On 5/21/2021 4:59 PM, WM wrote:
> Jim Burns schrieb
> am Freitag, 21. Mai 2021 um 19:26:24 UTC+2:

>> Otherwise, we would contradict
>> | Two continuous curves which cross must intersect.
>
> Digit sequences are not curves.

Any infinite digit sequence
(i) determines no more than one point in the geometric line, and
(ii) determines no less than one point in the geometric line.

You didn't realize that we have been talking about a
geometric line? How unexpected.

----
For any infinite digit sequence,
a finite initial segment of digits of length k partitions
the geometric line into three parts, L[k], H[k], and D[k].

A point x is in L[k] iff it is too _low_ for it to be possible
that the whole infinite digit sequence determines x.

For example, no matter how an infinite digit sequence beginning
1.4142 continues, any point x < 1.4142 cannot possibly be
determined by the whole sequence.

A point x is in H[k] iff it is too _high_ for it to be possible
that the whole infinite digit sequence determines x.

A point x is in D[k] iff it is not in L[k] or in H[k].
In general, D[k] is a line segment of length 1/10^k.

For any infinite digit sequence,
_all_ the finite initial segments of digits jointly partition
the geometric line into three parts, L, H, and D.

A point x is in L iff a finite initial segment of digits
of length k exists such that x is in L[k].

A point x is in H iff a finite initial segment of digits
of length k exists such that x is in H[k].

A point x is in D iff x is not in L or in H.

We will say of a point x in D that the infinite digit sequence
_determines_ x.

This does not say, so far, whether there is exactly one point
in D, or infinitely many points in D, or no points in D.
It only establishes what I mean, what is widely meant, by
"This infinite digit sequence determines that point."

> Stop your futile attempts to prove nonsense.

Any digit sequence _with or without_ a defining formula
_determines_ not less than one point.
Keep in mind what "determines" means here.

Otherwise, we would contradict
| Two continuous curves which cross must intersect.

----
Assume an example exists of an infinite digit sequence
which does not determine any point.

No points are in D.
All points are in either L or H.

Define f(x) such that
f(x) = 0 if x is in L
f(x) = 1 if x is in H

f(x) is not defined for points in D, but, by assumption,
there are no points in D. f(x) is defined everywhere.

For each too-low point x in L, there are too-low points y
to its left and its right, and f(y) = 0 for all too-low points.
f(x) is continuous at x, for each too-low point x in L.

Similarly,
f(x) is continuous at x, for each too-high point x in H.

f(x) is continuous everywhere.

Define g(x) such that g(x) = 1/2.
g(x) is defined everywhere and continuous everywhere.

f(x) and g(x) are continuous.
f(x) and g(x) cross.
f(x) and g(x) do not intersect.

That contradicts
| Two continuous curves which cross must intersect.

Therefore,
| if two continuous curves which cross must intersect,
| then any infinite digit sequence with or without a
| defining formula defines not less than one point.

> Try nonsense, namely to determine a real number
> by an infinite digit sequence.
> Then you will easier comprehend that it is nonsense.

Do you have an example of an infinite digit sequence that
determines less than one point?

No, you don't. I showed you why.

>> Assume an example exists of an infinite digit sequence
>> such that
>
> No, I am tired of your trash. Do what I said.

On 5/21/2021 4:21 PM, FromTheRafters wrote:
> WM expressed precisely :
>> Jim Burns schrieb am Freitag, 21. Mai 2021 um 19:26:24 UTC+2:
>>
>>
>>> Otherwise, we would contradict
>>>> Two continuous curves which cross must intersect.
>>
>> Digit sequences are not curves.
>> Stop your futile attempts to prove nonsense.
>> Try nonsense, namely to determine a real number by an infinite digit
>> sequence. Then you will easier comprehend that it is nonsense.
>>>
>>> Assume an example exists of an infinite digit sequence such that
>>
>> No, I am tired of your trash. Do what I said.
>
> Ah, comedy.

....and a nice change too!
Bet he comprehends that nonsense now!

