On Sunday, 27 February 2022 at 06:05:46 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Sonntag, 27. Februar 2022 um 03:49:17 UTC+1:
>
> > Good! So you actually have thought about the sequence. In every position other than the first column, you will get an increasing sequence of integers, so the structure of M(oo) is
> >
> > 1/1, oo, oo, oo, oo, oo, oo, ...
> > 1/2, oo, oo, oo, oo, oo, oo, ...
> > 2/1, oo, oo, oo, oo, oo, oo, ...
> > 1/3, oo, oo, oo, oo, oo, oo, ...
> > 2/2, oo, oo, oo, oo, oo, oo, ...
> > 3/1, oo, oo, oo, oo, oo, oo, ...
> > ..., ..., ..., ..., ..., ..., ...
> >
> > All fractions are in the first column!
> Which position is the first to get the "fraction" oo? Which column is the first to be conquered by oo's completely? Note that counting is the topic here. Counting excludes that all positions become oo simultaneously.

What an utterly ridiculous response. You still haven't grasped the concept of a limit. No element of any sequence of the 1x1-submatrices m_{i,j,}(k) is actually ever infinite, but the entries in each sequence diverge monotonically towards infinity. For that I use the shorthand entry "oo" in the matrix M(oo), which likewise is not actually an element of the sequence of matrices M(k). That you cannot grasp this simple idea does not speak favourably of your reasoning skills.

söndag 27 februari 2022 kl. 11:31:34 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Sonntag, 27. Februar 2022 um 11:23:31 UTC+1:
> > fredag 25 februari 2022 kl. 12:34:38 UTC+1 skrev WM:
>
> > > You can avoid the problem by gathering the fractions
> > > 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
> > > somewhere else, in another place, outside of the matrix.
> >
> > That works because that is a SEQUENCE, not a fucking matrix.
> >
> > > Then you get the impression that nothing will remain in the matrix. Apparently.
> > > If you don't look too close, Cantor seems right. But is he right then? Are the remaining fractions, like electrons in quantum theory, not there if not being observed? That is not what a mathematician should believe.
> >
> > 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!

>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.

>Dark numbers are real if infinity is actual.

Yet you cannot define "dark" in a meaningful mathematical way.

>Here you may find the most comprehensive essay about dark numbers presently available: https://www.hs-augsburg.de/~mueckenh/Transfinity/ choose Dark Numbers.

it is a nonsense book

>
> Further discussion will be irrelevant. Either you can understand or you you cannot.
>
> Regards, WM

FromTheRafters schrieb am Sonntag, 27. Februar 2022 um 12:28:13 UTC+1:
> on 2/27/2022, WM supposed :
> > horand....@gmail.com schrieb am Sonntag, 27. Februar 2022 um 03:49:17 UTC+1:
> >
> >> Good! So you actually have thought about the sequence. In every position
> >> other than the first column, you will get an increasing sequence of
> >> integers, so the structure of M(oo) is
> >>
> >> 1/1, oo, oo, oo, oo, oo, oo, ...
> >> 1/2, oo, oo, oo, oo, oo, oo, ...
> >> 2/1, oo, oo, oo, oo, oo, oo, ...
> >> 1/3, oo, oo, oo, oo, oo, oo, ...
> >> 2/2, oo, oo, oo, oo, oo, oo, ...
> >> 3/1, oo, oo, oo, oo, oo, oo, ...
> >> ..., ..., ..., ..., ..., ..., ...
> >>
> >> All fractions are in the first column!
> >
> > Which position is the first to get the "fraction" oo? Which column is the
> > first to be conquered by oo's completely? Note that counting is the topic
> > here.
> It is not about counting, it is about being "countable" where countable

means well-ordered like the natural numbers. If infinitely many places get oo simultaneously, then their fractions have not been well-ordered. But this idea is a bogus claim anyway. Matrix places are never empty and never carry a symbol oo.

Regards, 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.
> >Dark numbers are real if infinity is actual.
> Yet you cannot define "dark" in a meaningful mathematical way.

Dark numbers are not defined. You cannot subtract all natural numbers individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .
But you can collectively.
ℕ \ {1, 2, 3, ...} = { }.
That's the difference.

> >Here you may find the most comprehensive essay about dark numbers presently available: https://www.hs-augsburg.de/~mueckenh/Transfinity/ choose Dark Numbers.
> it is a nonsense book

It is not a book but only few pages.
> >
> > Further discussion will be irrelevant. Either you can understand or you you cannot.

Obviously you cannot.

