On 9/4/2021 5:23 PM, Chris M. Thomasson wrote:
> On 9/4/2021 3:20 PM, Gus Gassmann wrote:
>> On Saturday, 4 September 2021 at 18:15:29 UTC-3, WM wrote:
>>> markus...@gmail.com schrieb am Samstag, 4. September 2021 um 16:49:16 UTC+2:
>>>> lördag 4 september 2021 kl. 15:19:53 UTC+2 skrev WM:
>>>
>>>>> We don't have two sets of indices but only one. That is of relevance. This set is exhausted by indexing the first fractions of the unit intervals.
>>>>>
>>>> You have to be a bit more clear. We can construct a bijection between (n, n+1] and N. This gives a bijection from Q to N.
>>> You believe that you can construct the bijection. But you forget or have never realized that most nunbers are dark. This is proven here:
>>>
>>> Before ( in the sequential sense of 1, 2, 3, ...) all fractions of all unit intervals have been enumerated, at least one fraction must have been
>>> enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available.
>>
>> OK. At least you now use the word "before" correctly in your statement. However, your conclusion is still gibberish and belies a total and complete
>> failure to grasp Cantor's argument, which bijects the product space {1, 2, 3...} x (1, 2, 3, ...} (or IN x IN, if you prefer) with the set {1, 2, 3,
>> ...} = IN. The whole point of the diagonal argument (obviously utterly lost on you and clearly no longer within reach of your demented mind) is that
>> the enumeration works on *ALL* intervals simultaneously, albeit at different speeds. In time, *EVERY* rational number in *EVERY* unit interval (n-1,
>> n] will have been mapped, and more is not required.
>>
>> But a guy who is delusional enough to call Galileo's enumeration of the square numbers "inefficient", obviously can't be expected to process any of
>> the previous paragraph. You'd be better off putting all your poofs into a pipe and smoke them...
>>
>
> Here it is:
>
> https://youtu.be/xrsbjjuDTzU
>
> lol!

the follow up where they discover they are parked
https://www.youtube.com/watch?v=ZEavqjHWOv0

On 9/4/2021 4:40 PM, Serg io wrote:
> On 9/4/2021 5:23 PM, Chris M. Thomasson wrote:
>> On 9/4/2021 3:20 PM, Gus Gassmann wrote:
>>> On Saturday, 4 September 2021 at 18:15:29 UTC-3, WM wrote:
>>>> markus...@gmail.com schrieb am Samstag, 4. September 2021 um 16:49:16 UTC+2:
>>>>> lördag 4 september 2021 kl. 15:19:53 UTC+2 skrev WM:
>>>>
>>>>>> We don't have two sets of indices but only one. That is of relevance. This set is exhausted by indexing the first fractions of the unit intervals.
>>>>>>
>>>>> You have to be a bit more clear. We can construct a bijection between (n, n+1] and N. This gives a bijection from Q to N.
>>>> You believe that you can construct the bijection. But you forget or have never realized that most nunbers are dark. This is proven here:
>>>>
>>>> Before ( in the sequential sense of 1, 2, 3, ...) all fractions of all unit intervals have been enumerated, at least one fraction must have been
>>>> enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available.
>>>
>>> OK. At least you now use the word "before" correctly in your statement. However, your conclusion is still gibberish and belies a total and complete
>>> failure to grasp Cantor's argument, which bijects the product space {1, 2, 3...} x (1, 2, 3, ...} (or IN x IN, if you prefer) with the set {1, 2, 3,
>>> ...} = IN. The whole point of the diagonal argument (obviously utterly lost on you and clearly no longer within reach of your demented mind) is that
>>> the enumeration works on *ALL* intervals simultaneously, albeit at different speeds. In time, *EVERY* rational number in *EVERY* unit interval (n-1,
>>> n] will have been mapped, and more is not required.
>>>
>>> But a guy who is delusional enough to call Galileo's enumeration of the square numbers "inefficient", obviously can't be expected to process any of
>>> the previous paragraph. You'd be better off putting all your poofs into a pipe and smoke them...
>>>
>>
>> Here it is:
>>
>> https://youtu.be/xrsbjjuDTzU
>>
>> lol!
>
>
>
> the follow up where they discover they are parked
> https://www.youtube.com/watch?v=ZEavqjHWOv0
>

ROFL!!!!

