On 2/27/2022 10:56 PM, Chris M. Thomasson wrote:
> On 2/27/2022 1:14 AM, FromTheRafters wrote:
>> Chris M. Thomasson submitted this idea :
>>> On 2/25/2022 3:49 AM, FromTheRafters wrote:
>>>> Chris M. Thomasson pretended :
>>>>> On 2/24/2022 4:54 AM, WM wrote:
>>>>>> Fritz Feldhase schrieb am Mittwoch, 23. Februar 2022 um 16:18:03
>>>>>> UTC+1:
>>>>>>
>>>>>>> Can you actually PROVE your claim? (I got the impression that
>>>>>>> actually, say, (M_k)_1,2 --> oo. So far I see (M_k)_1,2 = (2/1,
>>>>>>> 3/1, 6/1, ...) and a little consideration shows that (M_k)_1,2
>>>>>>> diverges, no?)
>>>>>>
>>>>>> Like the sequence 1, 2, 3, ... . Important is only that never an
>>>>>> empty place can occur when two fractions are exchanged.
>>>>>
>>>>> 1, 2, 3, ... can be used to index every fraction.
>>>>
>>>> I disagree. We should be talking about rational numbers, not fractions.
>>>
>>> I am just thinking that 1, 2, 3... is infinite. Therefore it can be
>>> used to index any other infinite set?
>>
>> It is only countably so. I retract my statement about fractions
>> though. It is just that I was taught long ago that fractions are part
>> of a whole, not just simply a whole number ratio as is evidently the
>> case now. In fact 'whole numbers' weren't used at all after about the
>> third grade. The naturals are 'positive integers' and the naturals
>> with zero are non-negative integers.
>>
>> If you take a circle's diameter as the unit, and wrap that unit around
>> the circumference you get three 'whole units' as an 'integer part' and
>> a 'fractional part' of that whole which can be expressed in decimal
>> radix as <integer part>.<fractional part> which fractional part is not
>> even necessarily in Q.
>>
>> The Continued Fractional Expansion of the value can sometimes tell us
>> more about it being irrational by revealing an unending pattern such
>> as with the noble numbers and other clear unending patterns. The
>> rational fractions will terminate.
>
> The rational fractions will repeat, and then can be terminated. But, how
> many times do we have to iterate the digits in order find a point of
> repetition?

Think of a decimal whose digits were created by a TRNG from [0...9].
There would be no repeat, and no pattern.

Toss a ten sided die for each digit... When does it perfectly repeat itself?

Chris M. Thomasson explained :
> On 2/27/2022 1:14 AM, FromTheRafters wrote:
>> Chris M. Thomasson submitted this idea :
>>> On 2/25/2022 3:49 AM, FromTheRafters wrote:
>>>> Chris M. Thomasson pretended :
>>>>> On 2/24/2022 4:54 AM, WM wrote:
>>>>>> Fritz Feldhase schrieb am Mittwoch, 23. Februar 2022 um 16:18:03 UTC+1:
>>>>>>
>>>>>>> Can you actually PROVE your claim? (I got the impression that
>>>>>>> actually, say, (M_k)_1,2 --> oo. So far I see (M_k)_1,2 = (2/1, 3/1,
>>>>>>> 6/1, ...) and a little consideration shows that (M_k)_1,2 diverges,
>>>>>>> no?)
>>>>>>
>>>>>> Like the sequence 1, 2, 3, ... . Important is only that never an empty
>>>>>> place can occur when two fractions are exchanged.
>>>>>
>>>>> 1, 2, 3, ... can be used to index every fraction.
>>>>
>>>> I disagree. We should be talking about rational numbers, not fractions.
>>>
>>> I am just thinking that 1, 2, 3... is infinite. Therefore it can be used
>>> to index any other infinite set?
>>
>> It is only countably so. I retract my statement about fractions though. It
>> is just that I was taught long ago that fractions are part of a whole, not
>> just simply a whole number ratio as is evidently the case now. In fact
>> 'whole numbers' weren't used at all after about the third grade. The
>> naturals are 'positive integers' and the naturals with zero are
>> non-negative integers.
>>
>> If you take a circle's diameter as the unit, and wrap that unit around the
>> circumference you get three 'whole units' as an 'integer part' and a
>> 'fractional part' of that whole which can be expressed in decimal radix as
>> <integer part>.<fractional part> which fractional part is not even
>> necessarily in Q.
>>
>> The Continued Fractional Expansion of the value can sometimes tell us more
>> about it being irrational by revealing an unending pattern such as with the
>> noble numbers and other clear unending patterns. The rational fractions
>> will terminate.
>
> The rational fractions will repeat, and then can be terminated.

