Le 17/02/2024 à 13:15, Richard Damon a écrit :
> On 2/17/24 4:56 AM, WM wrote:
>> Mike Terry schrieb am Freitag, 16. Februar 2024 um 22:45:48 UTC+1:
>>
>>> Yes, it's not clear what "infinitely close" means
>>
>> It means dark numbers.
>> The function Number of Unit Fractions between (0, and x) has
>> the following properties:
>> (1) An increase from NUF(0) = 0 to NUF(x>0) > 0 cannot happen unless
>> NUF(x) increases at some x.
>> (2) NUF(x) cannot increase other than when passing unit fractions at some
>> x = 1/n.
>> (3) NUF(x) cannot pass more than one unit fraction at a single point x
>> because
>> ∀n ∈ ℕ: 1/n =/= 1/(n-1).
>> (4) This requires a first unit fraction, if all are there in actual
>> infinity.
>>
> In other words, "Dark Numbers" are made up numbers that try to patch the
> holes in your logic

There are no holes in my logic. There is nonsense in your belief.

> Of course, since your premises are just wrong,

My premises are (1) to (3). Nothing wrong.

Regards, WM

On 2/16/2024 12:49 PM, Chris M. Thomasson wrote:
> Take a number that wants to get close to zero. Say:
>
> [0] = 1
> [1] = .1
> [2] = .01
> [3] = .001
> [...] = [...]
>
> This gets close to zero, yet never will equal zero. Okay so:
>
> arbitrarily close seems to be the accepted term.
>
> infinitely close is the wrong wording?
>
> The function f(n) = 10^(-n) gets "infinitely close" to 0... lol. Using
> the "metaphysical formation" of arbitrarily close... ;^)

A fun summation:

10^(-0) + 10^(-2) + 10^(-4) + 10^(-6) + ...

1.0101010101...

or:

10^(-0)*1 + 10^(-2)*2 + 10^(-4)*3 + 10^(-6)*4 + ...

1.020304 ...

On 2/17/24 5:03 PM, WM wrote:
> Le 17/02/2024 à 13:15, Richard Damon a écrit :
>> On 2/17/24 4:56 AM, WM wrote:
>>> Mike Terry schrieb am Freitag, 16. Februar 2024 um 22:45:48 UTC+1:
>>>
>>>> Yes, it's not clear what "infinitely close" means
>>>
>>> It means dark numbers.
>>> The function Number of Unit Fractions between (0, and x) has
>>> the following properties:
>>> (1) An increase from NUF(0) = 0 to NUF(x>0) > 0 cannot happen unless
>>> NUF(x) increases at some x.
>>> (2) NUF(x) cannot increase other than when passing unit fractions at
>>> some
>>> x = 1/n.
>>> (3) NUF(x) cannot pass more than one unit fraction at a single point x
>>> because
>>> ∀n ∈ ℕ: 1/n =/= 1/(n-1).
>>> (4) This requires a first unit fraction, if all are there in actual
>>> infinity.
>>>
>> In other words, "Dark Numbers" are made up numbers that try to patch
>> the holes in your logic
>
> There are no holes in my logic. There is nonsense in your belief.

You just can't see the holes, because you close your eyes.

You seem to think that unbounded sets have there bounds in them.

>
>> Of course, since your premises are just wrong,
>
> My premises are (1) to (3). Nothing wrong.
>

You assume that your NUF exists, and that assumption requires that there
be a "first" (lowest) Unit Fraction which exists.

Since, no such number can exist, as if x is a unit fraction x/2 will be
too, and will be lower than it, you logic is fallacious.

So, your assumptions are incompatible with the definition of the Natural
Numbers, so make your system inconsistent, and thus worthless.

> Regards, WM
>
>

On 2/17/24 4:59 PM, WM wrote:
> Le 17/02/2024 à 13:15, Richard Damon a écrit :
>> On 2/17/24 3:23 AM, WM wrote:
>>>
>>> Take the function Number of Unit Fractions between (0, and x > 0). It
>>> has the following properties:
>>> (1) An increase from NUF(0) = 0 to NUF(x>0) > 0 cannot happen unless
>>> NUF(x) increases at some x.
>>> (2) NUF(x) cannot increase other than when passing unit fractions at
>>> some x = 1/n.
>>> (3) NUF(x) cannot pass more than one unit fraction at a single point
>>> x because
>>> ∀n ∈ ℕ: 1/n =/= 1/(n-1).
>>> (4) This requires a first unit fraction, if all are there in actual
>>> infinity.
>
>> Or, that such a function can't actually be defined, because it assumes
>> that there IS a "smallest unit fraction".
>
> This well-defined function proves its existence.
>
> Regards, WM
>
>

NOT well defined.

Your assumptions that define it are inconsistant with the definition of
Natural Numbers.

Your NUF(x) has an output range that is outside the range of the Natual
Numbers, and thus not "Well Defined" on that set (or the rationals, or
the Reals).

So, all you are showing is that you don't understand what a "well
defined" function means.

It is easy to create a word salad that seems to fully define a function,
that actually described a not possible to exist thing, and your above is
just one example of that.