Regards, WM

horand....@gmail.com schrieb am Sonntag, 27. Februar 2022 um 12:43:45 UTC+1:
> On Sunday, 27 February 2022 at 06:05:37 UTC-4, WM wrote:
> > horand....@gmail.com schrieb am Samstag, 26. Februar 2022 um 23:13:46 UTC+1:

> > > Where did I say anything about enumerating?
> > It is the topic here.
> No, it's not.
You are wrong: In oder to index all positive fractions

> I could open a new thread concerning the convergence behavior of the elements of the matrices M(k) as k -> oo,

That would be bogus. There is no convergence. Matrix places are never empty and never carry a symbol oo. That is provable.

Regards, WM

WM brought next idea :
> FromTheRafters schrieb am Sonntag, 27. Februar 2022 um 12:28:13 UTC+1:
>> on 2/27/2022, WM supposed :
>>> horand....@gmail.com schrieb am Sonntag, 27. Februar 2022 um 03:49:17
>>> UTC+1:
>>>
>>>> Good! So you actually have thought about the sequence. In every position
>>>> other than the first column, you will get an increasing sequence of
>>>> integers, so the structure of M(oo) is
>>>>
>>>> 1/1, oo, oo, oo, oo, oo, oo, ...
>>>> 1/2, oo, oo, oo, oo, oo, oo, ...
>>>> 2/1, oo, oo, oo, oo, oo, oo, ...
>>>> 1/3, oo, oo, oo, oo, oo, oo, ...
>>>> 2/2, oo, oo, oo, oo, oo, oo, ...
>>>> 3/1, oo, oo, oo, oo, oo, oo, ...
>>>> ..., ..., ..., ..., ..., ..., ...
>>>>
>>>> All fractions are in the first column!
>>>
>>> Which position is the first to get the "fraction" oo? Which column is the
>>> first to be conquered by oo's completely? Note that counting is the topic
>>> here.
>> It is not about counting, it is about being "countable" where countable
>
> means well-ordered like the natural numbers.

https://en.wikipedia.org/wiki/Countable_set

horand....@gmail.com schrieb am Sonntag, 27. Februar 2022 um 12:51:58 UTC+1:

> > Which position is the first to get the "fraction" oo? Which column is the first to be conquered by oo's completely? Note that counting is the topic here. Counting excludes that all positions become oo simultaneously.
> You still haven't grasped the concept of a limit.

There is no limit when counting. Cantor does not count ω. This "concept" is only the excuse for not knowing the dark numbers which should be counted before any limit, but cannot.

However, your splendid idea

1/1, oo, oo, oo, oo, oo, oo, ...
1/2, oo, oo, oo, oo, oo, oo, ...
2/1, oo, oo, oo, oo, oo, oo, ...
1/3, oo, oo, oo, oo, oo, oo, ...
2/2, oo, oo, oo, oo, oo, oo, ...
3/1, oo, oo, oo, oo, oo, oo, ...
..., ..., ..., ..., ..., ..., ...

proves that for every place in every column except the first one you cannot find out the last ℵo fractions leaving there. For all these ℵo places we get ℵo unknown (i.e. dark) fractions because for all indexes k of definable fractions we have
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .

Congratulations to your first proof of dark numbers!

Regards, WM

On 2/27/2022 4:15 AM, WM wrote:
> FromTheRafters schrieb am Sonntag, 27. Februar 2022 um 10:30:47 UTC+1:
>
>> Fuggetaboutit -- As Fritz Feldhase pointed out, I missed that WM was
>> talking about a superset of positive rationals
>
> That is what Cantor did
> k = (m + n - 1)(m + n - 2)/2 + m
> 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, ...
>
> I only mimic him. But instead of bringing the naturals to the serpent I bring the serpent to the naturals. This shows that never all fractions can be indexed. In appealing and easily understandable form you can read it here. https://www.hs-augsburg.de/~mueckenh/Transfinity/ choose Countability Contradicted. It is only one page, smashing 150 years of set theory.

No. You do not mimic Cantor at all.

You only continue to demonstrate your frustration, and total failure, to be unable to disprove such a simple and elegant proof by Cantor which is well
over 130 years ago.