In CDE not in CFE.

> But, how many times do we have to iterate the digits in order find a point of
> repetition?

'Therein lies the rub' so they say. The 'tailing off' may be far far
down the road but not infinitely far.

Chris M. Thomasson expressed precisely :
> On 2/27/2022 10:56 PM, Chris M. Thomasson wrote:
>> On 2/27/2022 1:14 AM, FromTheRafters wrote:
>>> Chris M. Thomasson submitted this idea :
>>>> On 2/25/2022 3:49 AM, FromTheRafters wrote:
>>>>> Chris M. Thomasson pretended :
>>>>>> On 2/24/2022 4:54 AM, WM wrote:
>>>>>>> Fritz Feldhase schrieb am Mittwoch, 23. Februar 2022 um 16:18:03
>>>>>>> UTC+1:
>>>>>>>
>>>>>>>> Can you actually PROVE your claim? (I got the impression that
>>>>>>>> actually, say, (M_k)_1,2 --> oo. So far I see (M_k)_1,2 = (2/1, 3/1,
>>>>>>>> 6/1, ...) and a little consideration shows that (M_k)_1,2 diverges,
>>>>>>>> no?)
>>>>>>>
>>>>>>> Like the sequence 1, 2, 3, ... . Important is only that never an empty
>>>>>>> place can occur when two fractions are exchanged.
>>>>>>
>>>>>> 1, 2, 3, ... can be used to index every fraction.
>>>>>
>>>>> I disagree. We should be talking about rational numbers, not fractions.
>>>>
>>>> I am just thinking that 1, 2, 3... is infinite. Therefore it can be used
>>>> to index any other infinite set?
>>>
>>> It is only countably so. I retract my statement about fractions though. It
>>> is just that I was taught long ago that fractions are part of a whole, not
>>> just simply a whole number ratio as is evidently the case now. In fact
>>> 'whole numbers' weren't used at all after about the third grade. The
>>> naturals are 'positive integers' and the naturals with zero are
>>> non-negative integers.
>>>
>>> If you take a circle's diameter as the unit, and wrap that unit around the
>>> circumference you get three 'whole units' as an 'integer part' and a
>>> 'fractional part' of that whole which can be expressed in decimal radix as
>>> <integer part>.<fractional part> which fractional part is not even
>>> necessarily in Q.
>>>
>>> The Continued Fractional Expansion of the value can sometimes tell us more
>>> about it being irrational by revealing an unending pattern such as with
>>> the noble numbers and other clear unending patterns. The rational
>>> fractions will terminate.
>>
>> The rational fractions will repeat, and then can be terminated. But, how
>> many times do we have to iterate the digits in order find a point of
>> repetition?
>
> Think of a decimal whose digits were created by a TRNG from [0...9]. There
> would be no repeat, and no pattern.

No formula either. Can we say it 'represents' a number? IMO we can say
it has a place on the real number line if we can just 'nail down' which
number it represents. Since we cannot do that with 'random' sequences I
don't consider them.

> Toss a ten sided die for each digit... When does it perfectly repeat itself?