Richard Damon used his keyboard to write :
> On 2/17/24 4:59 PM, WM wrote:
>> Le 17/02/2024 à 13:15, Richard Damon a écrit :
>>> On 2/17/24 3:23 AM, WM wrote:
>>>>
>>>> Take the function Number of Unit Fractions between (0, and x > 0). It has
>>>> the following properties:
>>>> (1) An increase from NUF(0) = 0 to NUF(x>0) > 0 cannot happen unless
>>>> NUF(x) increases at some x.
>>>> (2) NUF(x) cannot increase other than when passing unit fractions at some
>>>> x = 1/n.
>>>> (3) NUF(x) cannot pass more than one unit fraction at a single point x
>>>> because
>>>> ∀n ∈ ℕ: 1/n =/= 1/(n-1).
>>>> (4) This requires a first unit fraction, if all are there in actual
>>>> infinity.
>>
>>> Or, that such a function can't actually be defined, because it assumes
>>> that there IS a "smallest unit fraction".
>>
>> This well-defined function proves its existence.
>>
>> Regards, WM
>>
>>
>
>
> NOT well defined.
>
> Your assumptions that define it are inconsistant with the definition of
> Natural Numbers.
>
> Your NUF(x) has an output range that is outside the range of the Natual
> Numbers, and thus not "Well Defined" on that set (or the rationals, or the
> Reals).
>
> So, all you are showing is that you don't understand what a "well defined"
> function means.

Or what a set is.

> It is easy to create a word salad that seems to fully define a function, that
> actually described a not possible to exist thing, and your above is just one
> example of that.

Indeed! He'll never learn though, he's completely stuck.

On 02/18/2024 08:49 AM, FromTheRafters wrote:
> Richard Damon used his keyboard to write :
>> On 2/17/24 4:59 PM, WM wrote:
>>> Le 17/02/2024 à 13:15, Richard Damon a écrit :
>>>> On 2/17/24 3:23 AM, WM wrote:
>>>>>
>>>>> Take the function Number of Unit Fractions between (0, and x > 0).
>>>>> It has the following properties:
>>>>> (1) An increase from NUF(0) = 0 to NUF(x>0) > 0 cannot happen
>>>>> unless NUF(x) increases at some x.
>>>>> (2) NUF(x) cannot increase other than when passing unit fractions
>>>>> at some x = 1/n.
>>>>> (3) NUF(x) cannot pass more than one unit fraction at a single
>>>>> point x because
>>>>> ∀n ∈ ℕ: 1/n =/= 1/(n-1).
>>>>> (4) This requires a first unit fraction, if all are there in actual
>>>>> infinity.
>>>
>>>> Or, that such a function can't actually be defined, because it
>>>> assumes that there IS a "smallest unit fraction".
>>>
>>> This well-defined function proves its existence.
>>>
>>> Regards, WM
>>>
>>>
>>
>>
>> NOT well defined.
>>
>> Your assumptions that define it are inconsistant with the definition
>> of Natural Numbers.
>>
>> Your NUF(x) has an output range that is outside the range of the
>> Natual Numbers, and thus not "Well Defined" on that set (or the
>> rationals, or the Reals).
>>
>> So, all you are showing is that you don't understand what a "well
>> defined" function means.
>
> Or what a set is.
>
>> It is easy to create a word salad that seems to fully define a
>> function, that actually described a not possible to exist thing, and
>> your above is just one example of that.
>
> Indeed! He'll never learn though, he's completely stuck.

The iota-values the range of the "natural/unit equivalency function",
not-a-real-function, in the usual and standard sense of the word,
but standardly modeled by real functions, as a continuum limit,
are rather well-defined.

Will he ever learn?

On 2/17/2024 2:14 PM, WM wrote:
> Le 17/02/2024 à 19:35, Jim Burns a écrit :
>> On 2/17/2024 4:56 AM, WM wrote:

>>> (4) This requires a first unit fraction,
>>> if all are there in actual infinity.
>>
>> Each final.ordinal.reciprocal
>> is preceded by
>> another final.ordinal.reciprocal [⅟n]

⅟n⋅n = 1 ∧ {<n} ⃒⇇ {<n}∪{n⁺¹}

> No, this axiom must be given up.

One of these claim must be given up:
| a final.ordinal.reciprocal.free zone (0,δ)
| exists
or
| a skipping.function isn't all.continuous
or
| for final.ordinal.reciprocal ⅟m
| ⅟(4⋅m) is a final.ordinal.reciprocal.

Suppose we accept the second and third claims.

Then, the existence of
a final.ordinal.reciprocal.free zone (0,δ)
is contradictory.

With
a final.ordinal.reciprocal.free zone (0,δ)
and the second claim,
we must also have a maximal
final.ordinal.reciprocal.free zone (0,β) or (0,β]
and points 2β and β/2 such that both
there is
a final.ordinal.reciprocal < 2β
and
there isn't
a final.ordinal.reciprocal < β/2

However,
that contradicts the third claim, whereby
either both or neither
final.ordinal.reciprocal ⅟m < 2β
and
final.ordinal.reciprocal ⅟(4⋅m) < β/2

On the other hand,
suppose you (WM) give up
the second or third claims,
| a skipping.function isn't all.continuous
or
| for final.ordinal.reciprocal ⅟m
| ⅟(4⋅m) is a final.ordinal.reciprocal.