>
> Regards, WM
>
>
>

On 2/27/2022 3:21 AM, FromTheRafters wrote:
> Chris M. Thomasson laid this down on his screen :
>> On 2/15/2022 1:03 AM, FromTheRafters wrote:
>>> Chris M. Thomasson submitted this idea :
>>>> On 2/13/2022 3:15 AM, WM wrote:
>>>>> Chris M. Thomasson schrieb am Samstag, 12. Februar 2022 um 10:15:42 UTC+1:
>>>>>
>>>>>> think of Cantor Pairing creating an infinite number of single numbers
>>>>>> from two numbers, and vise versa.
>>>>>
>>>>> All positive fractions m/n are contained in the matrix
>>>>>
>>>>> 1/1, 1/2, 1/3, 1/4, ...
>>>>> 2/1, 2/2, 2/3, 2/4, ...
>>>>> 3/1, 3/2, 3/3, 3/4, ...
>>>>> 4/1, 4/2, 4/3, 4/4, ...
>>>>> 5/1, 5/2, 5/3, 5/4, ...
>>>>> ...
>>>>>
>>>>> Cantor claims that all can be indexed by natural numbers, i.e., written in a sequence, when he uses the sequence
>>>> [...]
>>>>
>>>> The set of natural numbers is infinite. Therefore, it can be used to index any other infinite set. Period.
>>>
>>> Countably infinite.
>>
>> It sure seems like every infinite set can be counted wrt iteration, on a per-step basis.
>
> How do you interate the reals? I give you a real number and you go look for the successor to that number and label it with its natural index?
>
> If I give you the eighth real number (or whichever I say it is) can you give me the ninth or next real number?

its just the one that is almost a smidgen over to the right.

remember no rounding

On 2/27/2022 9:49 AM, WM wrote:
> 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.

out of context, and misused here.

>>> 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.

no need, it is meaningless, as you use your bogus "defined" in it.

>>> Dark numbers are real if infinity is actual.
>> Yet you cannot define "dark" in a meaningful mathematical way.
>
> Dark numbers are not defined. You cannot subtract all natural numbers individually
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .
> But you can collectively.
> ℕ \ {1, 2, 3, ...} = { }.
> That's the difference.

No. There are no dark numbers, you have been unable to prove that.

>
>>> Here you may find the most comprehensive essay about dark numbers presently available: https://www.hs-augsburg.de/~mueckenh/Transfinity/ choose Dark Numbers.
>> it is a nonsense book
>
> It is not a book but only few pages.
>>>
>>> Further discussion will be irrelevant. Either you can understand or you you cannot.
>
> Obviously you cannot.

You present imagination and a spoof, no proof.

>
> Regards, WM

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

On 2/27/2022 4:57 AM, Effego Banos wrote:
> zelos...@gmail.com wrote:
>
>>> There is no final matrix visible. You can find every lit fraction in
>>> the first column. You will never find a lit fraction residing
>>> permanently outside of the first column. But you will never find an
>>> empty place in the matrix.
>>
>> That would be because by definition THERE IS NOTHING EMPTY IN A MATRIX!
>
> The CIA – 70 years in Ukraine
> https://www.youtube.com/embed/jRicZc-cZ0I
> https://theduran.com/the-cia-70-years-in-ukraine/

good links!

On 2/27/2022 10:31 AM, WM wrote:
>
> horand....@gmail.com schrieb am Sonntag, 27. Februar 2022 um 12:51:58 UTC+1:
>
>>> Which position is the first to get the "fraction" oo? Which column is the first to be conquered by oo's completely? Note that counting is the topic here. Counting excludes that all positions become oo simultaneously.
>> You still haven't grasped the concept of a limit.
>
> There is no limit when counting. Cantor does not count ω. This "concept" is only the excuse for not knowing the dark numbers which should be counted before any limit, but cannot.
>
> However, your splendid idea
>
> 1/1, oo, oo, oo, oo, oo, oo, ...
> 1/2, oo, oo, oo, oo, oo, oo, ...
> 2/1, oo, oo, oo, oo, oo, oo, ...
> 1/3, oo, oo, oo, oo, oo, oo, ...
> 2/2, oo, oo, oo, oo, oo, oo, ...
> 3/1, oo, oo, oo, oo, oo, oo, ...
> ..., ..., ..., ..., ..., ..., ...
>
> proves that for every place in every column except the first one you cannot find out the last ℵo fractions leaving there. For all these ℵo places we get ℵo unknown (i.e. dark) fractions because for all indexes k of definable fractions we have
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .
>
> Congratulations to your first proof of dark numbers!
>
> Regards, WM

Your matrix switcharoo TRANSswapping is degenerated beyond dumb and dumber, its dumber and dumbist.

When do this ideas come to you ? in the early morning when you are taking a dump?