Chris M. Thomasson laid this down on his screen :
> On 2/27/2022 10:30 AM, Fritz Feldhase wrote:
>> On Sunday, February 27, 2022 at 4:20:33 AM UTC+1, Chris M. Thomasson wrote:
>>
>>> I am just thinking that 1, 2, 3... is infinite. Therefore it can be used
>>> to index any other infinite set?
>>
>> No, it can not be used to index, say, IR. (You will always miss some
>> elements in IR.)
>>
>> See: https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument
>
> The the natural numbers should be able to be used to index any irrational.
> Think of the digits of pi:
>
> [0] = 3
> [1] = 1
> [2] = 4
> [3] = 1
> [4] = 5
> [5] = 9
> [...] = [...]
>
>
> index 3 = digit 1 of the irrational in question. It should work. PI has
> infinite digits, and the natural numbers are infinite in nature, so it can be
> used to index an irrational?

That's not how I see it, that is not indexing. Say, for the sake of
argument, that pi is the twenty-first irrational or transcendental
number. By indexing, what is the twenty-sixth transcendental number in
that ordering?

Hint: Iterate your successor function.

Jim Burns schrieb am Sonntag, 27. Februar 2022 um 23:03:25 UTC+1:
> On 2/27/2022 5:21 AM, WM wrote:
> > Chris M. Thomasson schrieb
> > am Sonntag, 27. Februar 2022 um 04:43:25 UTC+1:
> >> On 2/25/2022 3:24 AM, WM wrote:
>
> >> You are thinking to finite here. An infinite set
> >> goes on forever.
> >>
> > If it goes on forever, then you will never know whether
> > two sets are in bijection or whether one of them is
> > larger than the other.
> No.
>
> We can describe one of infinitely-many and then,
> taking only reliable statment-steps, arriveat
> further claims which we will _know_ are true of
> each individual described, even if there are
> infinitely-many of those individuals.
>
> The infinite set doesn't end,
> but _our reasoning_ can end.

It does. It cannot avoid that every definable natnumber has infinitely many successors, infinitely many of which cannot become defined:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo . (*)

Either they do not exist at all. Then we have potential infinity, no completeness, only the collection ℕ_def. Or we have completenes such that
ℕ \ {1, 2, 3, ...} = { }
but then this subtraction can only be performed collectively, not individually as in (*).

Regards, WM

Fritz Feldhase schrieb am Sonntag, 27. Februar 2022 um 23:23:03 UTC+1:
> On Sunday, February 27, 2022 at 11:03:25 PM UTC+1, Jim Burns wrote:
> > On 2/27/2022 5:21 AM, WM wrote:
> > > Chris M. Thomasson schrieb
> > > am Sonntag, 27. Februar 2022 um 04:43:25 UTC+1:
> > >> On 2/25/2022 3:24 AM, WM wrote:
> >
> > >> You are thinking to finite here. An infinite set
> > >> goes on forever.
> > >>
> > > If it goes on forever, then you will never know whether
> > > two sets are in bijection or whether one of them is
> > > larger than the other.
> Nonsense!
>
> We KNOW, since we can PROVE it, that that there is a bijection from {2*n : n e IN} onto {2*n + 1 : n e IN}, for example.

You can prove it for every definable natnumber has infinitely many successors, infinitely many of which cannot become defined:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo . (*)

Either they do not exist at all. Then we have potential infinity, no completeness, only the collection ℕ_def. Or we have completenes such that
ℕ \ {1, 2, 3, ...} = { }
but then this subtraction can only be performed collectively, not individually as in (*).