On Saturday, September 4, 2021 at 5:50:50 PM UTC-4, Dan Christensen wrote:
> On Saturday, September 4, 2021 at 5:15:29 PM UTC-4, WM wrote:
> > markus...@gmail.com schrieb am Samstag, 4. September 2021 um 16:49:16 UTC+2:
> > > lördag 4 september 2021 kl. 15:19:53 UTC+2 skrev WM:
> >
> > > > We don't have two sets of indices but only one. That is of relevance. This set is exhausted by indexing the first fractions of the unit intervals.
> > > >
> > > You have to be a bit more clear. We can construct a bijection between (n, n+1] and N. This gives a bijection from Q to N.
> > You believe that you can construct the bijection. But you forget or have never realized that most nunbers are dark.
> There are no "dark numbers," Mucke. Recall that, by YOUR OWN formal definition, EVERY natural number is defined.

In case you forgot, you wrote:

"ℕ_def = {k ∈ ℕ : |∩{E(1), E(2), ..., E(k)}|= ℵo}"
--sci.math 2021-08-11

So, after decades of fruitless effort, you STILL have no inconsistency in ZFC Must be frustrating as hell for you!
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Gus Gassmann schrieb am Sonntag, 5. September 2021 um 00:20:51 UTC+2:
> On Saturday, 4 September 2021 at 18:15:29 UTC-3, WM wrote:

> > Before ( in the sequential sense of 1, 2, 3, ...) all fractions of all unit intervals have been enumerated, at least one fraction must have been enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available.
> OK. At least you now use the word "before" correctly in your statement. However, your conclusion is still gibberish and belies a total and complete failure to grasp Cantor's argument, which bijects the product space {1, 2, 3...} x (1, 2, 3, ...} (or IN x IN, if you prefer) with the set {1, 2, 3, ....} = IN.

That is irrelevant since Cantor here uses only {1, 2, 3, ...}.

> The whole point of the diagonal argument is that the enumeration works on *ALL* intervals simultaneously, albeit at different speeds. In time, *EVERY* rational number in *EVERY* unit interval (n-1, n] will have been mapped, and more is not required.

Firstly, this is wrong, since the lower part of the matrix is never reached, but secondly it is irrelevant, since I show that Cantor would have used up all the indices for the first rationals of all intervals already.
>
> But a guy who is delusional enough to call Galileo's enumeration of the square numbers "inefficient",

Galilei had no grasp of the difference between potential and actual infinity. His treatment of the problem is as insufficient as would be Newton's treatment of a quantum computer.

Regards, WM

Jim Burns schrieb am Sonntag, 5. September 2021 um 00:58:57 UTC+2:
> On 9/4/2021 5:07 PM, WM wrote:
> > Jim Burns schrieb
> > am Samstag, 4. September 2021 um 21:21:26 UTC+2:
> >> One consequence of these claims is that,
> >> before all rationals have been enumerated, at least one
> >> unit interval does not have any rationals enumerated in it.
> >
> > Immediately before all rationals have been enumerated,
> I see that you have accidentally typed "Immediately" in front of
> what I actually claim.wrote you that

So you did not mean immediately? Why then wrote you that sentence? Of course there are many intervals not enumeretaed at the beginning. That is trivial and not of interest.
>
> Before all rationals have been enumerated, *anywhere* before
> all rationals have been enumerated, there are infinitely-many
> rationals which aren't enumerated. There is no "immediately before".

Of course there are many rationals not enumerated. But Cantor and you claim that there is an instance when all rationals have been enumerated. Hence there must be an instance way before, where all intervals have at least one rational enumerated.
>
> This is a consequence of there being infinitely-many rationals,
> each with a finite index.

Cantor's and your claim are wrong. This is a consequence of there being infinitely many rationals. Neverftheless you will continue to claim that all will be enumerated at some instance. But you will deny that before that has happened, all intervals will have received at least one hit? Then my "immediately" was appropriate.

