Synopsis

[WM]

If
{1, 2, 3, 4, 5, ...} <--> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
is a bijection (i.e. surjective and injective) and M is an infinite set, then
{1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}
is not a bijection (surjectivity is missing).

Hilbert's hotel works only in potential infinity. The fixed quantity (Cantor) of actualy infinite sets does not allow for expansion.

Between every definable natural number and ω there are infinitely many dark natnumbers
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
while between all natnumbers and ω there is nothing
|ℕ \ {1, 2, 3, ...}| = 0 .

(A) There are not more than four endsegments containing the natural numbers {4, 5, 6, ...}, namely E(1), E(2), E(3), and E(4). E(5) fails.
(B) There are not more than n endsegments containing the natnumbers {n, n+1, n+2, ...}.
(C) There are not more than finitely many infinite endsegments because the contents, infinitely many natural numbers in every infinite endsegment, cannot be used as indices and there are not two consecutive infinite sets in the natural order of ℕ possible. Hence the intersection of infinite endsegments is infinite as also inclusion monotony shows.

Regards, WM

[Gus Gassmann]

On Monday, 18 October 2021 at 05:25:41 UTC-3, WM wrote:
[Bullshit removed]

You're in fine form this morning. Nothing but bullshit, nonsense, non sequiturs. All this shit of yours has been repeatedly disproven and debunked. Collecting your garbage into one big pile of dung doesn't make it any better. Your cognitive powers are rapidly converging to zero.

[Wayde Ring]

but if Fully Vaccinated Colin Powell Has Died From COVID, how on earth
this imbecile became a 4 star general?? What is goeng on with that
country??

[Greg Cunt]

On Monday, October 18, 2021 at 10:25:41 AM UTC+2, WM wrote:

> [For each and every natural number n:] Between [n] and ω there are infinitely many [...] natnumbers

Indeed!

Proof: An e IN: card({m e IN: n < m < omega}) = aleph_0

> (B) There are not more than n endsegments containing the natnumbers {n, n+1, n+2, ...}.

I guess this should read (after all {n, n+1, n+2, ...} is a set, not "natnumbers"):

| (B) There are not more than n endsegments containing the natnumbers *in* {n, n+1, n+2, ...}.

Well, actually, there are no natural numbers in {n, n+1, n+2, ...}.

Remember, you claimed that "n isn't a natural number", as well as "n+1 is not a natural number too."

I guess that will hold for all elements in {n, n+1, n+2, ...}. Hence there are no natural numbers in {n, n+1, n+2, ...}.

[Timothy Golden]

It would be great to see you move on from this system.
For me, and I mean no proof by it, the usage of infinity versus the calculus usage of infinity within a limit is that the calculus works just fine on say a million segments; better on two million segments; and if possible through algebra perfectly on the formal limit, which allows the limit to dissolve. That said if back at a million segments the error in computation on say the integral of sine(x) from 0.3 to 0.4 is negligible, and even back at a hundred segments we had quite a good answer, then this need to address infinity so carefully goes away. By induction, if your argument does not work at n=100, and it does not work an 101, nor 102, then in all likelihood it does not work at infinity. I think induction can be applied inverse here. Nextly though the consequences of your theory seem not to be extensive. Possibly though you have gotten us thinking here. I have developed gray numbers; so named after your own dark numbers. That the continuum under epsilon/delta universally is gray is an acceptable feature. This then demolishes the mathematicians sense of perfection of number; say 4. When you state four upon the continuum to presume infinite precision is a deep mistake. In fact this value has terrible precision under the gray system. Rational values come under attack as do the irrational values; for both are values mixed with operators. This then reverts us back to the integer as fundamental and as we study decimal representations such as
4, 4.00001, 4.00002
we see that a mere dot separates these values from their integer counterparts. These are not rational values. There is no need to re-radix these values. These are gray values. Physicists use them all the time, and how the mathematician has settled into the 'real' value as fundamental is a fraud. There are operators and there are values. To claim that a value contains an operator is a conflicted position, such as the treatment by Dedekind of the square root of two and its interpretation. This awareness of operator versus value had to be disabled since the rational value was already known by mass hypnosis, which includes the division operator, which is not even a fundamental operator. While these statements are merely an interpretation they are as well a falsification of the philosophy of mathematics and its accumulated form.