Jim Burns schrieb am Samstag, 22. Mai 2021 um 02:09:12 UTC+2:

> Do you have an example of an infinite digit sequence that
> determines less than one point?

An undestructible random number generator generates such a digit sequence. But before it has finished, what never will happen, there is no result.

Regards, WM

William schrieb am Freitag, 21. Mai 2021 um 23:57:31 UTC+2:
> On Friday, May 21, 2021 at 11:59:47 AM UTC-3, WM wrote:

> > The definable naturals are well-ordered up to every k
> Yes, the set |N (the set that you call the definable naturals) is well ordered up to every k. This follows trivially from the fact that the Peano set |N is well ordered.

But hitherto you did not distunguish between completeness of infinite sets and their well-order. Cantor simply assumed that if all elements exist, then they all are well-ordered. He argued: "es erfährt daher der aus unsrer Regel resultierende Zuordnungsprozeß keinen Stillstand." [Cantor, gesammelte Werke, p. 239]. (There is no obstacle in the process of mapping.)

This argument has turned out insufficicient. When we are counting the unit fractions 1/1, 1/2, 1/3, ..., then there is no obstacle and no end, but we never reach zero. At every step ℵo unit fractions are before us and only finitely many are behind us. From this fact we can conclude that always ℵo unit fractions will be before us that means that always ℵo unit fractions will be before us that means that always ℵo unit fractions will be before us and after a while of pondering about that we see that ℵo unit fractions cannot be removed or counted. If ℵo exist, what we assume in order to have no gaps in the interval (0, 1] then they are existing but cannot be well-ordered I call that they are dark. They accomplish the complete covering of the interval (0, 1] such that there is not the least gap but nevertheless no last unit fraction can be recognized.

Example Zenos paradox: The tortoise leads at every step but after infinitely many steps Achilles has overtaken. That is impossible without dark steps. Therefore: If we can divide their way into infinitely many steps, then dark steps are necessary. The alternative is: We cannot divide their way so far.

Regards, WM

On Saturday, May 22, 2021 at 8:03:57 AM UTC-3, WM wrote:
> William schrieb am Freitag, 21. Mai 2021 um 23:57:31 UTC+2:
> > On Friday, May 21, 2021 at 11:59:47 AM UTC-3, WM wrote:
>
> > > The definable naturals are well-ordered up to every k
> > Yes, the set |N (the set that you call the definable naturals) is well ordered up to every k. This follows trivially from the fact that the Peano set |N is well ordered.
> But hitherto you did not distunguish between completeness of infinite sets and their well-order.

So? An infinite set may or may not be well-ordered. For example, the rational numbers are infinite but not (under the usual order) well-ordered. The rational numbers are complete because every rational number is an element of the rationals. |N (the set that you call the definable naturals) is well-ordered and, since every definable natural is an element of |N, |N is complete.

--
William Hughes

William schrieb am Samstag, 22. Mai 2021 um 17:21:36 UTC+2:
> On Saturday, May 22, 2021 at 8:03:57 AM UTC-3, WM wrote:
> > William schrieb am Freitag, 21. Mai 2021 um 23:57:31 UTC+2:
> > > On Friday, May 21, 2021 at 11:59:47 AM UTC-3, WM wrote:
> >
> > > > The definable naturals are well-ordered up to every k
> > > Yes, the set |N (the set that you call the definable naturals) is well ordered up to every k. This follows trivially from the fact that the Peano set |N is well ordered.
> > But hitherto you did not distunguish between completeness of infinite sets and their well-order.
> So? An infinite set may or may not be well-ordered. For example, the rational numbers are infinite but not (under the usual order) well-ordered. The rational numbers are complete because every rational number is an element of the rationals. |N (the set that you call the definable naturals) is well-ordered and, since every definable natural is an element of |N, |N is complete.

Have you read and understood my text?