> > Immediately before all rationals have been enumerated,
> > at least one unit interval does not have any rationals
> > enumerated in it?

> No. There is no "immediately before".

There is a sequence. Each index can well be distinguished.

> Your confusion is probably caused by assuming that I also
> claim that there is a last natural number.
> I don't claim that.

I don't claim that. But you claim that the indexing is completed somewhere.

> > I say:
> > Between the complete enumeration of all rationals of all intervals
> > and the enumeration of at least one rational in every interval
> > there are infinitely many further steps.

> I'll take a moment to point out that, while you say things,
> I and others prove things.

My statement is easily proved: Every interval has aleph_0 rationals. So for every interval between enumerating the first rational and completing it, there are aleph_0 steps of the sequence.
>
> There are infinitely many unit intervals (k,k+1].
>
> If at least one rational in each of infinitely-many intervals
> has been indexed, each finitely-indexed rational has been
> indexed.

So it is.
>
> All rationals are finitely-indexed rationals.

That is refuted by my proof.
>
> If at least one rational in each of infinitely-many intervals
> has been indexed, each rational has been indexed.

Impossible, because then aleph_0 indices have been applied, but at least one interval has been not enumerated completely.
>
> There are no steps, zero, none between the complete enumeration of
> all rationals of all intervals and the enumeration of at least
> one rational in every interval.

Herewith you prove that it is impossible to enumerate all rationals which lie between the first enumerated of every interval and the completion.
> Such a step would require a finitely-indexed rational to
> come after infinitely-many rationals.

Now you've got it! Yes, that is the core of my proof. in order to avoid it you prove that it is impossible to enumerate all rationals which lie between the first enumerated of every interval and the completion.
>
> Please review "finite", "infinite", and "index".

Why? I know that. Otherwise I would not have been able to refute Cantor's claim. But you can't believe it, can you?

Regards, WM

Serg io schrieb am Sonntag, 5. September 2021 um 01:36:34 UTC+2:

> r * ℵ0 = ℵ0 for r =/= 0 where r is your number of intervals, more is always available.

Cantor applied 1*ℵo. That is what I refuted.

Regrads, WM

On Sunday, 5 September 2021 at 09:07:53 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 5. September 2021 um 00:20:51 UTC+2:
> > On Saturday, 4 September 2021 at 18:15:29 UTC-3, WM wrote:
>
> > > Before ( in the sequential sense of 1, 2, 3, ...) all fractions of all unit intervals have been enumerated, at least one fraction must have been enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available.
> > OK. At least you now use the word "before" correctly in your statement. However, your conclusion is still gibberish and belies a total and complete failure to grasp Cantor's argument, which bijects the product space {1, 2, 3...} x (1, 2, 3, ...} (or IN x IN, if you prefer) with the set {1, 2, 3, ...} = IN.
> That is irrelevant since Cantor here uses only {1, 2, 3, ...}.
As usual you have zero understanding of the issues. A positive fraction is a ratio of two integers, and the correspondence between n/m (with n,m > 0) and the ordered pair (n,m) (which *is* an element of IN x IN ought to be even within your limited grasp.

> > The whole point of the diagonal argument is that the enumeration works on *ALL* intervals simultaneously, albeit at different speeds. In time, *EVERY* rational number in *EVERY* unit interval (n-1, n] will have been mapped, and more is not required.
>
> Firstly, this is wrong, since the lower part of the matrix is never reached,
I have no idea what image you have in your birdsized brain when you talk about the "lower" (and presumably "upper") part of a matrix not previously mentioned.
> but secondly it is irrelevant, since I show that Cantor would have used up all the indices for the first rationals of all intervals already.
This idiotic remark is exactly why I said that you have no clue how the diagonal method works. The unit intervals are *NOT* enumerated one at a time. I believe you even knew this at a time when you made your density argument. So you are either too dense to remember that or lying in the hopes of getting away with a knowingly false statement. (So maybe you *DO* know how the diagonal method works but pretend not to; it makes no difference.) Either way, this point is no longer worth discussing.
> > But a guy who is delusional enough to call Galileo's enumeration of the square numbers "inefficient",
> Galilei had no grasp of the difference between potential and actual infinity.

