FromTheRafters schrieb am Dienstag, 18. Mai 2021 um 15:41:03 UTC+2:
> on 5/18/2021, WM supposed :
> > Meritocracy schrieb am Montag, 17. Mai 2021 um 18:51:28 UTC+2:
> >
> >
> >> no knowledge of recursion, nor using limits...
> >
> > Recursion proves that every defined natnumber belongs to a FISON and has ℵo
> > successors.
> ITYM "induction" not recursion.

Not real a difference.
A recurrence relation is an equation that expresses each element of a sequence as a function of the preceding ones.
Induction expresses each element of a sequence as a function of the preceding one.

> Besides, you have *each* natural number
> having *one* successor then you need induction to *prove* the base case
> applies to *all*.

This case applies to all definable natnumbers n:

|ℕ \ (1, 2, 3, ..., n)| = ℵo.

> > Using a limit is not processing all elements individually but it
> > is processing them collectively.
> ITYM *induction* rather than *limit* here as well.

No.

Regards, WM

Jim Burns schrieb am Dienstag, 18. Mai 2021 um 17:17:18 UTC+2:

> The sequence determines a unique point, just not the way
> you want it to be determined. Get over it. Accept reality.

Gladly. Give a real example which describes a unique point. (Here we talk about sequences of digits without defining formula.)

Regards, WM

Meritocracy schrieb am Dienstag, 18. Mai 2021 um 19:53:02 UTC+2:
> On 5/16/2021 3:32 PM, WM wrote:
> > A formula can define an infinite digit sequence and its limit. But the digit sequences cannot do so.
> nope 0.999... = 1

"0.999..." is a formula. An infinite digit sequence has infinitely many digits, not only three or four.

Regards, WM

William schrieb am Dienstag, 18. Mai 2021 um 17:54:18 UTC+2:
> On Tuesday, May 18, 2021 at 8:59:52 AM UTC-3, WM wrote:
> > William schrieb am Montag, 17. Mai 2021 um 18:51:23 UTC+2:
> > > On Monday, May 17, 2021 at 12:19:34 PM UTC-3, WM wrote:
> > > > William schrieb am Montag, 17. Mai 2021 um 00:11:02 UTC+2:
> > > > > The fact that every element of a set is *definable* does not mean that there is a set of *defined* elements that contains every element.
> > > > That is wrong.
> > > No. It is easy to see that there is a infinite set of *definable* elements, but we can only chose a finite set to actually *define*
> > You cannot choose any natnumber with less than ℵo successors.
> Correct.

Therefore it is wrong when you say that you can choose all such that none remains as you can do collectively.

> Every element of the chosen finite (thus not containing every natural number) defined set has an infinite number of successors (a triviality). So what? You can chose any finite subset, but the subset you choose must be finite.

Yes, you can choose any FISON. Then its n and all smaller numbers are regarded as defined. But you cannot choose any number such that nothing remains. This however can be done collectively. Therefore ther exists a difference.

ℕ is larger than all FISONs:

|ℕ \ (1, 2, 3, ..., n)| = ℵo

much, much larger.

Regards, WM

Meritocracy schrieb am Dienstag, 18. Mai 2021 um 19:56:14 UTC+2:
> On 5/18/2021 7:08 AM, WM wrote:
> > Jim Burns schrieb am Dienstag, 18. Mai 2021 um 02:14:44 UTC+2:
> >> On 5/17/2021 11:22 AM, WM wrote:
> >
> >> My claim is that the points which we may or may not be
> >> determining exist in a line with the _Archimedean_ property.
> >
> > If it is assumed, then most of the elements is dark.
> unknown yes, dark no.

Unknown or dark can be used synonymously.

Regards, WM

WM formulated the question :
> FromTheRafters schrieb am Dienstag, 18. Mai 2021 um 15:41:03 UTC+2:
>> on 5/18/2021, WM supposed :
>>> Meritocracy schrieb am Montag, 17. Mai 2021 um 18:51:28 UTC+2:
>>>
>>>
>>>> no knowledge of recursion, nor using limits...
>>>
>>> Recursion proves that every defined natnumber belongs to a FISON and has ℵo
>>> successors.
>> ITYM "induction" not recursion.
>
> Not real a difference.
> A recurrence relation is an equation that expresses each element of a
> sequence as a function of the preceding ones.

ITYM an iterative function here. Recursion is self-referent.

> Induction expresses each
> element of a sequence as a function of the preceding one.

