorem is in agreement with out set-theoretic intuitions.

> Thus you [are convvinced] that infinite inclusion-monotonic sets [can] have an empty intersection.

Right. You know, contemporary mathematics is based on proofs.

In contrast to that your chatter is just

> Blatant nonsense.

Agree!

On Thursday, January 19, 2023 at 6:22:41 PM UTC+1, Jim Burns wrote:

> | For every Peano unit fraction

Could you please define the term /Peano unit fraction/?

What is the difference between a /Peano unit fraction/ and a "standard" unit fraction?

On 1/19/2023 11:44 AM, Fritz Feldhase wrote:
> On Thursday, January 19, 2023 at 6:22:41 PM UTC+1, Jim Burns wrote:
>
>> | For every Peano unit fraction
>
> Could you please define the term /Peano unit fraction/?
>
> What is the difference between a /Peano unit fraction/ and a "standard" unit fraction?
>

I have been informed by The Ant Group, and repost here;

Peano Unit Ants

Standard Unit Ants

Difference Ants without a unit

Asexual Difference between Peano Ants and Standard Ants

On Thursday, January 19, 2023 at 11:45:16 AM UTC+1, WM wrote:

> Can it distinguish this?

Some notions:

| U := {1/n : n e IN} ...the set of "unit fractions".
| "E^ℵo" ...a special quantifier, meaning: there are ℵo many <such and such>

> No further unit fraction fits between all unit fractions and zero.

~Ex e U: Au e U: 0 < x < u.
"There is no unit fraction x such that x is between 0 and all unit fractions."

Undoubtedly true, since no unit fraction can be smaller than itself.

[Actually, there's not even a/any _real number_ x such that x is between 0 and all unit fractions:
~Ex e IR: Au e U: 0 < x < u.]

> For every [...] unit fraction ℵo smaller unit fractions fit between it and zero.

Au e U: E^ℵox e U: 0 < x < u.
"For all unit fractions x: there are countably infinitely many unit fractions between 0 and x."

Again undoubtedly true, since for each and every natural number n the -countably infinitely many- unit fractions 1/(n+1), 1/(n+2), 1/(n+3), ... are bigger than 0 but smaller than 1/n.

A slightly weaker claim: "For all unit fractions u: there is a unit fraction between 0 and u."
Au e U: Ex e U: 0 < x < u.

Again undoubtedly true, since for each and every natural number n the unit fraction 1/(n+1) is bigger than 0 but smaller than 1/n.

Now it seems to me that Mr. Mückenheim is applying the (in)famous /quantifier shift/ again. Its application would allow to "conclude" Ex e U: Au e U: 0 < x < u (false) from Au e U: Ex e U: 0 < x < u (true). Seems to be the reason why he has problems to accept BOTH ~Ex e U: Au e U: 0 < x < u _and_ Au e U: E^ℵox e U: 0 < x < u (or Au e U: Ex e U: 0 < x < u).

For Mückenheim the claims

"For all unit fractions u there is a unit fraction x such that: 0 < x < u."
and
"There is a unit fraction x such that for all unit fractions u: 0 < x < u."

seem to be "equivalent". (See: quantifier dyslexia)

Fritz Feldhase schrieb am Donnerstag, 19. Januar 2023 um 18:39:21 UTC+1:
> On Thursday, January 19, 2023 at 11:37:23 AM UTC+1, WM wrote:
>
> > You claim that all endegments are infinite.
> Yes, we claim that, because it can be proved from the axioms and definitions in set theory (especially from the definitios of /endsegement/ and /infinite/).

Then you should consider ∀k ∈ ℕ: E(k+1) = E(k) \ {k}. No natnumber remains.

> > But it is not true that their intersection is empty.
> Nope. It's equally "true" in the context of set theory (say ZFC), since we can _prove_ it too.

So numbers remain in endsegments but do not remain in their intersection. What is the first / smallest number that remains in all endsegments?

Regards, WM

On Thursday, January 19, 2023 at 8:48:00 PM UTC+1, WM wrote:

> So numbers remain in endsegments but do not remain in their intersection.