He may not have stated it explicitly, but this passage from "Two New Sciences" (taken from Wikipedia) is pretty clear on the subject:

Salviati: So far as I see we can only infer that the totality of all numbers is infinite, that the number of squares is infinite, and that the number of their roots is infinite; neither is the number of squares less than the totality of all the numbers, nor the latter greater than the former; and finally the attributes "equal," "greater," and "less," are not applicable to infinite, but only to finite, quantities. When therefore Simplicio introduces several lines of different lengths and asks me how it is possible that the longer ones do not contain more points than the shorter, I answer him that one line does not contain more or less or just as many points as another, but that each line contains an infinite number.

There are infinitely many positive integers, and there are infinitely many squares, and he talks about the totality of *ALL THE [positive] NUMBERS*. This is very clearly completed, i.e., actual, infinity. (Note also that this is the *LAST* book that he was working on and the ideas expressed therein presumably supersede any previous contradictory opinions.) That you, as a self-proclaimed "expert" on the History of the Infinite, do not understand this, is telling. At this point you should not be left into a toilet unsupervised; you might be inclined to think it is for drinking.

On 9/5/2021 7:32 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 5. September 2021 um 00:58:57 UTC+2:
>> On 9/4/2021 5:07 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 4. September 2021 um 21:21:26 UTC+2:
>>>> One consequence of these claims is that,
>>>> before all rationals have been enumerated, at least one
>>>> unit interval does not have any rationals enumerated in it.
>>>
>>> Immediately before all rationals have been enumerated,
>> I see that you have accidentally typed "Immediately" in front of
>> what I actually claim.wrote you that
>
> So you did not mean immediately? Why then wrote you that sentence? Of course there are many intervals not enumeretaed at the beginning. That is trivial and not of interest.
>>
>> Before all rationals have been enumerated, *anywhere* before
>> all rationals have been enumerated, there are infinitely-many
>> rationals which aren't enumerated. There is no "immediately before".
>
> Of course there are many rationals not enumerated. But Cantor and you claim that there is an instance when all rationals have been enumerated. Hence there must be an instance way before, where all intervals have at least one rational enumerated.

wrong, that is your conjecture, you do not specify an interval size either.

You need to Prove that first, "there must be an instance way before, where all intervals have at least one rational enumerated."

On 9/5/2021 7:34 AM, WM wrote:
> Serg io schrieb am Sonntag, 5. September 2021 um 01:36:34 UTC+2:
>
>> r * ℵ0 = ℵ0 for r =/= 0 where r is your number of intervals, more is always available.
>
> Cantor applied 1*ℵo. That is what I refuted.
>
> Regrads, WM
>

no, you stated that "More are not available." which is false on its face.

This is what you snipped and are trying to hide;

"... at least one fraction must have been enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available."

On Sunday, September 5, 2021 at 4:03:40 PM UTC+2, Serg io wrote:
> On 9/5/2021 7:34 AM, WM wrote:
> > Serg io schrieb am Sonntag, 5. September 2021 um 01:36:34 UTC+2:
> >
> This is what you snipped and are trying to hide:
>
> "... at least one fraction must have been enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available." [WM]

I guess, the reason for this is that his claim is nonsensical. For example the fractions n/1 might be "enumerated" by, say, the natural numbers 2, 4, 6, 8, ... Then at least one fraction is enumerated in every unit fraction. And right, for this sake "ℵo natural numbers are required". But there are STILL ℵo natural numbers "available", namely 1, 3, 5, 7, ..... Hence WM's claim is WRONG.

On 9/5/2021 8:32 AM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 5. September 2021 um 00:58:57 UTC+2:
>> On 9/4/2021 5:07 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 4. September 2021 um 21:21:26 UTC+2:

>>>> One consequence of these claims is that,
>>>> before all rationals have been enumerated, at least one
>>>> unit interval does not have any rationals enumerated in it.
>>>
>>> Immediately before all rationals have been enumerated,
>>
>> I see that you have accidentally typed "Immediately" in front of
>> what I actually claim.wrote you that
>
> So you did not mean immediately? Why then wrote you that
> sentence?