The rational value, such as 1/3 is merely a radix three rewrite within decimal notation. This is a re-radix, and should we return to the radix form as fundamental, and take these thirds seriously then the representation of 1/3 is simply 0.1 within the radix three system, and this value will as well have the gray problem of precision. We work in the decimal system; our values even with decimal point in them are merely radix ten values from the basis 1234567890. There is no need to deny their integer quality. Simply remove the decimal place and replace it with another small whole value. We can compute the product without division now and that is good, for those who claim the rational value as fundamental have claimed division to be more fundamental than their product, and this structural failure is an instance of poor philosophy propagated for too many generations. It is proof of the human condition. Philosophy and mathematics as distinct are a false divide. Physics and the first two as well falsely divided. Academia needs these rooms yet their divisions as corrupt can be established as they readily divorce themselves from each other. From my position unification will not be had in a system which acts this way. Certainly there is room for a new philosophy yet to build it as an accumulation from the old system is a stale start. Polysign numbers are such an instance which confront all sects. How they could go overlooked for so long is beyond my understanding. Finally they have led me to a new interpretation (through operator theory) of the continuum that holds to physical correspondence. While the dissonance to standing systems can be argued as minor, there are formal falsifications that collapse much of mathematics. We can now face geometry as a modulo system. General dimensional terms have been engaged both geometrically and algebraically from such simple rules that beget support for emergent spacetime. As our representation of the continuum is likewise a modulo representation, then our work is likewise entrapped there.

Those who attempt to go beyond yet cannot utter an instance of which they speak have gone into a strange and endless territory. Minor ambiguities will form major blunders. We are all subject to this phenomenon. We are all subject to mass hypnosis way beyond any national identity. Our mimicry instinct develops our social abilities regardless of coherency. Certainly our native languages are evidence, and we have no ability to perceive those mechanisms of learning from within. They are automatic in nature. Often enough within analysis one misplaced assertion is merely followed by another weaker assertion such that the first goes obeyed. Within politics those offending some principal accuse their competitor of that very transgression. The weakness of the human form in my time and my nation is best instanced with Russiagate as perpetrated onto Donald Trump; an outsider to the political system; the first to make it through the gauntlet only to be set on fire; at least his pants. Not my favorite person, either. Yet the human minds that have been swayed by repetition without integrity and provably false tales make me ashamed to identify as on the left. By game theory the destruction of the republican party via Obama somewhat ensured the destruction of the democratic party as well. Now we sit in a time of realignment and how long will it be... and until what? Until our stomachs settle, I suppose? Until the US divides in concession just as the soviets did? When human identity tops out in nationality then the whole world loses. Our identity as citizens of the world will establish shortly.

We have a problem absorbing falsifications. Clearly the sheer repetition of the original argument has us habituated. Supposedly such entrenched positions are so strong that they are beyond falsification. Clearly however the scientific position has to declare all such problems open. Otherwise mathematics lands in the same ground as religion, and politics for that matter.

[Dan Christensen]

On Monday, October 18, 2021 at 4:25:41 AM UTC-4, WM wrote:

You recently wrote:

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

We have:

E(n) = {n, n+1, n+2, ... } for all n in N.

|E(n)|= ℵo for all n in N

∩{E(1), E(2), ..., E(k)} = E(k) for all k in N

|∩{E(1), E(2), ..., E(k)}| = |E(k)| = ℵo for all k in N.

Substituting in your "definition":

N_def = {k ∈ ℕ : |∩{E(1), E(2), ..., E(k)}|= ℵo} = {k ∈ ℕ : |E(k)}|= ℵo} = {k ∈ ℕ : ℵo = ℵo} = N.

N_def = N

Don't you feel silly? After all those years, you are reduced to this. All that wasted time and effort for NOTHING! Very sad indeed. Time to cut your losses and move on, Mucke.

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Serg io]

On 10/18/2021 3:25 AM, WM wrote:
> If
> {1, 2, 3, 4, 5, ...} <--> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
> is a bijection (i.e. surjective and injective) and M is an infinite set, then
> {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}
> is not a bijection (surjectivity is missing).

that is pointless.