On 2/27/2022 4:21 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:
>
>>> It has no limit. It vanishes in darkness.
>> Why does it vanish? 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. Cantor completes the indexing, so he believes at least. But that is impossible.

no Cantor is correct.

>You can read it here: https://www.hs-augsburg.de/~mueckenh/Transfinity/ choose Countability Contradicted. One page.

that is a troll book, written by a crank.

>
> Regards, WM

On Sunday, February 27, 2022 at 11:15:59 AM UTC+1, WM wrote:

> [...] This shows that never all fractions can be indexed. In appealing and easily understandable form you can read it here: 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?

"but it is not." Indeed! And no one (except you) claimed that it should.

EOD

Fritz Feldhase schrieb am Sonntag, 27. Februar 2022 um 21:20:03 UTC+1:
> On Sunday, February 27, 2022 at 11:15:59 AM UTC+1, WM wrote:
>
> > [...] This shows that never all fractions can be indexed. In appealing and easily understandable form you can read it here: 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?

No, 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, ...
it is possible and 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, ...
.... ... ... ...

There is simply no obstacle and there is no reason why there should appear an obstacle.
>
> "but it is not."

It is! Find the first failure. Fail.

Regards, WM

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.

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.

And we KNOW, since we can PROVE it, that card(IN) = card(Q) and card(IN) < card(IR).

> No.

Right.

> 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.

Yeah. But WM is all talk, no reasoning.

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?

On 2/27/2022 1:21 AM, FromTheRafters wrote:
> Chris M. Thomasson laid this down on his screen :
>> On 2/15/2022 1:03 AM, FromTheRafters wrote:
>>> Chris M. Thomasson submitted this idea :
>>>> On 2/13/2022 3:15 AM, WM wrote:
>>>>> Chris M. Thomasson schrieb am Samstag, 12. Februar 2022 um 10:15:42
>>>>> UTC+1:
>>>>>
>>>>>> think of Cantor Pairing creating an infinite number of single numbers
>>>>>> from two numbers, and vise versa.
>>>>>
>>>>> All positive fractions m/n are contained in the matrix
>>>>>
>>>>> 1/1, 1/2, 1/3, 1/4, ...
>>>>> 2/1, 2/2, 2/3, 2/4, ...
>>>>> 3/1, 3/2, 3/3, 3/4, ...
>>>>> 4/1, 4/2, 4/3, 4/4, ...
>>>>> 5/1, 5/2, 5/3, 5/4, ...
>>>>> ...
>>>>>
>>>>> Cantor claims that all can be indexed by natural numbers, i.e.,
>>>>> written in a sequence, when he uses the sequence
>>>> [...]
>>>>
>>>> The set of natural numbers is infinite. Therefore, it can be used to
>>>> index any other infinite set. Period.
>>>
>>> Countably infinite.
>>
>> It sure seems like every infinite set can be counted wrt iteration, on
>> a per-step basis.
>
> How do you interate the reals? I give you a real number and you go look
> for the successor to that number and label it with its natural index?

I can take the floor and ceiling of the real number to get a range, and
iterate the range using any scale I want. This would have to be iterated
infinitely many times to get at every real in between them.

> If I give you the eighth real number (or whichever I say it is) can you
> give me the ninth or next real number?

Give me r = 3.17664238, okay:

floor(r) = 3
ceil(r) = 4

We have a range that we have to iterate through. When we find
3.17664238, we can say that the next one is:

3.17664238 + .00000001 = 3.17664239

At that scale. Its basically nested infinities. We have to start a new
range between:

3.17664238 and 3.17664239

Start again, for infinity. So, it depends on how much precision you want
to get wrt a finite answer. Now, _all_ of these steps can be indexed by
the natural numbers, because they are infinite in nature as well...

On 2/27/2022 2:15 AM, WM wrote:
> FromTheRafters schrieb am Sonntag, 27. Februar 2022 um 10:30:47 UTC+1:
>
>> Fuggetaboutit -- As Fritz Feldhase pointed out, I missed that WM was
>> talking about a superset of positive rationals
>
> That is what Cantor did
> k = (m + n - 1)(m + n - 2)/2 + m
> 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, ...
>
> I only mimic him. But instead of bringing the naturals to the serpent I bring the serpent to the naturals.

> This shows that never all fractions can be indexed.

They can all be indexed by the naturals because the naturals are
infinite in nature.