Regards, WM

On 5/22/2021 6:03 AM, WM wrote:
> William schrieb am Freitag, 21. Mai 2021 um 23:57:31 UTC+2:
>> On Friday, May 21, 2021 at 11:59:47 AM UTC-3, WM wrote:
>
>>> The definable naturals are well-ordered up to every k
>> Yes, the set |N (the set that you call the definable naturals) is well ordered up to every k. This follows trivially from the fact that the Peano set |N is well ordered.
>
> But hitherto you did not distunguish between completeness of infinite sets and their well-order. Cantor simply assumed that if all elements exist, then they all are well-ordered. He argued: "es erfährt daher der aus unsrer Regel resultierende Zuordnungsprozeß keinen Stillstand." [Cantor, gesammelte Werke, p. 239]. (There is no obstacle in the process of mapping.)

no. translation is "therefore the assignment process resulting from our
rule does not come to a standstill."

>
> This argument has turned out insufficicient. When we are counting the unit fractions 1/1, 1/2, 1/3, ..., then there is no obstacle and no end, but we never reach zero.

that is what Cantor said!

>At every step ℵo unit fractions are before us and only finitely many are behind us. From this fact we can conclude that always ℵo unit fractions will be before us that means that always ℵo unit fractions will be before us that means that always ℵo unit fractions will be before us and after a while of pondering about that we see that ℵo unit fractions cannot be removed or counted.

You keep repeating this stuff, as if you do not understand it.

it is trivial. oo - k = oo for k finite

> If ℵo exist, what we assume in order to have no gaps in the interval (0, 1] then they are existing but cannot be well-ordered

you need to show that unit fractions are not well ordered.
1/1, 1/2, 1/3, 1/4,...

> I call that they are dark. They accomplish the complete covering of the interval (0, 1] such that there is not the least gap but nevertheless no last unit fraction can be recognized.

you stop in your sequence at k, and say the rest are dark, when Cantor
says do not stop.

>
> Example Zenos paradox: The tortoise leads at every step but after infinitely many steps Achilles has overtaken. That is impossible without dark steps.

how and why are your dark steps needed.

Therefore: If we can divide their way into infinitely many steps, then
dark steps are necessary. The alternative is: We cannot divide their way
so far.
>
> Regards, WM
>

Meritocracy schrieb am Samstag, 22. Mai 2021 um 22:30:15 UTC+2:
> On 5/22/2021 6:03 AM, WM wrote:

> > Example Zenos paradox: The tortoise leads at every step but after infinitely many steps Achilles has overtaken. That is impossible without dark steps.
> how and why are your dark steps needed.

Because visible steps are not reaching the point where Achilles overtakes.

> Therefore: If we can divide their way into infinitely many steps, then
> dark steps are necessary. The alternative is: We cannot divide their way
> so far.

Yes, but that kills set theory.

Regards, WM

On 5/22/2021 6:43 AM, WM wrote:
> Jim Burns schrieb
> am Samstag, 22. Mai 2021 um 02:09:12 UTC+2:

>> Do you have an example of an infinite digit sequence
>> that determines less than one point?
>
> An undestructible random number generator generates
> such a digit sequence.
> But before it has finished, what never will happen,
> there is no result.

Sorry, that appears to be a sincere attempt. But, no.

In order to be an example of an infinite digit sequence
with the requested property, that would need to be an
infinite digit sequence with-or-without that property.
You have not described an infinite digit sequence.

I gave some partial description of an infinite digit
sequence, back when I first decided to address your question
about infinite digit sequences.
( Maybe you remember that post. You complained about how
( long it was.

An infinite digit sequence _which I am talking about_
is a crowd of ordered pairs (d,n) such that
d is one of 0,1,2,...,9
n is a _countable-to_ and _countable-beyond_ index
for all countable-to and countable-beyond indices n,
one and only one d in 0,1,...,9 exists such that
(d,n) is in the infinite-digit-sequence/crowd-of-pairs.

Consider an indestructible random number generator.
But before it has finished, which never will happen,
there is no infinite digit sequence. Not any infinite digit
sequence about which I am making claims.