No, it goes further than that.

>> Besides, you have *each* natural number
>> having *one* successor then you need induction to *prove* the base case
>> applies to *all*.
>
> This case applies to all definable natnumbers n:
>
>> ℕ \ (1, 2, 3, ..., n)| = ℵo.
>
>>> Using a limit is not processing all elements individually but it
>>> is processing them collectively.
>> ITYM *induction* rather than *limit* here as well.
>
> No.

Well, you should.

On Tuesday, May 18, 2021 at 4:41:56 PM UTC-3, WM wrote:
> William schrieb am Dienstag, 18. Mai 2021 um 17:54:18 UTC+2:
> > On Tuesday, May 18, 2021 at 8:59:52 AM UTC-3, WM wrote:
> > > William schrieb am Montag, 17. Mai 2021 um 18:51:23 UTC+2:
> > > > On Monday, May 17, 2021 at 12:19:34 PM UTC-3, WM wrote:
> > > > > William schrieb am Montag, 17. Mai 2021 um 00:11:02 UTC+2:
> > > > > > The fact that every element of a set is *definable* does not mean that there is a set of *defined* elements that contains every element.
> > > > > That is wrong.
> > > > No. It is easy to see that there is a infinite set of *definable* elements, but we can only chose a finite set to actually *define*
> > > You cannot choose any natnumber with less than ℵo successors.
> > Correct.
> Therefore it is wrong when you say that you can choose all such that none remains as you can do collectively.
> > Every element of the chosen finite (thus not containing every natural number) defined set has an infinite number of successors (a triviality). So what? You can chose any finite subset, but the subset you choose must be finite.
> Yes, you can choose any FISON. Then its n and all smaller numbers are regarded as defined. But you cannot choose any number such that nothing remains.
Of course not, you are choosing a finite set. However, you are not trying to choose "all such that none remains".

--
William Hughes

William schrieb am Dienstag, 18. Mai 2021 um 22:21:01 UTC+2:
> On Tuesday, May 18, 2021 at 4:41:56 PM UTC-3, WM wrote:
> > William schrieb am Dienstag, 18. Mai 2021 um 17:54:18 UTC+2:
> > > On Tuesday, May 18, 2021 at 8:59:52 AM UTC-3, WM wrote:
> > > > William schrieb am Montag, 17. Mai 2021 um 18:51:23 UTC+2:
> > > > > On Monday, May 17, 2021 at 12:19:34 PM UTC-3, WM wrote:
> > > > > > William schrieb am Montag, 17. Mai 2021 um 00:11:02 UTC+2:
> > > > > > > The fact that every element of a set is *definable* does not mean that there is a set of *defined* elements that contains every element.
> > > > > > That is wrong.
> > > > > No. It is easy to see that there is a infinite set of *definable* elements, but we can only chose a finite set to actually *define*
> > > > You cannot choose any natnumber with less than ℵo successors.
> > > Correct.
> > Therefore it is wrong when you say that you can choose all such that none remains as you can do collectively.
> > > Every element of the chosen finite (thus not containing every natural number) defined set has an infinite number of successors (a triviality). So what? You can chose any finite subset, but the subset you choose must be finite.
> > Yes, you can choose any FISON. Then its n and all smaller numbers are regarded as defined. But you cannot choose any number such that nothing remains.
> Of course not, you are choosing a finite set. However, you are not trying to choose "all such that none remains".

And even if you tried, you would fail. Simplest example:

U_{n∈ℕ} [1/(n+1), 1/n] = (0, 1] .

But for *every* chosen 1/n:

[1/n, 1] =/= (0, 1] .

The rest is darkness. (After William S.)