>
> Hilbert's hotel works only in potential infinity. The *fixed quantity* (Cantor) of actualy infinite sets does not allow for expansion.

please correct the errors before posting, it does not make any sense.

>
> Between every definable natural number and ω there are infinitely many dark natnumbers
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

your "definable" is rejected as totally unworkable, it is Troll Math.

> while between all natnumbers and ω there is nothing
> |ℕ \ {1, 2, 3, ...}| = 0 .

pointless.

>
> (A) There are not more than four endsegments containing the natural numbers {4, 5, 6, ...}, namely E(1), E(2), E(3), and E(4). E(5) fails.
> (B) There are not more than n endsegments containing the natnumbers {n, n+1, n+2, ...}.

the number of endsegments containg k, is k. SO WHAT ?

> (C) There are not more than finitely many infinite endsegments because the contents, infinitely many natural numbers in every infinite endsegment, cannot be used as indices and there are not two consecutive infinite sets in the natural order of ℕ possible. Hence the intersection of infinite endsegments is infinite as also inclusion monotony shows.

no.


>
> Regards, WM
>

[Michael Moroney]

Even though he's dead, he is still smarter than you, nymshifter. Nobody
claims vaccines are 100%. Even though Powell had a 11 times better
chance of surviving covid than unvaccinated people, he happened to roll
snakeeyes. Delta is nasty.

[Wayde Ring]

Michael Moroney wrote:

>> but if Fully Vaccinated Colin Powell Has Died From COVID, how on earth
>> this imbecile became a 4 star general?? What is goeng on with that
>> country??
>
> Even though he's dead, he is still smarter than you, nymshifter. Nobody
> claims vaccines are 100%. Even though Powell had a 11 times better
> chance of surviving covid than unvaccinated people, he happened to roll
> snakeeyes. Delta is nasty.

and I was about asking about the idiots, got above another imbecile.

[Jim Burns]

And it's nastier for some than for others.

https://news.yahoo.com/colin-powell-dead-84-covid-122356856.html
|
| Powell was fully vaccinated, his family said.
| However, he had previously been diagnosed with
| multiple myeloma, a blood cancer that weakens
| the body’s ability to fight infections, CNBC reported.

[Wayde Ring]

Jim Burns wrote:

>> Even though he's dead, he is still smarter than you, nymshifter. Nobody
>> claims vaccines are 100%.  Even though Powell had a 11 times better
>> chance of surviving covid than unvaccinated people,
>> he happened to roll snakeeyes. Delta is nasty.
>
> And it's nastier for some than for others.
>
> https://news.yahoo.com/colin-powell-dead-84-covid-122356856.html |
> | Powell was fully vaccinated, his family said.
> | However, he had previously been diagnosed with | multiple myeloma, a
> blood cancer that weakens | the body’s ability to fight infections, CNBC
> reported.

that's exactly what you get from the "vaccines" which are not vaccines,
fucking stupid. Amazing this "math" group here.

[Serg io]

go post in a vaccine group then.

[zelos...@gmail.com]

>is a bijection (i.e. surjective and injective) and M is an infinite set, then

False, just because one bijection exist doesn't that mean anotehr does not.

>Hilbert's hotel works only in potential infinity.

There are no such things in mathematics.

>The fixed quantity (Cantor) of actualy infinite sets does not allow for expansion.

There is no need for expansion so not an issue.

>Between every definable natural number and ω there are infinitely many dark natnumbers

Empty assertion.

[WM]

zelos...@gmail.com schrieb am Dienstag, 19. Oktober 2021 um 11:50:10 UTC+2:
> >is a bijection (i.e. surjective and injective) and M is an infinite set, then
> False, just because one bijection exist doesn't that mean anotehr does not.

But both must be surjective. Therefore if one set is increased, the other must be increased too.

> >Hilbert's hotel works only in potential infinity.
> There are no such things in mathematics.

There are the known prime numbers as an example. Another example is the number of born humans.

> >The fixed quantity (Cantor) of actualy infinite sets does not allow for expansion.
> There is no need for expansion so not an issue.