<WM>
| Before all fractions have been enumerated there, at least one
| fraction must have been enumerated in every unit interval.

<JB>
| Before all rationals have been enumerated, at least one
| unit interval does not have any rationals enumerated in it.

Do you want me to explain the difference between these claims?

> Of course there are many intervals not enumeretaed at the
> beginning. That is trivial and not of interest.

Before all rationals have been enumerated, *anywhere* before
all rationals have been enumerated, there are infinitely-many
rationals which aren't enumerated. There is no "immediately before".

>> Before all rationals have been enumerated, *anywhere* before
>> all rationals have been enumerated, there are infinitely-many
>> rationals which aren't enumerated. There is no "immediately before".
>
> Of course there are many rationals not enumerated.

For each p in Q+, there are j,k in N+, j/k = p.
For each j,k in N+, there are steppable {0,...,j}, {0,...,k}
in which, for adjacent h,i, i = h+1.

Each j, each k in N+ is countable to, in principle.

_All the positive rationals_ are in Q+

----
There are many schemes for indexing _all the positive rationals_

Cantor mapped each pair j,k to a single natural m.
m = j + (j+k-1)*(j+k-2)/2

That scheme maps each rational p to infinitely-many indexes m,
which seems more than sufficient for our purposes here.

Each j and each k in N+ is countable to, in principle.
Two naturals which are countable to, in principle,
have a sum which is a natural countable to, in principle.
Two naturals which are countable to, in principle,
have a product which is a natural countable to, in principle.
Since m = j + (j+k-1)*(j+k-2)/2
m is a natural countable to, in principle.

In Cantor's scheme, _all the positive rationals_ are
enumerated by an index countable to, in principle.
(Actually, each rational is enumerated infinitely-many times.)

> But Cantor and you claim that there is an instance when
> all rationals have been enumerated.

That's a point of contention between us. I have been trying to
work my way around that point to the conclusions I want.

Can we agree that there are instances when NOT all rationals
have been enumerated? Can we agree that that is what
"before all rationals have been enumerated" means?

> Hence there must be an instance way before, where all
> intervals have at least one rational enumerated.

You keep saying that, but it doesn't get any more correct.

Each rational has a finite index, countable to, in principle.
For each m, there is a steppable {0,...,m} in which,
for adjacent h,i, i = h+1.

For each rational, there are no more than finitely-many
rationals before it in the enumeration.
Finitely-many aren't enough for one in all unit intervals.
Finitely-many aren't enough to "complete" any unit interval.

( Cantor's way, each rational has infinitely-many indexes,
( which can make expressing things a bit tricky, a bit messy.
( ( This is why I prefer mapping the _unique prime factorization_
( of a rational to the unique prime factorization of its
( index. For each rational, a unique index, and, for each index
( a unique rational.
( ( Of course, Cantor's scheme has more historical resonance,
( for those who care about that sort of thing.

Before all the rationals have been enumerated, no more than
finitely-many rationals have been enumerated

>> This is a consequence of there being infinitely-many rationals,
>> each with a finite index.
>
> Cantor's and your claim are wrong.

Do you mean this claim?
| If steppable {0,...,j} and {0,...,k} exist in which,
| for adjacent h,i, i = h+1,
| then steppable {0,...,j+k} and {0,...,j*k} exist in which,
| for adjacent h,i, i = h+1.

Does your "intuition" disagree with that claim?
Do you want me to prove it for you?

> Cantor's and your claim are wrong.
> This is a consequence of there being infinitely many rationals.
> Neverftheless you will continue to claim that all will be
> enumerated at some instance. But you will deny that before
> that has happened, all intervals will have received at least
> one hit?

Before that has happened, no more than finitely-many rationals
have been enumerated. So, yes, I deny it.

> Then my "immediately" was appropriate.

Your "immediately" is incorrect. There is no "immediately".
You claim there is.

What was _inappropriate_ was your putting _your_ claim in
_my_ mouth. One might think that someone holding onto his
classroom because of "academic freedom" would understand that.

Of course that assumes a lot about the honesty and intelligence
of such a person.