Regards, WM

zelos...@gmail.com schrieb am Montag, 28. Februar 2022 um 07:45:18 UTC+1:
> söndag 27 februari 2022 kl. 16:49:41 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Sonntag, 27. Februar 2022 um 13:18:15 UTC+1:
> > > söndag 27 februari 2022 kl. 11:31:34 UTC+1 skrev WM:
> >
> > > > > They are all in the fucking sequence, what are you on about you fucking crank?
> > > > All fractions which you can define
> > > What does "define" mean here? It is meaningless!
> > It is what Poincaré defined: Can be expressed with a finite amount of words.
> > > >are in the first column.
> > > and? Your "matrix" is not a mathematical object.
> > > >But every such fraction has ℵo successors which you will never have defined completely.
> > > Meaningless nonsense.
> > Fact. Try to find a counter example.
> Of WHAT?

Of the fact that every definable natnumber has infinitely many successors, infinitely many of which cannot become defined:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .
This defines ℕ_def.

These however are not all natnumbers because collectively all can be handled such that not infinitely many successors remain, not even one:
ℕ \ {1, 2, 3, ...} = { } .
This defines ℕ.

Regards, WM

måndag 28 februari 2022 kl. 11:54:07 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 28. Februar 2022 um 07:45:18 UTC+1:
> > söndag 27 februari 2022 kl. 16:49:41 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Sonntag, 27. Februar 2022 um 13:18:15 UTC+1:
> > > > söndag 27 februari 2022 kl. 11:31:34 UTC+1 skrev WM:
> > >
> > > > > > They are all in the fucking sequence, what are you on about you fucking crank?
> > > > > All fractions which you can define
> > > > What does "define" mean here? It is meaningless!
> > > It is what Poincaré defined: Can be expressed with a finite amount of words.
> > > > >are in the first column.
> > > > and? Your "matrix" is not a mathematical object.
> > > > >But every such fraction has ℵo successors which you will never have defined completely.
> > > > Meaningless nonsense.
> > > Fact. Try to find a counter example.
> > Of WHAT?
> Of the fact that every definable natnumber has infinitely many successors, infinitely many of which cannot become defined:

What does "DEFINED" EVEN MEAN!?

You use it but it means nothing here!

> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .
> This defines ℕ_def.

This makes N_def = N

>
> These however are not all natnumbers because collectively all can be handled such that not infinitely many successors remain, not even one:
> ℕ \ {1, 2, 3, ...} = { } .
> This defines ℕ.

No, what you say is A/A={}

Which is the definition of set subtraction, but it means nothing in defining N!

You are saying shit that means nothing!

>
> Regards, WM

On Monday, February 28, 2022 at 12:38:30 AM UTC+1, Chris M. Thomasson wrote:
> On 2/27/2022 10:30 AM, Fritz Feldhase wrote:
> > On Sunday, February 27, 2022 at 4:20:33 AM UTC+1, Chris M. Thomasson wrote:
> >
> >> I am just thinking that 1, 2, 3... is infinite. Therefore it can be used
> >> to index any other infinite set?
> >
> > No, it can not be used to index, say, IR. (You will always miss some elements in IR.)
> >
> > See: https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument
> >
> The the natural numbers should be able to be used to index any
> irrational. Think of the digits of pi:
>
> [0] = 3
> [1] = 1
> [2] = 4
> [3] = 1
> [4] = 5
> [5] = 9
> [...] = [...]
>
> PI has infinite digits, and the natural numbers are infinite in nature, so it
> can be used to index an irrational?

THIS wasn't your question, idiot. [Now you are just "indexing" the "digits" of pi.]

Your question was:
> >> I am just thinking that 1, 2, 3... is infinite. Therefore it can be used
> >> to index any other infinite set?

Answer: No, it can't.

EOD.