That is strange. The hotel |N is complete and in bijection with its guests G. When another guest enters, there must be another room not available before (because before there was a bijection).

Regards, WM

[WM]

timba...@gmail.com schrieb am Montag, 18. Oktober 2021 um 17:11:51 UTC+2:

> We have a problem absorbing falsifications. Clearly the sheer repetition of the original argument has us habituated. Supposedly such entrenched positions are so strong that they are beyond falsification.

Set theory is beyond, at least in the opinions of their adherents. For instance we know in mathematics that inclusion monotony reaches as far as no empty set is encountered. But set theorists declare that it does not cover infinite sets.

The pigeon hole principle holds in all cases where elements have to be distinguished. Set theorists declare that it is invalid for infinite sets.

> Clearly however the scientific position has to declare all such problems open. Otherwise mathematics lands in the same ground as religion, and politics for that matter.

Set theory aka matheology is in the same league.

Regards, WM

[zelos...@gmail.com]

That is not how infinite sets work

sets can be subsets of each other and be of equal cardinality. How is it you have yet to understand this!?

>There are the known prime numbers as an example. Another example is the number of born humans.

THe number of primes known in 2021 is a finite number, it does not change.
The primes known in 2022 is finite and does not change.
Sets do not change.

>That is strange.

There is nothing strange with this, this is basic fucking infinity properties.

>The hotel |N is complete and in bijection with its guests G. When another guest enters, there must be another room not available before (because before there was a bijection).

Incorrect. You tell everyone already there to move to the next room above, and let the new one get room #1.

All were occupied yet we could make room for more.

[zelos...@gmail.com]

The issue is not mathematics WM, it is that you are a complete fucking idiot.

[WM]

zelos...@gmail.com schrieb am Dienstag, 19. Oktober 2021 um 13:14:00 UTC+2:
> tisdag 19 oktober 2021 kl. 12:01:44 UTC+2 skrev WM:

> >But both must be surjective. Therefore if one set is increased, the other must be increased too.
> That is not how infinite sets work

It is. But it is not how matheologians can think.

>
> sets can be subsets of each other and be of equal cardinality.

No, only potentially infinite set can. But they have no cardinality.

> >There are the known prime numbers as an example. Another example is the number of born humans.
> THe number of primes known in 2021 is a finite number, it does not change.

But the set of numbers describing them over the time is potentially infinite.

> The primes known in 2022 is finite and does not change.

It belongs to the potentially infinite set.

> Sets do not change.

Wrong.

> >The hotel |N is complete and in bijection with its guests G. When another guest enters, there must be another room not available before (because before there was a bijection).
> Incorrect. You tell everyone already there to move to the next room above, and let the new one get room #1.
>
> All were occupied yet we could make room for more.

Then it turns out that before there was no bijection. The new room was empty. Or has it been created in a set which does not change?

Regards, WM

[zelos...@gmail.com]

>It is. But it is not how matheologians can think.

It isn't, if you do you are too stupid to do mathematics and should butt out.

>No, only potentially infinite set can. But they have no cardinality.

False, ALL sets have cardinality and sets can be subsets of each other and be same cardinality

N and 2N are such. 2N is a subset of N, but same cardinality. My god you are retarded.

>But the set of numbers describing them over the time is potentially infinite.

Time does not exist in mathematics.

>It belongs to the potentially infinite set.

There exists no such sets in mathematics.

>Wrong.

Correct, sets do not change. They are static. Show me 1 set that "changes". And no "set of known primes" is shorthand for "set of primes known
at the date and time of now" which is constant as the next time we say it, it is a different set.

>Then it turns out that before there was no bijection.

All rooms were occupied before. Get over it you retard.

>The new room was empty. Or has it been created in a set which does not change?

The new room was MADE empty by having the person MOVE OUT and into the NEXT room along with EVERY OTHER INHABITANT!

The amount of rooms have not changed, still aleph_0, but we show that 2 different sets fit into it.

[Greg Cunt]

On Wednesday, October 20, 2021 at 12:24:01 PM UTC+2, WM wrote:

> potentially infinite set <bla>

There are no such entities in mathematics, Mückenheim.