Which do you (WM) give up?
How do you justify giving it up?

Or,
do you (WM) embrace the contradiction and
use Ex Falso Quodlibet to "prove"
whatever you (WM) feel like "proving"
at the moment?

>> The first final.ordinal.reciprocal not.exists.
>
> The alternative would be
> an increase of NUF(x) to infinity at zero.

For each β > 0
for each final.ordinal n
from point 0-β to point 0+β
there are more.than.n final.ordinal.reciprocals
β > ⅟1⁺ᵐᵝ > ... > ⅟n⁺ᵐᵝ > ⅟(n+1)⁺ᵐᵝ > 0
for
0 =< mᵦ =< ⅟β < mᵦ+1 = 1⁺ᵐᵝ

> Not acceptable.

Why?

Le 18/02/2024 à 19:16, Jim Burns a écrit :
>
> Which do you (WM) give up?
> How do you justify giving it up?

I will never give up the following self-evidence:
If there are ℵo unit fractions in the interval (0, eps), then there is
an x with only a finite number of unit fractions in (0, x).

Why? Because unit fractions are real points on the real line. They cannot
appear as an infinite swarm without a finite start.

The intersection of all intervals (0, eps) that can be chosen by anybody
in eternity however contains ℵo unit fractions.

Regards, WM

Le 18/02/2024 à 13:36, Richard Damon a écrit :

> Your assumptions that define it are inconsistant with the definition of
> Natural Numbers.
>
If there are ℵo unit fractions in the interval (0, eps), then there is
an x with only a finite number of unit fractions in (0, x).

Why? Because unit fractions are real points on the real line. They cannot
appear as an infinite swarm without a finite start.

The intersection of all intervals (0, eps) that can be chosen by anybody
in eternity however contains ℵo unit fractions.

Regards, WM

On 2/18/24 4:00 PM, WM wrote:
> Le 18/02/2024 à 19:16, Jim Burns a écrit :
>>
>> Which do you (WM) give up?
>> How do you justify giving it up?
>
> I will never give up the following self-evidence:
> If there are ℵo unit fractions in the interval (0, eps), then there is
> an x with only a finite number of unit fractions in (0, x).
> Why? Because unit fractions are real points on the real line. They
> cannot appear as an infinite swarm without a finite start.

But the start was at 1/1

Remember "real points" take up no spacd, so we can always pack more of
them into any finite space, so there doesn't actually need to be a "first"

If there WAS a "finite first" unit fraction, then there couldn't be an
infinite swarm of them, because then you have a finite length divided in
to segments with a finite lower bound of size, and thus have a finite
count of how many can fit.

But if there was a finite first unit fraction, then there would also be
a finite maximum Natural Number, which is a contradiction of definitions.

So, your logic is just backwards, based on embedded your misconceptions
into a made up function that can't actually exist.

>
> The intersection of all intervals (0, eps) that can be chosen by anybody
> in eternity however contains ℵo unit fractions.
>
> Regards, WM

Right, because infinity can never be decreased by finite operations. So
you can not expect it to.

If there was a "first", then you never had an infinite set.

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

> On 2/16/2024 1:07 PM, Ben Bacarisse wrote:
>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>
>>> Take a number that wants to get close to zero.
>> This makes no sense. "a number" is one number. And numbers don't want
>> anything.
>
> That was designed to raise a laugh or two. I guess it bombed. Yikes!

Read the room. There are lots of posts here that would be laughed at in
any maths department common room, but the posters make them in all
seriousness. In this context it's hard to make jokes.

>>> Say:
>>>
>>> [0] = 1
>>> [1] = .1
>>> [2] = .01
>>> [3] = .001
>>> [...] = [...]
>> Strange notation. [0] = 1. Eh? Why not just use a more conventional
>> notation for a s sequence:
>
> Too used to a programming language wrt indexing arrays I guess. :^)

What programming language allows you to index nothing?

>> s_0 = 1
>> s_1 = 0.1
>> etc.
>> You can, if you prefer, write it as a function: s(n) = 10^-n (as you do
>> later).
>>
>>> This gets close to zero, yet never will equal zero. Okay so:
>>>
>>> arbitrarily close seems to be the accepted term.
>> But it's not a very good one. For example, one could say that
>> p(n) = 2^n when n is even
>> p(n) = 2^-n when n is odd
>> gets arbitrarily close to zero but also arbitrarily far away from zero.
>
> That's fine with me. I can see it wrt your logic.
>
>
>>> infinitely close is the wrong wording?
>> I would not know what you mean if you said that, so I would say it's the
>> wrong wording. The best wording is to say
>> lim_{n->oo} s(n) = 0.
>> which you can read as "the limit, as n tends to infinity, if s(n) is
>> zero".
>
> The limit of f(n) = 10^(-n) is zero. However, none of the iterates equal
> zero. They just get closer and closer to it...