On Monday, February 28, 2022 at 11:49:25 AM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 27. Februar 2022 um 23:23:03 UTC+1:
> > On Sunday, February 27, 2022 at 11:03:25 PM UTC+1, Jim Burns wrote:
> > > On 2/27/2022 5:21 AM, WM wrote:
> > > > Chris M. Thomasson schrieb
> > > > am Sonntag, 27. Februar 2022 um 04:43:25 UTC+1:
> > > >> On 2/25/2022 3:24 AM, WM wrote:
> > > >>
> > > >> You are thinking to finite here. An infinite set
> > > >> goes on forever.
> > > >>
> > > > If it goes on forever, then you will never know whether
> > > > two sets are in bijection or <bla bla bla>
> > > >
> > Nonsense!
> >
> > We KNOW, since we can PROVE it, that there is a bijection from {2*n : n e IN} onto {2*n + 1 : n e IN}, for example.
> >
> You can prove it for <nonsense deleted>

Yeah, Mückenheim, whatever.

On 2/28/2022 4:53 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 28. Februar 2022 um 07:45:18 UTC+1:
>> söndag 27 februari 2022 kl. 16:49:41 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Sonntag, 27. Februar 2022 um 13:18:15 UTC+1:
>>>> söndag 27 februari 2022 kl. 11:31:34 UTC+1 skrev WM:
>>>
>>>>>> They are all in the fucking sequence, what are you on about you fucking crank?
>>>>> All fractions which you can define
>>>> What does "define" mean here? It is meaningless!
>>> It is what Poincaré defined: Can be expressed with a finite amount of words.
>>>>> are in the first column.
>>>> and? Your "matrix" is not a mathematical object.
>>>>> But every such fraction has ℵo successors which you will never have defined completely.
>>>> Meaningless nonsense.
>>> Fact. Try to find a counter example.
>> Of WHAT?
>
> Of the fact that every definable natnumber has infinitely many successors, infinitely many of which cannot become defined:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .
> This defines ℕ_def.

def, defined, as you use it, is meaningless. Just look at your own definition of it, it is meaningless.
what you call as a fact above, and your equation is bogus.

>
> These however are not all natnumbers because collectively all can be handled such that not infinitely many successors remain, not even one:
> ℕ \ {1, 2, 3, ...} = { } .
> This defines ℕ.

No. that is a fib. You are defining { }.

"collectively", "handled", "remain" are not mathematical terms

>
> Regards, WM
>

Your post is rejected for grave intentional errors.

WM presented the following explanation :
> zelos...@gmail.com schrieb am Montag, 28. Februar 2022 um 07:45:18 UTC+1:
>> söndag 27 februari 2022 kl. 16:49:41 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Sonntag, 27. Februar 2022 um 13:18:15 UTC+1:
>>>> söndag 27 februari 2022 kl. 11:31:34 UTC+1 skrev WM:
>>>>>> They are all in the fucking sequence, what are you on about you fucking
>>>>>> crank?
>>>>> All fractions which you can define
>>>> What does "define" mean here? It is meaningless! It is what Poincaré
>>>> defined: Can be expressed with a finite amount of words.
>>>>> are in the first column. and? Your "matrix" is not a mathematical
>>>>> object. But every such fraction has ℵo successors which you will never
>>>>> have defined completely.
>>>> Meaningless nonsense.
>>> Fact. Try to find a counter example.
>> Of WHAT?
>
> Of the fact that every definable natnumber has infinitely many successors,
> infinitely many of which cannot become defined: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3,
> ..., k}| = ℵo . This defines ℕ_def.
>
> These however are not all natnumbers because collectively all can be handled
> such that not infinitely many successors remain, not even one: ℕ \ {1, 2, 3,
> ...} = { } . This defines ℕ.

It says that there is no difference between the set of natural numbers
and the set of natural numbers.

On Monday, February 28, 2022 at 5:36:17 PM UTC+1, FromTheRafters wrote:
> WM presented the following explanation :
> >
> > collectively all [natural numbers] can be handled such that not infinitely many successors remain,
> > not even one: ℕ \ {1, 2, 3, ...} = { } . This defines ℕ.
> >
> It says that there is no difference between the set of natural numbers and the set of natural numbers.

Yeah. A fascinating statement, indeed.