In appealing and easily understandable form you can read it here.
https://www.hs-augsburg.de/~mueckenh/Transfinity/ choose Countability
Contradicted. It is only one page, smashing 150 years of set theory.
>
> Regards, WM
>
>
>

On 2/27/2022 3:49 PM, Chris M. Thomasson wrote:
> On 2/27/2022 1:21 AM, FromTheRafters wrote:
>> Chris M. Thomasson laid this down on his screen :
>>> On 2/15/2022 1:03 AM, FromTheRafters wrote:
>>>> Chris M. Thomasson submitted this idea :
>>>>> On 2/13/2022 3:15 AM, WM wrote:
>>>>>> Chris M. Thomasson schrieb am Samstag, 12. Februar 2022 um
>>>>>> 10:15:42 UTC+1:
>>>>>>
>>>>>>> think of Cantor Pairing creating an infinite number of single
>>>>>>> numbers
>>>>>>> from two numbers, and vise versa.
>>>>>>
>>>>>> All positive fractions m/n are contained in the matrix
>>>>>>
>>>>>> 1/1, 1/2, 1/3, 1/4, ...
>>>>>> 2/1, 2/2, 2/3, 2/4, ...
>>>>>> 3/1, 3/2, 3/3, 3/4, ...
>>>>>> 4/1, 4/2, 4/3, 4/4, ...
>>>>>> 5/1, 5/2, 5/3, 5/4, ...
>>>>>> ...
>>>>>>
>>>>>> Cantor claims that all can be indexed by natural numbers, i.e.,
>>>>>> written in a sequence, when he uses the sequence
>>>>> [...]
>>>>>
>>>>> The set of natural numbers is infinite. Therefore, it can be used
>>>>> to index any other infinite set. Period.
>>>>
>>>> Countably infinite.
>>>
>>> It sure seems like every infinite set can be counted wrt iteration,
>>> on a per-step basis.
>>
>> How do you interate the reals? I give you a real number and you go
>> look for the successor to that number and label it with its natural
>> index?
>
> I can take the floor and ceiling of the real number to get a range, and
> iterate the range using any scale I want. This would have to be iterated
> infinitely many times to get at every real in between them.
>
>> If I give you the eighth real number (or whichever I say it is) can
>> you give me the ninth or next real number?
>
> Give me r = 3.17664238, okay:
>
> floor(r) = 3
> ceil(r) = 4
>
> We have a range that we have to iterate through. When we find
> 3.17664238, we can say that the next one is:
>
> 3.17664238 + .00000001 = 3.17664239
>
> At that scale. Its basically nested infinities. We have to start a new
> range between:
>
> 3.17664238 and 3.17664239

Basically, this seems very similar to arithmetic encoding. ;^)

>
> Start again, for infinity. So, it depends on how much precision you want
> to get wrt a finite answer. Now, _all_ of these steps can be indexed by
> the natural numbers, because they are infinite in nature as well...

On 2/27/2022 3:34 PM, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 27. Februar 2022 um 21:20:03 UTC+1:
>> On Sunday, February 27, 2022 at 11:15:59 AM UTC+1, WM wrote:
>>
>>> [...] This shows that never all fractions can be indexed. In appealing and easily understandable form you can read it here: 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?
>
> No, 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, ...

there is no reason at all for your matrixs below, none at all.

Cantors enumeration is already done, you listed it in the line above.

the rest of what you post is diversion

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

No, there is no reason to do any of that at all. The only reason you do it is to obscure and confuse.

>>
>> "but it is not."
>
> It is! Find the first failure. Fail.

It is your continued failure to disprove Cantor's Enumeration. And no wonder, Cantors Enumeration is clear, simple, and bullet proof.

>
> Regards, WM
>

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? You are asking me to find a sklroskrli and won't even define what it is!

> > >Dark numbers are real if infinity is actual.
> > Yet you cannot define "dark" in a meaningful mathematical way.
> Dark numbers are not defined. You cannot subtract all natural numbers individually

WHAT DOES THAT MEAN!?

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

WHAT IS N_DEF!? You haven't defined it meaningfully!

> But you can collectively.
> ℕ \ {1, 2, 3, ...} = { }.
> That's the difference.

Of what?

You are doing different set subtractions that is meaningless!

> > >Here you may find the most comprehensive essay about dark numbers presently available: https://www.hs-augsburg.de/~mueckenh/Transfinity/ choose Dark Numbers.
> > it is a nonsense book
> It is not a book but only few pages.

It is all garbage.

> > >
> > > Further discussion will be irrelevant. Either you can understand or you you cannot.
> Obviously you cannot.

I can understand mathematics, I do not understand your nonsense because it is NONSENSE!

>
> Regards, WM

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?