For all n, s(n) != 0. Is there a point in re-wording things from how a
mathematician would write it? I ask, because most cranks are just
playing language games so re-wording things in some metaphorical way can
suggest you are disputing something in conventional mathematics.

--
Ben.

On 2/18/2024 4:01 PM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>
>> On 2/16/2024 1:07 PM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> Take a number that wants to get close to zero.
>>> This makes no sense. "a number" is one number. And numbers don't want
>>> anything.
>>
>> That was designed to raise a laugh or two. I guess it bombed. Yikes!
>
> Read the room. There are lots of posts here that would be laughed at in
> any maths department common room, but the posters make them in all
> seriousness. In this context it's hard to make jokes.
>
>>>> Say:
>>>>
>>>> [0] = 1
>>>> [1] = .1
>>>> [2] = .01
>>>> [3] = .001
>>>> [...] = [...]
>>> Strange notation. [0] = 1. Eh? Why not just use a more conventional
>>> notation for a s sequence:
>>
>> Too used to a programming language wrt indexing arrays I guess. :^)
>
> What programming language allows you to index nothing?

I forgot a name for the indexing.

f[n] = 10(^n)

therefore:

f[0] = 10(-0) = 1
f[1] = 10(-1) = .1
f[2] = 10(-2) = .01
....

That is easy for me to read.

>
>>> s_0 = 1
>>> s_1 = 0.1
>>> etc.
>>> You can, if you prefer, write it as a function: s(n) = 10^-n (as you do
>>> later).
>>>
>>>> This gets close to zero, yet never will equal zero. Okay so:
>>>>
>>>> arbitrarily close seems to be the accepted term.
>>> But it's not a very good one. For example, one could say that
>>> p(n) = 2^n when n is even
>>> p(n) = 2^-n when n is odd
>>> gets arbitrarily close to zero but also arbitrarily far away from zero.
>>
>> That's fine with me. I can see it wrt your logic.
>>
>>
>>>> infinitely close is the wrong wording?
>>> I would not know what you mean if you said that, so I would say it's the
>>> wrong wording. The best wording is to say
>>> lim_{n->oo} s(n) = 0.
>>> which you can read as "the limit, as n tends to infinity, if s(n) is
>>> zero".
>>
>> The limit of f(n) = 10^(-n) is zero. However, none of the iterates equal
>> zero. They just get closer and closer to it...
>
> For all n, s(n) != 0. Is there a point in re-wording things from how a
> mathematician would write it? I ask, because most cranks are just
> playing language games so re-wording things in some metaphorical way can
> suggest you are disputing something in conventional mathematics.
>

Le 19/02/2024 à 00:12, Richard Damon a écrit :
> On 2/18/24 4:00 PM, WM wrote:
>> Le 18/02/2024 à 19:16, Jim Burns a écrit :
>>>
>>> Which do you (WM) give up?
>>> How do you justify giving it up?
>>
>> I will never give up the following self-evidence:
>> If there are ℵo unit fractions in the interval (0, eps), then there is
>> an x with only a finite number of unit fractions in (0, x).
>> Why? Because unit fractions are real points on the real line. They
>> cannot appear as an infinite swarm without a finite start.
>
> But the start was at 1/1
>
> Remember "real points" take up no space,

But unit fractions have internal distances and they take up space.

If there is a set of real points with distances at the real axis, then
every point can be considered as the border between two subsets. If it is
impossible to reduce the left-hand subset to a finite amount, then there
is no point available dividing infinitely many unit fractions. Then they
sit at one point. That is impossible by ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n
> 0 .

Regards, WM

On 2/19/24 3:14 AM, WM wrote:
> Le 19/02/2024 à 00:12, Richard Damon a écrit :
>> On 2/18/24 4:00 PM, WM wrote:
>>> Le 18/02/2024 à 19:16, Jim Burns a écrit :
>>>>
>>>> Which do you (WM) give up?
>>>> How do you justify giving it up?
>>>
>>> I will never give up the following self-evidence:
>>> If there are ℵo unit fractions in the interval (0, eps), then there
>>> is an x with only a finite number of unit fractions in (0, x).
>>> Why? Because unit fractions are real points on the real line. They
>>> cannot appear as an infinite swarm without a finite start.
>>
>> But the start was at 1/1
>>
>> Remember "real points" take up no space,
>
> But unit fractions have internal distances and they take up space.

But a space that gets vanishingly small, and thus we CAN fit an infinite
number of them in a finite space.

>
> If there is a set of real points with distances at the real axis, then
> every point can be considered as the border between two subsets. If it
> is impossible to reduce the left-hand subset to a finite amount, then
> there is no point available dividing infinitely many unit fractions.
> Then they sit at one point. That is impossible by ∀n ∈ ℕ: 1/n - 1/(n+1)
> = d_n

Nope. False conclusion. Why must the left side every become finite?

>> 0 .
>
> Regards, WM
>
>
>

I guess you agree that Achilles can't pass the Tortoise, or maybe only
at some "dark" time that we can not see.