Look, dumbo: For _each and every_ endsegment there is a number which "is in it". But there is no number which is in _all_ endsegments.

You know:

(*) As in ENDSEG: En e IN: n in s

does not imply

(**) En e IN: As in ENDSEG: n in s

Actually the invalid "step" from (*) to (**) is a logical fallacy known as /quantifier shift/.
See: https://en.wikipedia.org/wiki/Quantifier_shift

> What is the first / smallest number that remains in all endsegments?

Huh?! This is a nonsensical question. Since there is _no_ number "that remains in" (read: is in) all endsegments, there can't be a first/smallest such number, you silly crank.

On 1/19/2023 1:47 PM, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 19. Januar 2023 um 18:39:21 UTC+1:
>> On Thursday, January 19, 2023 at 11:37:23 AM UTC+1, WM wrote:
>>
>>> You claim that all endegments are infinite.
>> Yes, we claim that, because it can be proved from the axioms and definitions in set theory (especially from the definitios of /endsegement/ and /infinite/).
>
> Then you should consider ∀k ∈ ℕ: E(k+1) = E(k) \ {k}. No natnumber remains.

You should consider you stopped at k. and Endsegment(k+1) remains

>
>>> But it is not true that their intersection is empty.
>> Nope. It's equally "true" in the context of set theory (say ZFC), since we can _prove_ it too.
>
> So numbers remain in endsegments but do not remain in their intersection.

yes, intersection is a set operator, google INTERSECTION

and the intersection of all endsegments is not an endsegment.

> What is the first / smallest number that remains in all endsegments?

for some reason, you think it is k

but that is wrong.

>
> Regards, WM

On 1/19/2023 5:37 AM, WM wrote:
> Jim Burns schrieb am Mittwoch,
> 18. Januar 2023 um 23:37:28 UTC+1:
>> On 1/18/2023 5:16 AM, WM wrote:

>>>>> I assert that
>>>>> mathematics and logic are true.

>>> Your chain of claims does not reach
>>> until infinity. But Cantors sets do.
>>
>> You are saying that you disagree with
>> mathematics and logic
>> without saying that you disagree with
>> mathematics and logic.
>
> I adhere to logic.
> You adhere to nonsense called mathematical logic.

Changing the label on a box
does not change what's in the box.
You assert that mathematics and logic
is not true,
in the widely-used sense.

Give your argument for your assertion, then.
But it is a lie that you (WM) assert that
mathematics and logic are true.

>>> Your chain of claims does not reach
>>> until infinity.
>>
>> No, that's wrong.
>
> You confuse the infinities.
>
> The set of unit fractions 1/n
> has cardinality ℵo.
> No further unit fraction fits between
> all unit fractions and zero.
>
> There are less Peano unit fractions
> because for every Peano unit fraction
> ℵo smaller unit fractions fit between it and zero.
>
> That is a difference, a huge difference,
> an infinite difference.

If
we define ℕ_Peano and ℕ_Cantor such that,
for each FISON F and each inductive M
F ⊆ ℕ_Peano ⊆ ℕ_Cantor ⊆ M
then,
no, there is no difference between
ℕ_Peano and ℕ_Cantor

For each FISON F,
F⊕⟨F⌝⁺⁺⟩ is a FISON too,

F⊕⟨F⌝⁺⁺⟩ is
F with
the successor F⌝⁺⁺ of
its end F⌝
appended.

Thus,
for each k in FISON F ⊆ ℕ_Peano,
k⁺⁺ is in F⊕⟨F⌝⁺⁺⟩ ⊆ ℕ_Peano
and so
ℕ_Peano is inductive.

By definition,
for each inductive M
ℕ_Peano ⊆ ℕ_Cantor ⊆ M

ℕ_Peano ⊆ ℕ_Cantor ⊆ ℕ_Peano

ℕ_Peano = ℕ_Cantor = ℕ_Peano

No difference.

>> There is "all" and there is "more than all".
>> You (WM) tend to read claims as though they are
>> about more than all.
>
> More than all that are covered by your induction.