>>> Immediately before all rationals have been enumerated,
>>> at least one unit interval does not have any rationals
>>> enumerated in it?
>
>> No. There is no "immediately before".
>
> There is a sequence. Each index can well be distinguished.

Only a last enumerated rational can be immediately before
all the rationals being enumerated. There isn't any.
There is no "immediately before".

>> Your confusion is probably caused by assuming that I also
>> claim that there is a last natural number.
>> I don't claim that.
>
> I don't claim that.
> But you claim that the indexing is completed somewhere.

I've avoided claiming that.
Whatever it means for the indexing to be completed somewhere,
it should be clear what it means for the indexing to NOT-be
complete. As in: "before all rationals have been enumerated".

Before all rationals have been enumerated, some rational and
all the rationals enumerated after it have not been enumerated.

Before all rationals have been enumerated, no more than
finitely-many rationals have been enumerated.

Before all rationals have been enumerated, no unit interval
has had all of its rationals enumerated.

Before all rationals have been enumerated, not all
unit intervals have at least one rational enumerated.

Jim Burns schrieb am Sonntag, 5. September 2021 um 20:32:21 UTC+2:

> Before all rationals have been enumerated, not all
> unit intervals have at least one rational enumerated.

So, do you claim that all rationals have been enumerated before every interval has at least one rational enumerated?

Regards, WM

Greg Cunt schrieb am Sonntag, 5. September 2021 um 16:32:42 UTC+2:

> > "... at least one fraction must have been enumerated in every unit interval. For this sake already ℵo natural numbers are required. More are not available." [WM]
>
> I guess, the reason for this is that his claim is nonsensical. For example the fractions n/1 might be "enumerated" by, say, the natural numbers 2, 4, 6, 8, ...

Cantor enumerates by 1, 2, 3, ...

Do you claim that all rationals have been enumerated by 1, 2, 3, ... before every interval has at least one rational enumerated? Yes or no?

Regards, WM

Gus Gassmann schrieb am Sonntag, 5. September 2021 um 14:53:55 UTC+2:

> There are infinitely many positive integers, and there are infinitely many squares, and he talks about the totality of *ALL THE [positive] NUMBERS*. This is very clearly completed, i.e., actual, infinity.

Well, fine, irrelevant.

Do you understand that *before* having enumerated all positive fractions at least one fraction must have been enumerated in every unit interval (n-1, n], n ∈ ℕ?

Regards, WM

On 9/5/2021 4:14 PM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 5. September 2021 um 20:32:21 UTC+2:

>> Before all rationals have been enumerated, not all
>> unit intervals have at least one rational enumerated.
>
> So, do you claim that all rationals have been enumerated
> before every interval has at least one rational enumerated?

No infinite subset of finite indexes is all enumerated before
any other infinite subset of finite indexes is all enumerated.

For example,
is all of Primes = {2,3,5,7,11,...} enumerated
before all of Squares = {1,4,9,16,25,...} ?

No, because there is no square in Squares after all
the primes in Primes.

Is all of Squares = {1,4,9,16,25,...} enumerated
before all of Primes = {2,3,5,7,11,...} ?

No, because there is no prime in Primes after all
the squares in Squares.

Therefore, no,
all (infinitely-many) rationals will NOT have been enumerated
before every (infinitely-many) interval has at least one
rational enumerated.

Notice that "there is no ... after all ..." does not
refer to anything infinite, just primes and squares,
finite things which are all followed by more finite things.

On Sunday, 5 September 2021 at 17:20:56 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 5. September 2021 um 14:53:55 UTC+2:
>
> > There are infinitely many positive integers, and there are infinitely many squares, and he talks about the totality of *ALL THE [positive] NUMBERS*. This is very clearly completed, i.e., actual, infinity.
> Well, fine, irrelevant.
After you removed the context, of course. Can't admit even once that you may have been even a smidgen wrong? What a pathetic, narcissistic prick!

> Do you understand that *before* having enumerated all positive fractions at least one fraction must have been enumerated in every unit interval (n-1, n], n ∈ ℕ?

Why should *anyone* subscribe to your deluded version of reality?