Le 19/02/2024 à 13:33, Richard Damon a écrit :
> On 2/19/24 3:14 AM, WM wrote:
>> Le 19/02/2024 à 00:12, Richard Damon a écrit :
>>> On 2/18/24 4:00 PM, WM wrote:
>>>> Le 18/02/2024 à 19:16, Jim Burns a écrit :
>>>>>
>>>>> Which do you (WM) give up?
>>>>> How do you justify giving it up?
>>>>
>>>> I will never give up the following self-evidence:
>>>> If there are ℵo unit fractions in the interval (0, eps), then there
>>>> is an x with only a finite number of unit fractions in (0, x).
>>>> Why? Because unit fractions are real points on the real line. They
>>>> cannot appear as an infinite swarm without a finite start.
>>>
>>> But the start was at 1/1
>>>
>>> Remember "real points" take up no space,
>>
>> But unit fractions have internal distances and they take up space.
>
> But a space that gets vanishingly small,

Bot by the number of points - there are always infinitely many between two
adjacent unit fractions.

>> If there is a set of real points with distances at the real axis, then
>> every point can be considered as the border between two subsets. If it
>> is impossible to reduce the left-hand subset to a finite amount, then
>> there is no point available dividing infinitely many unit fractions.
>> Then they sit at one point. That is impossible by ∀n ∈ ℕ: 1/n - 1/(n+1)
>> = d_n
>
> Why must the left side every become finite?

If there are teally existing real points, then each one can be used, in
principle, as the border.
>
> I guess you agree that Achilles can't pass the Tortoise, or maybe only
> at some "dark" time that we can not see.

Irrelevant for the present topic. But true that he overtakes in darkness.

Regards, WM

On 02/19/2024 04:33 AM, Richard Damon wrote:
> On 2/19/24 3:14 AM, WM wrote:
>> Le 19/02/2024 à 00:12, Richard Damon a écrit :
>>> On 2/18/24 4:00 PM, WM wrote:
>>>> Le 18/02/2024 à 19:16, Jim Burns a écrit :
>>>>>
>>>>> Which do you (WM) give up?
>>>>> How do you justify giving it up?
>>>>
>>>> I will never give up the following self-evidence:
>>>> If there are ℵo unit fractions in the interval (0, eps), then there
>>>> is an x with only a finite number of unit fractions in (0, x).
>>>> Why? Because unit fractions are real points on the real line. They
>>>> cannot appear as an infinite swarm without a finite start.
>>>
>>> But the start was at 1/1
>>>
>>> Remember "real points" take up no space,
>>
>> But unit fractions have internal distances and they take up space.
>
> But a space that gets vanishingly small, and thus we CAN fit an infinite
> number of them in a finite space.
>
>>
>> If there is a set of real points with distances at the real axis, then
>> every point can be considered as the border between two subsets. If it
>> is impossible to reduce the left-hand subset to a finite amount, then
>> there is no point available dividing infinitely many unit fractions.
>> Then they sit at one point. That is impossible by ∀n ∈ ℕ: 1/n -
>> 1/(n+1) = d_n
>
> Nope. False conclusion. Why must the left side every become finite?
>
>>> 0 .
>>
>> Regards, WM
>>
>>
>>
>
> I guess you agree that Achilles can't pass the Tortoise, or maybe only
> at some "dark" time that we can not see.

How about if you take the turtle's velocity and subtract it
from Achilles' velocity, then just compute Achilles' time-to-travel
to the Tortoise, and add the Tortoise's progress, to where they meet.

Though, if the Tortoise is 0 m/s, that's infinity s/m, ....

Of course it assumes that the Arrow already has a
geometric series that has a sum that equals one, ....

Adding a "postulate of continuity" to geometry was
deemed the needful about a millenia or two after Euclid.

Again this is just the Sorites/Heap, then as with regards
to the "infinitely-divided" of the "infinitely divisible",
what you want is not samples of the "infinitely divisible",
that will never end, instead the "continuum limit" of
the function that is the "infinite limit" of the divisions.

Then the values in front start with finitely many instead,
and it's just either side of that "continuous" and "discrete",
it's about the most fundamental concept relating together
the "continuous" and "discrete", in one relation.

It is, ....

mitchr...@gmail.com wrote:

> On Saturday, February 17, 2024 at 2:00:16 AM UTC-8, WM wrote:
>> because ∀n ∈ ℕ: 1/n =/= 1/(n-1).
>> This requires a first unit fraction, if all are there in actual
>> infinity. Of course the first unit fractions cannot be seen. They are
>> dark. Regards, WM
>
> The first fraction is 1/infinity.That is dark. Zero is below it and
> can't be seen. It is not even dark.

yes, I can see that. Fuck you amrica. You are the dirt at the bottom of
the dirt. Dirty 𝗹𝗶𝗯𝗲𝗿𝗮𝗹_𝗰𝗮𝗽𝗶𝘁𝗮𝗹𝗶𝘀𝘁 bitches. You are going to suck large dicks, if
the morons of amrica are not waking up, very fast.