On the other hand, the claim

> > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo . This defines ℕ_def. [WM]

is nonsense too. Nothing is "defined" here. "ℕ" instead of "ℕ_def" would satisfy the formula too:

∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo ,

of course. In other words,

∀n ∈ ℕ: |{k e ℕ : k > n}| = ℵo .

What a surprise!

On Sunday, February 27, 2022 at 10:34:55 PM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 27. Februar 2022 um 21:20:03 UTC+1:
> >
> > https://www.hs-augsburg.de/~mueckenh/Transfinity/ choose Countability Contradicted
> >
> > There you write: "It should also be possible to model this configuration by exchanging pairs of fractions"
> >
> > Why? Because YOU SAY SO?

Hint: "to model" is mumbo-jumbo in this context.

I guess what you actually MEAN is:

| "It should also be possible to reach this configuration by exchanging pairs of fractions."

But "reach" in what sense exactly?

Using mathematical terms, by best guess would be: There has to be a sequence (M_i) of matrices, such that (1) for all i e IN M_i+1 results from M_i by an exchange of a pair of fractions AND (2) there is a k e IN such that M_k = the_target_matrix.

So my question is just: __WHY?__

No justification is given for this claim.

But it seems that this comes as a surprise to you: "...but it is not.".

The rest of your text consists of "even more nonsense".

| If clearing happens, then not by definable terms.

Luckily no "clearing happens".

| What causes this failure?

Which failure?

| "It is impossible to empty the matrix."

Did anyone CLAIM that this is (or has to be) possible (except you)?

_______________________________

Again, your claim is (in mathematical terms): There has to be a sequence (M_i) of matrices, such that (1) for all i e IN M_i+1 results from M_i by an exchange of a pair of fractions AND (2) there is a k e IN such that M_k = the_target_matrix.

My question is ___why___?

> simply because up to every term of Cantor's sequence
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
> [...] the corrresponding matrix exists
>
> 1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 5/1, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ... 4/1, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 2/2, 5/2, 5/3, 5/4, ...
> ... ... ... ...

So what? This is an EXTREMELY trivial fact which does NOT imply your claim.

> There is simply no obstacle and there is no reason why there should appear an obstacle.

No, there is no "no obstacle and there is no reason why there should appear an obstacle". Did anyone claim that there is an "obstacle"?

> > "but it is not."
> >
> It is!

Huh?!

THIS IS YOUR CAIM, IDIOT!

You wrote:

| "It should also be possible to model this configuration by exchanging pairs of fractions [...] but it is not."

It surprises you that something which does not have to be the case is not the case? FASCINATING!

> Find the first failure.

The first (serious) failure in your text is your claim that "[i]t should also be possible to reach this configuration by exchanging pairs of fractions."

No, it shouldn't AND it isn't.

| 'I know what you're thinking about,' said Tweedledum; 'but it isn't so, nohow.'
| | 'Contrariwise,' continued Tweedledee, 'if it was so, it might be; and if it were so, it would be; but as it isn't, it ain't. That's logic.'

On Monday, February 28, 2022 at 6:54:23 PM UTC+1, Fritz Feldhase wrote:

Seems that your "argument" is the following:

[WM:] "In my opinion..."

> | "it should also be possible to model this configuration by exchanging pairs of fractions [...]

Well, opinions aren't "good enough" in mathematics, Mückenheim.

> | but it is not."

Big surprise!

Your fellow crank: "In my opinion it should be possible to divide by 0. But it is not! Something's not right in mainstream math!"

Yeah, a convincing argument!

zelos...@gmail.com schrieb am Montag, 28. Februar 2022 um 12:32:16 UTC+1:
> måndag 28 februari 2022 kl. 11:54:07 UTC+1 skrev WM:

> > > > Fact. Try to find a counter example.
> > > Of WHAT?
> > Of the fact that every definable natnumber has infinitely many successors, infinitely many of which cannot become defined:
> What does "DEFINED" EVEN MEAN!?