Is that intended to be a criticism?

ℕ_Peano induction claims only cover ℕ_Peano
Yes!
Why would we want it otherwise?

I claim that
all duck-billed platypuses have webbed feet.

Do not read that as a more-than-all claim that
all natural numbers also have webbed feet.

>> I have no responsibility for more-than-all
>> readings of a claim of mine.
>
> You claim that all endegments are infinite.
> That is true as far as induction reaches.

As far as induction reaches == ℕ_Peano
ℕ_Peano = ℕ_Cantor = ℕ

It is true as far as the claim reaches.
which is as far as it needs to reach.

> But it is not true that
> their intersection is empty.
> You conclude that only from the fact that
> you cannot see the dark numbers.

You announce you're ignoring my proof and then
you tell me what I have in my proof --
incorrectly.

You are wrong.
I conclude that from the fact that
F⊕⟨F⌝⁺⁺⟩ is a FISON too.

Anything which is in all end segments
is after everything else.
Because F⊕⟨F⌝⁺⁺⟩ is a FISON too,
nothing exists which is after everything else.
(More-than-all claims are thrown out.)

Nothing exists which is in all end segments.
The intersection is empty.

>> in other words, that there isn't "more than all".
>>
>> Is that not always true?
>
> See the set of unit fractions above.
> Try to understand it.

For each finite ordinal,
for each split up to it,
some i is last-before and
some j is first-after
and i⁺⁺ = j

For each ordinal, finite or infinite,
for each split up to it,
---- - -- ----------- ---
some j is first-after
--- --- - -

Not all splits of ordinals
have a last-before == are _last-befored_

If all were last-befored, then
"infinity" (not-infinity) would only be
a reallyreallyreallyreallyreallyreally large
individually usable (finite) number.

However,
infinity is NOT
a reallyreallyreallyreallyreallyreally large
individually usable number.

Consider the split
{all-last-befored} {not-all-last-befored}

That split has a first-after.
That first-after is first in {not-all-last-befored}
That first-after is what we name ω omega.

Each split of {all-last-befored} is last-befored.
All the splits up to ω omega are
splits of {all-last-befored} and
{all-last-befored} {not-all-last-befored}

The only split up to ω omega which
could be not last-befored is
{all-last-befored} {not-all-last-befored}
and one split is not last-befored.

Therefore,
{all-last-befored} {not-all-last-befored}
is not last-befored.

ω omega is first-after that split.
Nothing exists, visible or dark,
which is last-before that split

.... because some splits are not last-befored,
and infinity is NOT
a reallyreallyreallyreallyreallyreally large
individually usable number.

On 1/19/2023 12:44 PM, Fritz Feldhase wrote:
> On Thursday, January 19, 2023
> at 6:22:41 PM UTC+1, Jim Burns wrote:

>> | For every Peano unit fraction
>
> Could you please define the term
> /Peano unit fraction/?

The Peano unit fractions are totally-ordered.
Splits of Peano unit fractions exist.

1/n is a Peano unit fraction
iff
for each split between 1/1 and 1/n
some 1/j is last-before
and 1/j⁺⁺ is first-after.

A successor j⁺⁺
is not 0
is not the successor of anything else
has a successor.

> What is the difference between
> a /Peano unit fraction/ and
> a "standard" unit fraction?

There is no true difference.

However,
WM thinks there is a difference between
ℕ_Peano and ℕ_Cantor

ℕ_Peano is the least upper bound of FISONs

ℕ_Cantor is the greatest lower bound of
inductive sets.

Dark numbers are elements of ℕ_Cantor\ℕ_Peano

For each FISON F and each inductive set M
F ⊆ ℕ_Peano ⊆ ℕ_Cantor ⊆ M

I give them separate names on my way to
proving that they are equal.[1]

However crank-infested this discussion may be,
(a point I won't argue)
showing that two definitions define the
same thing is respectable mathematics.

[1]
If F is a FISON
then F⊕⟨F⌝⁺⁺⟩ is a FISON too.