𝗛𝘂𝗻𝗴𝗮𝗿𝘆_𝘀𝗻𝘂𝗯𝘀_𝗨𝗦_𝘀𝗲𝗻𝗮𝘁𝗼𝗿𝘀_–_𝗮𝗺𝗯𝗮𝘀𝘀𝗮𝗱𝗼𝗿
A bipartisan delegation had sought to discuss Sweden’s NATO bid with
senior officials in Budapest
https://r%74.com/news/592675-hungary-boycotts-us-senators/

lol

Senior Hungarian officials have refused to meet four US senators who
arrived in Budapest on Sunday, Washington’s envoy to the country has said.
The American lawmakers are attempting to press Prime Minister Viktor Orban
into speeding up approval of Sweden’s accession to NATO.

Hungary is right to not meet the US war criminals.

The Mafia enforcers have arrived on tourist visas, not as an invited
political delegation. Wise to rebuff them.

fuck you amrica, a stolen territory ruled by 𝗸𝗵𝗮𝘇𝗮𝗿_𝗴𝗼𝘆𝘀. You are promoting,
committing and supporting genocide on planet Earth.

Refusing to receive US bullies seems to be trending.👍

"called the boycott “strange and concerning”" .. They might want to get
used to it. The US is the old kid in town. Even the US citizens just laugh
at their politicians.

On 2/18/2024 4:00 PM, WM wrote:
> Le 18/02/2024 à 19:16, Jim Burns a écrit :

>> One of these claim must be given up:
>> | a final.ordinal.reciprocal.free zone (0,δ)
>> | exists
>> or
>> | a skipping.function isn't all.continuous
>> or
>> | for final.ordinal.reciprocal ⅟m
>> | ⅟(4⋅m) is a final.ordinal.reciprocal.

>> Which do you (WM) give up?
>> How do you justify giving it up?
>
> I will never give up the following self-evidence:
> If
> there are ℵo unit fractions in
> the interval (0, eps),
> then
> there is an x with only a finite number of
> unit fractions in (0, x).
> Why?
> Because unit fractions are real points on
> the real line.
> They cannot appear as
> an infinite swarm without a finite start.

For each ⅟j in an uninterrupted sequence

each final.ordinal fits leftward
not.each final.ordinal fits rightward

In other words,
infinitely.many are leftward
finitely.many are rightward
for each ⅟j

No ⅟j is in a finite start
That is the reason that
an infinite swarm is the only possibility

----
A finite ordinal is final:
It is the last which has its cardinality.

Ordinal.finality is determined by not.fitting:
If
Bob is inserted into {<j} the ordinals before final j
then
{<j}⁺ᴮᵒᵇ not.fits {<j}
{<j} ⃒⇇ {<j}⁺ᴮᵒᵇ
not.exists 1.to.1 map to {<j} from {<j}⁺ᴮᵒᵇ

Visibleᵂᴹ or darkᵂᴹ,
not.exists final‖not.final j‖j⁺¹

If
j⁺¹ is not.final and
exists 1.to.1 map g to {<j⁺¹} from {<j⁺¹}⁺ᴮᵒᵇ
then
g can be edited[1] to
1.to.1 map f to {<j} from {<j}⁺ᴮᵒᵇ
and j is also not.final.

[1]
if g(i)=j⁺¹
then define f(i)=g(j⁺¹)
else define f(i)=g(i)

That is normal,
if it is normal to be an abstraction.
I speculate that
your thinking in your yet.unstated argument
leans on the normal.ness of that.

----
A set S⁺ᴮᵒᵇ with Bob inserted which
does not fit into S
is a _normal_ (finite) set S
normal S ⟺ S ⃒⇇ S⁺ᴮᵒᵇ

For each normal set S
exists a final ordinal j such that
S fits into {<j}
(normal) S ⃒⇇ S⁺ᴮᵒᵇ
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
∃j: S ⇉ {<j} ⃒⇇ {<j}⁺ᴮᵒᵇ

Not.fitting is essential to counting.

----
Define ℕ as the set of final ordinals.
∀j: j ∈ ℕ ⟺ {<j} ⃒⇇ {<j}⁺ᴮᵒᵇ

(normal) S ⃒⇇ S⁺ᴮᵒᵇ
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
∃j ∈ ℕ: S ⇉ {<j}

ℕ isn't normal.

For each j ∈ ℕ
exists {<j⁺¹} ⊆ ℕ: {<j} ⃒⇇ {<j⁺¹}

For each j ∈ ℕ
{<j} ⃒⇇ ℕ

| Assume {<j} ⇇ ℕ
| | {<j} ⇇ ℕ
| {<j} ⇇ ℕ ⇇ {<j⁺¹}
| {<j} ⇇ {<j⁺¹}
| | However,
| {<j} ⃒⇇ {<j⁺¹}
| Contradiction.