This is what I am doing, this is how I am arguing:
-- Make true statements about one of infinitely-many.
-- Preserve truth.
-- Make more true statements about the same infinitely-many.

A thing that we aren't making a claim about can't be a
counter-example to a claim we are making.
If I say "All crows are black",
you would be correct to say, "This canary is yellow",
but that is not a _counter-example_

Your indestructible generator before it has finished
is a yellow canary trying to be evidence that
not all crows are black.

----
I am reminded of True Scotsmen.

<wiki>
| | Person A: "No Scotsman puts sugar on his porridge."
| | Person B: "But my uncle Angus is a Scotsman and
| he puts sugar on his porridge."
| | Person A: "But no _true_ Scotsman puts sugar on
| his porridge."

I am taking the role of Person A here.
No _true_ infinite digit sequence is missing (d,n)
for any countable-to and countable-beyond index.

And, in that bit of dialog, Person A is committing a
fallacy, the No True Scotsman fallacy. However, I am not.

In the wikipage, "no sugar on his porridge" is an
arbitrary addition to _what we mean_ by "Scotsman".
If we allowed that, who knows what else we would need
to agree to?

However, I am not making arbitrary additions to
_what we mean_ by "infinite digit sequence". I told you
what I was talking about way back when I decided to
address your question. And I haven't changed it since.

It seems to me that your indestructible generator before
it has finished is an arbitrary _expansion_ of what we
are talking about, then inverse of the arbitrary
_contraction_ of what we are talking about which is the
No-True-Scotsman fallacy. I'd classify it as _equivocation_

Jim Burns schrieb am Sonntag, 23. Mai 2021 um 19:54:19 UTC+2:
> On 5/22/2021 6:43 AM, WM wrote:
> > Jim Burns schrieb
> > am Samstag, 22. Mai 2021 um 02:09:12 UTC+2:
>
> >> Do you have an example of an infinite digit sequence
> >> that determines less than one point?
> >
> > An undestructible random number generator generates
> > such a digit sequence.
> > But before it has finished, what never will happen,
> > there is no result.
> Sorry, that appears to be a sincere attempt. But, no.

Infinite means without end.
>
> Consider an indestructible random number generator.
> But before it has finished, which never will happen,
> there is no infinite digit sequence.

That is the typical property of infinite digit sequences.

> Not any infinite digit
> sequence about which I am making claims.

Then that sequence and your claims about it are pure nonsense. Therefore I do not bother to read these claims.

Regards, WM

On 5/23/2021 5:14 PM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 23. Mai 2021 um 19:54:19 UTC+2:
>> On 5/22/2021 6:43 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 22. Mai 2021 um 02:09:12 UTC+2:

>>>> Do you have an example of an infinite digit sequence
>>>> that determines less than one point?
>>>
>>> An undestructible random number generator generates
>>> such a digit sequence.
>>> But before it has finished, what never will happen,
>>> there is no result.
>>
>> Sorry, that appears to be a sincere attempt. But, no.
>
> Infinite means without end.

Perhaps you have a reason for pointing that out.

----
Finite crowds/collections/sets can be processed
(in principle) by finite beings
-- in the usual sense of "process".

That is a stronger condition than merely having an end.

We can process (count to) each natural number.
That's part of what we mean by natural number.

We cannot process every natural number.
That's another part of what we mean by natural number.

This can't be changed by adding an end (omega) to
the natural numbers. In order to process 0,1,2,3,...,omega,
one must be able to process 0,1,2,3,... too.

This is what I mean by completable.
A linearly-ordered crowd pp is completable iff
each non-empty sub-crowd qq of pp contains a first and a last.

This view of finiteness is due to is Paul Stäckel,
so far as I am aware.

>> Consider an indestructible random number generator.
>> But before it has finished, which never will happen,
>> there is no infinite digit sequence.
>
> That is the typical property of infinite digit sequences.
>
>> Not any infinite digit
>> sequence about which I am making claims.
>
> Then that sequence and your claims about it are pure nonsense.
> Therefore I do not bother to read these claims.
>
>
> Regards, WM
>