If k ∈ F ⊆ ℕ_Peano
then k⁺⁺ ∈ F⊕⟨F⌝⁺⁺⟩ ⊆ ℕ_Peano

ℕ_Peano is inductive

ℕ_Peano ⊆ ℕ_Cantor ⊆ ℕ_Peano

ℕ_Peano = ℕ_Cantor = ℕ_Peano

Dark numbers are in ℕ_Cantor\ℕ_Peano
Nothing is in ℕ_Cantor\ℕ_Peano

It is not that we merely do not see dark numbers.
Dark numbers do not exist
because
F⊕⟨F⌝⁺⁺⟩ is a FISON too.

On 1/19/2023 3:36 PM, Jim Burns wrote:
> On 1/19/2023 12:44 PM, Fritz Feldhase wrote:
>> On Thursday, January 19, 2023
>> at 6:22:41 PM UTC+1, Jim Burns wrote:
>
>>> | For every Peano unit fraction
>>
>> Could you please define the term
>> /Peano unit fraction/?
>
> The Peano unit fractions are totally-ordered.

you can't order just one..., you have to order the whole showboutle

> Splits of Peano unit fractions exist.

Split Peano Soup, exists but costs more.

( sorry for the bad jokes, not serious today, but hungry )

>
> 1/n is a Peano unit fraction
> iff
> for each split between 1/1 and 1/n
> some 1/j is last-before
> and 1/j⁺⁺ is first-after.
>
> A successor j⁺⁺
> is not 0
> is not the successor of anything else
> has a successor.
>
>> What is the difference between
>> a /Peano unit fraction/ and
>> a "standard" unit fraction?
>
> There is no true difference.
>
> However,
> WM thinks there is a difference between
> ℕ_Peano and ℕ_Cantor
>
> ℕ_Peano is the least upper bound of FISONs
>
> ℕ_Cantor is the greatest lower bound of
> inductive sets.
>
> Dark numbers are elements of ℕ_Cantor\ℕ_Peano
>
> For each FISON F and each inductive set M
> F ⊆ ℕ_Peano ⊆ ℕ_Cantor ⊆ M
>
> I give them separate names on my way to
> proving that they are equal.[1]
>
> However crank-infested this discussion may be,
> (a point I won't argue)
> showing that two definitions define the
> same thing is respectable mathematics.
>
> [1]
> If F is a FISON
> then F⊕⟨F⌝⁺⁺⟩ is a FISON too.
>
> If k ∈ F ⊆ ℕ_Peano
> then k⁺⁺ ∈ F⊕⟨F⌝⁺⁺⟩ ⊆ ℕ_Peano
>
> ℕ_Peano is inductive
>
> ℕ_Peano ⊆ ℕ_Cantor ⊆ ℕ_Peano
>
> ℕ_Peano = ℕ_Cantor = ℕ_Peano
>
> Dark numbers are in ℕ_Cantor\ℕ_Peano
> Nothing is in ℕ_Cantor\ℕ_Peano
>
> It is not that we merely do not see dark numbers.
> Dark numbers do not exist
> because
> F⊕⟨F⌝⁺⁺⟩ is a FISON too.
>
>

Fritz Feldhase schrieb am Donnerstag, 19. Januar 2023 um 21:03:05 UTC+1:
> On Thursday, January 19, 2023 at 8:48:00 PM UTC+1, WM wrote:
>
> > So numbers remain in endsegments but do not remain in their intersection.
> For _each and every_ endsegment there is a number which "is in it". But there is no number which is in _all_ endsegments.

Nonsense. All non-empty endsegments have a non-empty intersection by inclusion-monotony.

Regards, WM

William schrieb am Donnerstag, 19. Januar 2023 um 17:28:54 UTC+1:
> On Thursday, January 19, 2023 at 6:22:58 AM UTC-4, WM wrote:
> > The set of unit fractions 1/n has cardinality ℵo.
> > No further unit fraction fits between all unit fractions and zero.
> > There are less Peano unit fractions because for every Peano unit fraction ℵo smaller unit fractions fit between it and zero.
> Please rephrase without using the word "all".