Try to study mathematics. Then you know it. Here is a hint every mathematician knows: Henri Poincaré: "Über transfinite Zahlen" in "Sechs Vorträge über ausgewählte Gegenstände aus der reinen Mathematik und der mathematischen Physik", Teubner, Leipzig (1910) p. 37.

Here is the definition too:
>
> > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .
> > This defines ℕ_def.
> This makes N_def = N
> >
> > These however are not all natnumbers because collectively all can be handled such that not infinitely many successors remain, not even one:
> > ℕ \ {1, 2, 3, ...} = { } .
> > This defines ℕ.
> No, what you say is A/A={}

What I say above is that this subtraction is not possible with individually defined elements.
N \N_def =/= { }
>
Regards, WM

Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 14:57:30 UTC+1:
> On Monday, February 28, 2022 at 11:49:25 AM UTC+1, WM wrote:

> > > We KNOW, since we can PROVE it, that there is a bijection from {2*n : n e IN} onto {2*n + 1 : n e IN}, for example.
> > >
> > You can prove it for every definable natnumber has infinitely many successors, infinitely many of which cannot become defined:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .

You cannot prove it for all natnumbers.

Regards, WM

FromTheRafters schrieb am Montag, 28. Februar 2022 um 17:36:17 UTC+1:
> WM presented the following explanation :

> > Of the fact that every definable natnumber has infinitely many successors,
> > infinitely many of which cannot become defined: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3,
> > ..., k}| = ℵo . This defines ℕ_def.
> >
> > These however are not all natnumbers because collectively all can be handled
> > such that not infinitely many successors remain, not even one: ℕ \ {1, 2, 3,
> > ...} = { } . This defines ℕ.
> It says that there is no difference between the set of natural numbers
> and the set of natural numbers.

But the set of definable natural numbers differs:
ℕ \ ℕ_def =/= { }

Regards, WM

On Monday, February 28, 2022 at 7:20:55 PM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 14:57:30 UTC+1:
> >
> > We KNOW, since we can PROVE it, that there is a bijection from {2*n : n e IN} onto {2*n + 1 : n e IN}, for example.
> >
> You cannot prove it for all natnumbers.

We cannot prove WHAT, you silly asshole?

On 2/28/2022 5:46 AM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 27. Februar 2022 um 23:03:25 UTC+1:
>> On 2/27/2022 5:21 AM, WM wrote:
>>> Chris M. Thomasson schrieb
>>> am Sonntag, 27. Februar 2022 um 04:43:25 UTC+1:

>>>> You are thinking to finite here. An infinite set
>>>> goes on forever.
>>>
>>> If it goes on forever, then you will never know whether
>>> two sets are in bijection or whether one of them is
>>> larger than the other.
>>
>> No.
>>
>> We can describe one of infinitely-many and then,
>> taking only reliable statment-steps, arriveat
>> further claims which we will _know_ are true of
>> each individual described, even if there are
>> infinitely-many of those individuals.
>>
>> The infinite set doesn't end,
>> but _our reasoning_ can end.
>
> It does. It cannot avoid that every definable natnumber
> has infinitely many successors, infinitely many of which
> cannot become defined:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo . (*)
>
> Either they do not exist at all. Then we have
> potential infinity, no completeness,

For what you (WM) mean by completeness,
you require elements that are not equal to themselves.

We exclude from consideration elements which are not
equal to themselves.

We have all the collections ⟨1,...,𝑛⟩ which have a
counting-order which begins at 1 and ends somewhere.
Here, its _second end_ is called 𝑛

Each ⟨1,...,𝑛⟩ _cannot_ be matched to any of the
proper subsets or proper supersets of ⟨1,...,𝑛⟩.