¬∃j ∈ ℕ: ℕ ⇉ {<j}
¬(normal ℕ)
¬(ℕ ⃒⇇ ℕ⁺ᴮᵒᵇ)
ℕ⁺ᴮᵒᵇ ⇉ ℕ

On 2/19/2024 7:33 AM, Richard Damon wrote:
> On 2/19/24 3:14 AM, WM wrote:
>> Le 19/02/2024 à 00:12, Richard Damon a écrit :
>>> On 2/18/24 4:00 PM, WM wrote:
>>>> Le 18/02/2024 à 19:16, Jim Burns a écrit :

>>>>> Which do you (WM) give up?
>>>>> How do you justify giving it up?
>>>>
>>>> I will never give up the following self-evidence:
>>>> If there are
>>>> ℵo unit fractions in the interval (0, eps),
>>>> then there is an x with only a finite number of
>>>> unit fractions in (0, x).
>>>> Why? Because unit fractions are real points on
>>>> the real line.
>>>> They cannot appear as an infinite swarm without
>>>> a finite start.
>>>
>>> But the start was at 1/1
>>>
>>> Remember "real points" take up no space,
>>
>> But unit fractions have internal distances
>> and they take up space.
>
> But a space that gets vanishingly small,
> and thus
> we CAN fit an infinite number of them in
> a finite space.

One can avoid using the I.word by saying
| we CAN fit more than any finite number of them in
| a finite space.

On 2/19/24 7:59 AM, WM wrote:
> Le 19/02/2024 à 13:33, Richard Damon a écrit :
>> On 2/19/24 3:14 AM, WM wrote:
>>> Le 19/02/2024 à 00:12, Richard Damon a écrit :
>>>> On 2/18/24 4:00 PM, WM wrote:
>>>>> Le 18/02/2024 à 19:16, Jim Burns a écrit :
>>>>>>
>>>>>> Which do you (WM) give up?
>>>>>> How do you justify giving it up?
>>>>>
>>>>> I will never give up the following self-evidence:
>>>>> If there are ℵo unit fractions in the interval (0, eps), then there
>>>>> is an x with only a finite number of unit fractions in (0, x).
>>>>> Why? Because unit fractions are real points on the real line. They
>>>>> cannot appear as an infinite swarm without a finite start.
>>>>
>>>> But the start was at 1/1
>>>>
>>>> Remember "real points" take up no space,
>>>
>>> But unit fractions have internal distances and they take up space.
>>
>> But a space that gets vanishingly small,
>
> Bot by the number of points - there are always infinitely many between
> two adjacent unit fractions.
>

So?

Your mind just doesn't seem to be able to understand that fact.

>>> If there is a set of real points with distances at the real axis,
>>> then every point can be considered as the border between two subsets.
>>> If it is impossible to reduce the left-hand subset to a finite
>>> amount, then there is no point available dividing infinitely many
>>> unit fractions. Then they sit at one point. That is impossible by ∀n
>>> ∈ ℕ: 1/n - 1/(n+1) = d_n
>>
>> Why must the left side every become finite?
>
> If there are teally existing real points, then each one can be used, in
> principle, as the border.

No, only the one ON the border would be the border, but for every one
that you might want to think of as the border, there are more that are
closer to the border.

Thus, there is NOT one (in the set) on the border, and thus there is no
"first" (from the left) unit fraction, and thus NUF(x) isn't defined.

>>
>> I guess you agree that Achilles can't pass the Tortoise, or maybe only
>> at some "dark" time that we can not see.
>
> Irrelevant for the present topic. But true that he overtakes in darkness.

But he doesn't, he passes the Tortoise at a easily determined finite
time (if we know the actual speeds).

This shows that to you ALL numbers have become "dark", because you can't
actually use any of them without hitting a problem.

>
> Regards, WM
>
>

Le 19/02/2024 à 21:21, Jim Burns a écrit :
> On 2/18/2024 4:00 PM, WM wrote:
>> I will never give up the following self-evidence:
>> If
>> there are ℵo unit fractions in
>> the interval (0, eps),
>> then
>> there is an x with only a finite number of
>> unit fractions in (0, x).
>> Why?
>> Because unit fractions are real points on
>> the real line.
>> They cannot appear as
>> an infinite swarm without a finite start.

> In other words,
> infinitely.many are leftward
> finitely.many are rightward
> for each ⅟j

That correctly describes an evolving infinite collection, i.e., a
potentially infinite set where more and more elements are created which
initially have not existed.
A complete, i.e. actually infinite set of ℵo real fixed points on the
real axis can be subdivided by any of its elements (since all are
existing) such that the subsets have cardinalities from 0, ℵo over n,
ℵo to ℵo, n, and ℵo, 0.
>
> No ⅟j is in a finite start
> That is the reason that
> an infinite swarm is the only possibility

Then you are not talking about really existing invariable points. But that
is what I discuss.

Regards, WM

Le 19/02/2024 à 21:37, Jim Burns a écrit :

> One can avoid using the I.word by saying
> | we CAN fit more than any finite number of them in
> | a finite space.

But if all are there existing from the scratch as an actually infinite
set, then each one can be addressed as border between two subsets, in
principle. Even the smallest one.

Regards, WM

Le 20/02/2024 à 03:17, Richard Damon a écrit :
> On 2/19/24 7:59 AM, WM wrote:

Measured by the number of points - there are always infinitely many
between two adjacent unit fractions.