The word "all" is dictated by Cantor's actual infinity.
Do you doubt that there are all unit fractions?

Regards, WM

William schrieb am Donnerstag, 19. Januar 2023 um 15:21:05 UTC+1:
> On Thursday, January 19, 2023 at 6:22:58 AM UTC-4, WM wrote:
> > William schrieb am Mittwoch, 18. Januar 2023 um 15:53:39 UTC+1:
> > > On Wednesday, January 18, 2023 at 6:07:17 AM UTC-4, WM wrote:
> > > > William schrieb am Dienstag, 17. Januar 2023 um 17:08:13 UTC+1:
> > > >
> > > > > A Peano set has cardinality ℵo.
> > > > No.
> > > A Peano set does not have finite cardinality. A Peano set does not change so it cannot be finite but bounded.
> > A Peano set is not a set (in the meaning of ZFC).
> Piffle. Of course it is.

Then there must be all elements, not only for some considered one a next one.

Regards, WM

Jim Burns schrieb am Donnerstag, 19. Januar 2023 um 18:22:41 UTC+1:

> | For every Peano unit fraction
> | ℵo smaller unit fractions fit between it and zero.
> |
> is not first possibly-false.

Is it true for all Peano unit fractions too?

And what about the claim: No further unit fraction fits between all unit fractions and zero?

Regards, WM

Fritz Feldhase schrieb am Donnerstag, 19. Januar 2023 um 20:05:48 UTC+1:

> Now it seems to me that Mr. Mückenheim is applying the (in)famous /quantifier shift/ again.

No. There are two different statements:

For every Peano unit fraction u, ℵo smaller unit fractions fit between it and zero.
In fact there are more unit fractions between 0 and u than between u and 1..
Collect all unit fractions with this property into a set.
Does that change this property for any element?
Does it change this property for the set of all elements?
How could that be proven?

Regards, WM

WM expressed precisely :
> Fritz Feldhase schrieb am Donnerstag, 19. Januar 2023 um 21:03:05 UTC+1:
>> On Thursday, January 19, 2023 at 8:48:00 PM UTC+1, WM wrote:
>>
>>> So numbers remain in endsegments but do not remain in their intersection.
>> For _each and every_ endsegment there is a number which "is in it". But
>> there is no number which is in _all_ endsegments.
>
> Nonsense. All non-empty endsegments have a non-empty intersection by
> inclusion-monotony.

All non-empty endsegments? Did you validate this? How do you know that
they are each in their assigned place on your otherwise dark number
line?

On 1/19/2023 4:27 PM, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 19. Januar 2023 um 21:03:05 UTC+1:
>> On Thursday, January 19, 2023 at 8:48:00 PM UTC+1, WM wrote:
>>
>>> So numbers remain in endsegments but do not remain in their intersection.
>> For _each and every_ endsegment there is a number which "is in it". But there is no number which is in _all_ endsegments.
>
> Nonsense. All non-empty endsegments have a non-empty intersection by inclusion-monotony.

no, by delusion-montony

>
> Regards, WM

Jim Burns explained on 1/19/2023 :
> On 1/19/2023 12:44 PM, Fritz Feldhase wrote:
>> On Thursday, January 19, 2023
>> at 6:22:41 PM UTC+1, Jim Burns wrote:
>
>>> | For every Peano unit fraction
>>
>> Could you please define the term
>> /Peano unit fraction/?
>
> The Peano unit fractions are totally-ordered.
> Splits of Peano unit fractions exist.
>
> 1/n is a Peano unit fraction
> iff
> for each split between 1/1 and 1/n
> some 1/j is last-before
> and 1/j⁺⁺ is first-after.
>
> A successor j⁺⁺
> is not 0
> is not the successor of anything else
> has a successor.
>
>> What is the difference between
>> a /Peano unit fraction/ and
>> a "standard" unit fraction?
>
> There is no true difference.
>
> However,
> WM thinks there is a difference between
> ℕ_Peano and ℕ_Cantor
>
> ℕ_Peano is the least upper bound of FISONs
>
> ℕ_Cantor is the greatest lower bound of
> inductive sets.
>
> Dark numbers are elements of ℕ_Cantor\ℕ_Peano
>
> For each FISON F and each inductive set M
> F ⊆ ℕ_Peano ⊆ ℕ_Cantor ⊆ M

Can F actually be an improper subset of N_Peano here?