Crank Wolfgang Mueckenheim, aka WM wrote:
> Jim Burns schrieb am Sonntag, 5. September 2021 um 20:32:21 UTC+2:
>
>> Before all rationals have been enumerated, not all
>> unit intervals have at least one rational enumerated.
>
> So, do you claim that all rationals have been enumerated before every interval has at least one rational enumerated?

This is not Jim's claim. This small exchange is, nevertheless, a
blatant illustration of both a) your inability to deal with any
kind of simple logical statements and b) your profound intellectual
dishonesty, Crank Wolfgang Mueckenheim, from Hochschule Augsburg.

On 9/5/2021 3:14 PM, WM wrote:
> Jim Burns schrieb am Sonntag, 5. September 2021 um 20:32:21 UTC+2:
>
>> Before all rationals have been enumerated, not all
>> unit intervals have at least one rational enumerated.
>
> So, do you claim that all rationals have been enumerated before every interval has at least one rational enumerated?
>
> Regards, WM
>

Why did you snip out the word "Before", Miss Leader ? Munging peoples replies! Shame!

On 9/5/2021 4:37 PM, Gus Gassmann wrote:
> On Sunday, 5 September 2021 at 17:20:56 UTC-3, WM wrote:
>> Gus Gassmann schrieb am Sonntag, 5. September 2021 um 14:53:55 UTC+2:
>>
>>> There are infinitely many positive integers, and there are infinitely many squares, and he talks about the totality of *ALL THE [positive] NUMBERS*. This is very clearly completed, i.e., actual, infinity.
>> Well, fine, irrelevant.
>
> After you removed the context, of course. Can't admit even once that you may have been even a smidgen wrong? What a pathetic, narcissistic prick!
>
>> Do you understand that *before* having enumerated all positive fractions at least one fraction must have been enumerated in every unit interval (n-1, n], n ∈ ℕ?
>
> Why should *anyone* subscribe to your deluded version of reality?
>

An Enumerated Fractional Ant in Every unit interval Pot

>Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.

We can prove it for ALL rational numbers. So you are wrong.

zelos...@gmail.com schrieb am Montag, 6. September 2021 um 07:07:17 UTC+2:
> >Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.
>
> We can prove it for ALL rational numbers.

You believe that all fractions including all n/1 for n > 1 will become enumerated. They belong to the bijection, don't they? Would you also think so if we changed the bijection slightly such that all fractions n/1 for n > 1 come first? If all are get indexed anhow, why not first index them?

Do you accept that *before* having enumerated all positive fractions at least one fraction must be enumerated in every unit interval (n-1, n], n ∈ ℕ?

Regards, WM

Jim Burns schrieb am Sonntag, 5. September 2021 um 22:42:10 UTC+2:
> On 9/5/2021 4:14 PM, WM wrote:
> > Jim Burns schrieb
> > am Sonntag, 5. September 2021 um 20:32:21 UTC+2:
>
> >> Before all rationals have been enumerated, not all
> >> unit intervals have at least one rational enumerated.
> >
> > So, do you claim that all rationals have been enumerated
> > before every interval has at least one rational enumerated?
> No infinite subset of finite indexes is all enumerated before
> any other infinite subset of finite indexes is all enumerated.

So you cannot analyze the enumeration between having indexed every interval at least once and having enumerated all intervals completely? That is the more deplorable as only finite natural numbers are used for indexing. Where does the visible part of the sequence 1, 2, 3, ... cease?

> Therefore, no,
> all (infinitely-many) rationals will NOT have been enumerated
> before every (infinitely-many) interval has at least one
> rational enumerated.
>
> Notice that "there is no ... after all ..." does not
> refer to anything infinite, just primes and squares,
> finite things which are all followed by more finite things.

For intervals this is very clear. All are finite and all are in a well-order.
And here we have an "after all": It is impossible to complete the enumeration of all intervals unless all intervals have been deflowered before. That is a very simple chain of cause and result.

Bet perhaps you can imagine a slight correction of Cantor's sequence

1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...

Since you believe that all fractions are indexed, also all fractions n/1 for n > 0 will be indexed. Therefore it would do no harm if we indexed first als fractions of the form n/1 and then the remaining fractions. Note that this is what Cantor does, if he covers all these fractions. He does it only slightly later in order to confuse his audience. Every set theorist knows that my proposal will fail. But when these fractions are better hidden in the sequence, nobody will notice.