>>>> If there is a set of real points with distances at the real axis,
>>>> then every point can be considered as the border between two subsets.
>>>> If it is impossible to reduce the left-hand subset to a finite
>>>> amount, then there is no point available dividing infinitely many
>>>> unit fractions. Then they sit at one point. That is impossible by ∀n
>>>> ∈ ℕ: 1/n - 1/(n+1) = d_n
>>>
>>> Why must the left side every become finite?
>>
>> If there are really existing real points, then each one can be used, in
>> principle, as the border.
>
> No, only the one ON the border would be the border, but for every one
> that you might want to think of as the border, there are more that are
> closer to the border.

That means they appear onlylater. That is potential infinity.
>
> Thus, there is NOT one (in the set) on the border, and thus there is no
> "first" (from the left) unit fraction, and thus NUF(x) isn't defined.

It is very well defined.

Further, if
∃^ℵ y ∈ {1/n : n ∈ ℕ} ∀x ∈ (0, 1]: 0 < y < x
is false, then there must be x > 0 which make it false, i.e., which have
fewer smaller unit fractions. Therefore
∀x ∈ (0, 1]: ∃^ℵ y ∈ {1/n : n ∈ ℕ}: 0 < y < x
cannot be true for all x > 0.
>
>>> I guess you agree that Achilles can't pass the Tortoise, or maybe only
>>> at some "dark" time that we can not see.
>>
>> Irrelevant for the present topic. But true that he overtakes in darkness.
>
> But he doesn't, he passes the Tortoise at a easily determined finite
> time (if we know the actual speeds).

Yes, the limit is well known, like the limit 0 of the unit fractions or
the limit ω of the natural numbers.

Regards, WM

On 2/20/24 3:24 AM, WM wrote:
> Le 20/02/2024 à 03:17, Richard Damon a écrit :
>> On 2/19/24 7:59 AM, WM wrote:
>
> Measured by the number of points - there are always infinitely many
> between two adjacent unit fractions.
>
>>>>> If there is a set of real points with distances at the real axis,
>>>>> then every point can be considered as the border between two
>>>>> subsets. If it is impossible to reduce the left-hand subset to a
>>>>> finite amount, then there is no point available dividing infinitely
>>>>> many unit fractions. Then they sit at one point. That is impossible
>>>>> by ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n
>>>>
>>>> Why must the left side every become finite?
>>>
>>> If there are really existing real points, then each one can be used,
>>> in principle, as the border.
>>
>> No, only the one ON the border would be the border, but for every one
>> that you might want to think of as the border, there are more that are
>> closer to the border.
>
> That means they appear onlylater. That is potential infinity.

So, it wasn't on the border.

You don't seem to understand the meaning of the terms you use.

>>
>> Thus, there is NOT one (in the set) on the border, and thus there is
>> no "first" (from the left) unit fraction, and thus NUF(x) isn't defined.
>
> It is very well defined.

The WORDS have meaning. But the function isn't defined.

By YOUR logic, the square root of two can be shown to be rational.

Let us define Qn(x) to be the numerator of the simplest form of the
rational number that when squared equals x, and Qd(x) to be the
denominator of that same rational number that when squared equals x.

By DEFINITION, both Qn(x) and Qd(x) are natural numbers.

Thus Qn(2) / Qd(2) will be equal to the square root of 2, and has
expressed its value as the ratio of natural numbers, thus the square
root of two is rational.

>
> Further, if
> ∃^ℵ y ∈ {1/n : n ∈ ℕ} ∀x ∈ (0, 1]: 0 < y < x
> is false, then there must be x > 0 which make it false, i.e., which have
> fewer smaller unit fractions. Therefore
> ∀x ∈ (0, 1]: ∃^ℵ y ∈ {1/n : n ∈ ℕ}: 0 < y < x
> cannot be true for all x > 0.

You are just using broken logic.

You can't just swap the order of qualifiers like that.

>>
>>>> I guess you agree that Achilles can't pass the Tortoise, or maybe
>>>> only at some "dark" time that we can not see.
>>>
>>> Irrelevant for the present topic. But true that he overtakes in
>>> darkness.
>>
>> But he doesn't, he passes the Tortoise at a easily determined finite
>> time (if we know the actual speeds).
>
> Yes, the limit is well known, like the limit 0 of the unit fractions or
> the limit ω of the natural numbers.

But for Achilles, the limit is in the set of real values in the domain
of discourse. Not like 0 for the positive numbers, or infinity for the
finite numbers.

>
> Regards, WM

On 2/20/24 3:15 AM, WM wrote:
> Le 19/02/2024 à 21:37, Jim Burns a écrit :
>
>> One can avoid using the I.word by saying
>> | we CAN fit more than any finite number of them in
>> | a finite space.
>
> But if all are there existing from the scratch as an actually infinite
> set, then each one can be addressed as border between two subsets, in
> principle. Even the smallest one.
> Regards, WM

But there isn't a "Smallest One".

You just don't understand that fact, because you mind is just to filled
with Darkness.