On 1/19/2023 4:47 PM, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 19. Januar 2023 um 20:05:48 UTC+1:
>
>> Now it seems to me that Mr. Mückenheim is applying the (in)famous /quantifier shift/ again.
>
> No. There are two different statements:
>
> For every Peano unit fraction u, ℵo smaller unit fractions fit between it and zero.
> In fact there are more unit fractions between 0 and u than between u and 1.
> Collect all unit fractions with this property into a set.
> Does that change this property for any element?
> Does it change this property for the set of all elements?
> How could that be proven?
>
> Regards, WM
>

So, what is the second statement ?

What is a "Peano unit fraction" ?

The "property" you imply, does not belong to the "element".

What is the "set of all elements" ?

A Keeper!
"How could that be proven?" - WM

On 1/19/2023 4:29 PM, WM wrote:
> William schrieb am Donnerstag, 19. Januar 2023 um 17:28:54 UTC+1:
>> On Thursday, January 19, 2023 at 6:22:58 AM UTC-4, WM wrote:
>>> The set of unit fractions 1/n has cardinality ℵo.
>>> No further unit fraction fits between all unit fractions and zero.
>>> There are less Peano unit fractions because for every Peano unit fraction ℵo smaller unit fractions fit between it and zero.
>> Please rephrase without using the word "all".
>
> The word "all" is dictated by Cantor's actual infinity.
> Do you doubt that there are all unit fractions?
>
> Regards, WM

to be clear, use "each and every" instead of "all"

Cantor never was a dictator.

Utter failure.

On Thursday, January 19, 2023 at 6:32:21 PM UTC-4, WM wrote:
> William schrieb am Donnerstag, 19. Januar 2023 um 15:21:05 UTC+1:
> > On Thursday, January 19, 2023 at 6:22:58 AM UTC-4, WM wrote:
> > > William schrieb am Mittwoch, 18. Januar 2023 um 15:53:39 UTC+1:
> > > > On Wednesday, January 18, 2023 at 6:07:17 AM UTC-4, WM wrote:
> > > > > William schrieb am Dienstag, 17. Januar 2023 um 17:08:13 UTC+1:
> > > > >
> > > > > > A Peano set has cardinality ℵo.
> > > > > No.
> > > > A Peano set does not have finite cardinality. A Peano set does not change so it cannot be finite but unbounded.
> > > A Peano set is not a set (in the meaning of ZFC).
> > Piffle. Of course it is.
> Then there must be all elements,

Which, like for any set, there are. A set does not change, You cannot add elements. Note that every element of a Peano set (except the first) is the successor to some element of the Peano set. There are no other elements.

On Thursday, January 19, 2023 at 3:18:13 PM UTC-8, Sergi o wrote:
> On 1/19/2023 4:29 PM, WM wrote:
> > William schrieb am Donnerstag, 19. Januar 2023 um 17:28:54 UTC+1:
> >> On Thursday, January 19, 2023 at 6:22:58 AM UTC-4, WM wrote:
> >>> The set of unit fractions 1/n has cardinality ℵo.
> >>> No further unit fraction fits between all unit fractions and zero.
> >>> There are less Peano unit fractions because for every Peano unit fraction ℵo smaller unit fractions fit between it and zero.
> >> Please rephrase without using the word "all".
> >
> > The word "all" is dictated by Cantor's actual infinity.
> > Do you doubt that there are all unit fractions?
> >
> > Regards, WM
> to be clear, use "each and every" instead of "all"

That is a ridiculous request. He's been asked to do that for over 20 years now. He will never comply for two reasons:

1) He doesn't know how to do it
2) To do so would completely destroy the entire edifice of ambiguity and misdirection and gibberish upon which his magic kingdom is constructed.