The collection ⋃{⟨1,...,𝑘⟩} of all second ends of
the ⟨1,...,𝑛⟩ _can_ be matched to some of the
proper subsets and proper supersets of ⋃{⟨1,...,𝑘⟩}

> potential infinity, no completeness, only the
> collection ℕ_def. Or we have completenes such that
> ℕ \ {1, 2, 3, ...} = { }
> but then this subtraction can only be performed
> collectively, not individually as in (*).

We can _reason about_ ℕ \ {1, 2, 3, ...} by
describing one of the elements of ℕ \ {1, 2, 3, ...}
and then, taking only reliable statement-steps,
arrive at claims in addition to the description.

All of the elements are equal to themselves.
Apparently, things which are equal to themselves
are not dark, so none of them are dark.

On Monday, February 28, 2022 at 7:17:53 PM UTC+1, WM wrote:

> Here is [a] definition too:
> >
> > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
> > This defines ℕ_def.

No, this is NOT a definition and hence it does NOT define "ℕ_def".

It's just shit.

Hint: ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

Actually, for each and every X c IN we have:

∀n ∈ X: |ℕ \ {1, 2, 3, ..., n}| = ℵo. (*)

So (*) doesn't SPECIFY (or "define") a certain (i. e. unique) X, you silly crank.

> > > ℕ \ {1, 2, 3, ...} = { } .
> > > This defines ℕ.

No, it doesn't.

Hint: If ℕ were {1, 2, 3} it would be TRUE too, since

{1, 2, 3} \ {1, 2, 3, ...} = { } .

But actually,

> > what you say is A \ A={}

with A = ℕ.

> What I say above is

shit. As usual.

On Monday, February 28, 2022 at 11:54:07 AM UTC+1, WM wrote:
> zelos...@gmail.com schrieb am Montag, 28. Februar 2022 um 07:45:18 UTC+1:
> > söndag 27 februari 2022 kl. 16:49:41 UTC+1 skrev WM:
> >
> > Try to find a counter example.

OF WHAT, you silly crank?

> > Of WHAT? You are asking me to find a sklroskrli and won't even define what it is!
> >
> Of the fact that every [...] natnumber has infinitely many successors

Huh?! WHAT THE FUCK, ARE YOU TALKING ABOUT, YOU SILLY ASSHOLE?

Did anyone deny that fact?

> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
> This defines ℕ_def.

No, it doesn't. Hint: ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

> These however are not all natnumbers ...

Huh?! "These"?

Hint: {n e ℕ : |ℕ \ {1, 2, 3, ..., n}| = ℵo} = ℕ.

On Monday, February 28, 2022 at 10:23:29 AM UTC-8, WM wrote:

> But the set of definable natural numbers

There is no such set

eod

On Monday, 28 February 2022 at 14:20:55 UTC-4, WM wrote:
> Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 14:57:30 UTC+1:
> > On Monday, February 28, 2022 at 11:49:25 AM UTC+1, WM wrote:
>
> > > > We KNOW, since we can PROVE it, that there is a bijection from {2*n : n e IN} onto {2*n + 1 : n e IN}, for example.
> > > >
> > > You can prove it for every definable natnumber has infinitely many successors, infinitely many of which cannot become defined:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .
> You cannot prove it for all natnumbers.

Ah, the patented quantifier exchange is being set up. Subtle, Muckenheim, really subtle.

Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 18:54:23 UTC+1:
> On Sunday, February 27, 2022 at 10:34:55 PM UTC+1, WM wrote:

>
> | "It should also be possible to reach this configuration by exchanging pairs of fractions."
>
> But "reach" in what sense exactly?

In this sense: Every fraction of Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... can be found in the first column. No fraction of this sequence will stay permanently in any other column.
>
> So my question is just: __WHY?__
>
This is what Cantor prescribes. If the terms of the sequences are gathered elsewhere, then the remaining dark fractions are not visible. In my model their existence is proved by the fact that never an empty place appears.

> No justification is given for this claim.

It is enough justification to do exactly what Cantor does.

Regards, WM