Regards, WM

On Tuesday, May 18, 2021 at 5:45:32 PM UTC-3, WM wrote:
> William schrieb am Dienstag, 18. Mai 2021 um 22:21:01 UTC+2:
> > On Tuesday, May 18, 2021 at 4:41:56 PM UTC-3, WM wrote:
> > > William schrieb am Dienstag, 18. Mai 2021 um 17:54:18 UTC+2:
> > > > On Tuesday, May 18, 2021 at 8:59:52 AM UTC-3, WM wrote:
> > > > > William schrieb am Montag, 17. Mai 2021 um 18:51:23 UTC+2:
> > > > > > On Monday, May 17, 2021 at 12:19:34 PM UTC-3, WM wrote:
> > > > > > > William schrieb am Montag, 17. Mai 2021 um 00:11:02 UTC+2:
> > > > > > > > The fact that every element of a set is *definable* does not mean that there is a set of *defined* elements that contains every element.
> > > > > > > That is wrong.
> > > > > > No. It is easy to see that there is a infinite set of *definable* elements, but we can only chose a finite set to actually *define*
> > > > > You cannot choose any natnumber with less than ℵo successors.
> > > > Correct.
> > > Therefore it is wrong when you say that you can choose all such that none remains as you can do collectively.
> > > > Every element of the chosen finite (thus not containing every natural number) defined set has an infinite number of successors (a triviality). So what? You can chose any finite subset, but the subset you choose must be finite.
> > > Yes, you can choose any FISON. Then its n and all smaller numbers are regarded as defined. But you cannot choose any number such that nothing remains.
> > Of course not, you are choosing a finite set. However, you are not trying to choose "all such that none remains".
> And even if you tried
You are not trying, so what would happen if you tried is of no interest.
--
William Hughes

WM was thinking very hard :
> William schrieb am Dienstag, 18. Mai 2021 um 22:21:01 UTC+2:
>> On Tuesday, May 18, 2021 at 4:41:56 PM UTC-3, WM wrote:
>>> William schrieb am Dienstag, 18. Mai 2021 um 17:54:18 UTC+2:
>>>> On Tuesday, May 18, 2021 at 8:59:52 AM UTC-3, WM wrote:
>>>>> William schrieb am Montag, 17. Mai 2021 um 18:51:23 UTC+2:
>>>>>> On Monday, May 17, 2021 at 12:19:34 PM UTC-3, WM wrote:
>>>>>>> William schrieb am Montag, 17. Mai 2021 um 00:11:02 UTC+2:
>>>>>>>> The fact that every element of a set is *definable* does not mean that
>>>>>>>> there is a set of *defined* elements that contains every element.
>>>>>>> That is wrong.
>>>>>> No. It is easy to see that there is a infinite set of *definable*
>>>>>> elements, but we can only chose a finite set to actually *define*
>>>>> You cannot choose any natnumber with less than ℵo successors.
>>>> Correct. Therefore it is wrong when you say that you can choose all such
>>>> that none remains as you can do collectively. Every element of the chosen
>>>> finite (thus not containing every natural number) defined set has an
>>>> infinite number of successors (a triviality). So what? You can chose any
>>>> finite subset, but the subset you choose must be finite.
>>> Yes, you can choose any FISON. Then its n and all smaller numbers are
>>> regarded as defined. But you cannot choose any number such that nothing
>>> remains.
>> Of course not, you are choosing a finite set. However, you are not trying to
>> choose "all such that none remains".
>
> And even if you tried, you would fail. Simplest example:
>
> U_{n∈ℕ} [1/(n+1), 1/n] = (0, 1] .
>
> But for *every* chosen 1/n:
>
> [1/n, 1] =/= (0, 1] .
>
> The rest is darkness. (After William S.)

ITYM (2(B)) \/ ~(2(B))

On 5/18/2021 3:31 PM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 18. Mai 2021 um 17:17:18 UTC+2:

>> The sequence determines a unique point, just not the way
>> you want it to be determined. Get over it. Accept reality.
>
> Gladly.

Here's the reality:

If an infinite digit sequence permits more than one point,
then there are infinitely-distant points in the line.

If an infinite digit sequence permits less than one point,
then there are continuous curves which cross but
do not intersect.

Since the line does not have infinitely-distant points
or continuous curves which cross but do not intersect,
each infinite digit sequence determines a unique point.

> Give a real example which describes a unique point.
> (Here we talk about sequences of digits without
> defining formula.)

Suppose that ABC is a dark right triangle
-- meaning, a right triangle it is impossible to show you.

Is it still true that
the square of the hypotenuse of ABC is equal to
the sum of the squares of the two remaining sides of ABC?

Yes, it is still true, because the reasoning supporting
that claim about ABC says nothing about whether I can
show it to you.

But! Perhaps you'd say, where is ABC? Show it to me.

(i) I can't show ABC to you, and
(ii) it doesn't matter that I can't show you ABC,
the square of the hypotenuse of ABC is still equal to
the sum of the squares of the two remaining sides of ABC.

Each infinite digit sequence determines a unique point.
It doesn't matter whether there is a digit-formula or not.
Each infinite digit sequence determines a unique point.