Of course your agument with primes and squares is of same kind. If he really indexed all primes, then indexing them first would not change the whole story.

Fight under the banner of confusion!

Regards, WM

On 9/6/2021 7:56 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 5. September 2021 um 22:42:10 UTC+2:
>> On 9/5/2021 4:14 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Sonntag, 5. September 2021 um 20:32:21 UTC+2:
>>
>>>> Before all rationals have been enumerated, not all
>>>> unit intervals have at least one rational enumerated.
>>>
>>> So, do you claim that all rationals have been enumerated
>>> before every interval has at least one rational enumerated?
>> No infinite subset of finite indexes is all enumerated before
>> any other infinite subset of finite indexes is all enumerated.
>
> So you cannot analyze the enumeration between having indexed every interval at least once and having enumerated all intervals completely?

Since you say are a Math Professor, you should be the one providing answers along with the showing or Proof. But you do not.

> That is the more deplorable as only finite natural numbers are used for indexing.

100% wrong, did you stop at k again ?

> Where does the visible part of the sequence 1, 2, 3, ... cease?

at the 'last' one, silly.

>
>> Therefore, no,
>> all (infinitely-many) rationals will NOT have been enumerated
>> before every (infinitely-many) interval has at least one
>> rational enumerated.
>>
>> Notice that "there is no ... after all ..." does not
>> refer to anything infinite, just primes and squares,
>> finite things which are all followed by more finite things.
>
> For intervals this is very clear. All are finite and all are in a well-order.

wrong , you have brain fart. Intervals are infinite

> And here we have an "after all": It is impossible to complete the enumeration of all intervals unless all intervals have been deflowered before. That is a very simple chain of cause and result.

wrong, and wrong. Your flailing is noted

>
> Bet perhaps you can imagine a slight correction of Cantor's sequence
>
> 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...
>
> Since you believe that all fractions are indexed, also all fractions n/1 for n > 0 will be indexed. Therefore it would do no harm if we indexed first als fractions of the form n/1 and then the remaining fractions. Note that this is what Cantor does, if he covers all these fractions. He does it only slightly later in order to confuse his audience. Every set theorist knows that my proposal will fail. But when these fractions are better hidden in the sequence, nobody will notice.

no. Go back to the math and study, avoid jumping to conclusions

>
> Of course your agument with primes and squares is of same kind. If he really indexed all primes, then indexing them first would not change the whole story.

red herring

>
> Fight under the banner of confusion!

>
> Regards, WM
>

On 9/6/2021 7:40 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 6. September 2021 um 07:07:17 UTC+2:
>>> Cantor claims that all positive fractions can be enumerated although he can prove this only for the first (less than ℵo) fractions.
>>
>> We can prove it for ALL rational numbers.
>
> You believe that all fractions including all n/1 for n > 1 will become enumerated. They belong to the bijection, don't they? Would you also think so if we changed the bijection slightly such that all fractions n/1 for n > 1 come first? If all are get indexed anhow, why not first index them?

go ahead, its your idea, show us how...

>
> Do you accept that *before* having enumerated all positive fractions at least one fraction must be enumerated in every unit interval (n-1, n], n ∈ ℕ?

no, you have not provided a proof. where is it ?

>
> Regards, WM
>

>You believe that all fractions

We KNOW it because it can be PROVEN.

>Would you also think so if we changed the bijection slightly such that all fractions n/1 for n > 1 come first? If all are get indexed anhow, why not first index them?

You need to prove that the new function is a bijection then.

When it comes to the question of cardinality the actual bijection is not relevant only that ONE exists.

>Do you accept that *before* having enumerated all positive fractions at least one fraction must be enumerated in every unit interval (n-1, n], n ∈ ℕ?

It is entirely fucking irrelevant. All that matters is that the bijection exist.

If we can prove that there is a surjection Q->N and another surjection N->Q we know there exists a bijection N->Q and nothing else is needed and we can EASILY create a surjection in either direction.