They only exist in your delusion

[FromTheRafters]

on 10/20/2021, WM supposed :

> zelos...@gmail.com schrieb am Dienstag, 19. Oktober 2021 um 13:14:00 UTC+2:
>> tisdag 19 oktober 2021 kl. 12:01:44 UTC+2 skrev WM:
>
>>> But both must be surjective. Therefore if one set is increased, the other
>>> must be increased too.
>> That is not how infinite sets work
>
> It is. But it is not how matheologians can think.
>>
>> sets can be subsets of each other and be of equal cardinality.
>
> No, only potentially infinite set can. But they have no cardinality.

Stop lying.

[Serg io]

On 10/19/2021 5:09 AM, WM wrote:
> timba...@gmail.com schrieb am Montag, 18. Oktober 2021 um 17:11:51 UTC+2:
>
>> We have a problem absorbing falsifications. Clearly the sheer repetition of the original argument has us habituated. Supposedly such entrenched positions are so strong that they are beyond falsification.
>
> Set theory is beyond, at least in the opinions of their adherents. For instance we know in mathematics that inclusion monotony reaches as far as no empty set is encountered. But set theorists declare that it does not cover infinite sets.

Bullshitification of inclusion monotony is extensively used by fakemathethists

>
> The pigeon hole principle holds in all cases where elements have to be distinguished. Set theorists declare that it is invalid for infinite sets.

fakemathethists are infested with pigeons and distinguishing their holes

>
>> Clearly however the scientific position has to declare all such problems open. Otherwise mathematics lands in the same ground as religion, and politics for that matter.
>
> Set theory aka matheology is in the same league.
>
> Regards, WM
>

coming from someone that does not know basic algebra!

[WM]

zelos...@gmail.com schrieb am Mittwoch, 20. Oktober 2021 um 12:53:01 UTC+2:
> >It is. But it is not how matheologians can think.
> It isn't, if you do you are too stupid to do mathematics and should butt out.
> >No, only potentially infinite set can. But they have no cardinality.
> False, ALL sets have cardinality and sets can be subsets of each other and be same cardinality
>
> N and 2N are such. 2N is a subset of N, but same cardinality. My god you are retarded.
> >But the set of numbers describing them over the time is potentially infinite.
> Time does not exist in mathematics.

Wrong. Mathematics describes motion, velocity, acceleration, etc. I mean real mathematics.

> >It belongs to the potentially infinite set.
> There exists no such sets in mathematics.
>
> >Wrong.
>
> Correct, sets do not change. They are static. Show me 1 set that "changes". And no "set of known primes" is shorthand for "set of primes known
> at the date and time of now" which is constant as the next time we say it, it is a different set.

The sequence of these sets is often denoted as the potentially infinite set.

> >Then it turns out that before there was no bijection.
> All rooms were occupied before.

Then a new room has been created. Then sets are not invariable.

> >The new room was empty. Or has it been created in a set which does not change?
> The new room was MADE empty by having the person MOVE OUT and into the NEXT room along with EVERY OTHER INHABITANT!

And the number of persons has decreased? The number of occupied rooms has decreased by one if the number of rooms is the same.

>
> The amount of rooms have not changed, still aleph_0,

aleph_0 is not a number. It describes the prime numbers and the algebraic numbers. Do you claim these sets are one and the same?

> but we show that 2 different sets fit into it.

You do not show this but you believe it erroneously. You believe even that the number of points on the diameter of a quark is same as the number of points in the whole universe. Hardly possible to imagine a stronger perversion.

Regards, WM

[WM]

Greg Cunt schrieb am Mittwoch, 20. Oktober 2021 um 15:48:38 UTC+2:
> On Wednesday, October 20, 2021 at 12:24:01 PM UTC+2, WM wrote:
>
> > potentially infinite set
>

> There are no such entities in mathematics,

The set of prime gaps is potentially infinite.

Regards, WM

[FromTheRafters]

WM formulated on Wednesday :

Stop lying.

[Nikolai vdB]

WM schrieb am Montag, 18. Oktober 2021 um 10:25:41 UTC+2:
> If
> {1, 2, 3, 4, 5, ...} <--> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
> is a bijection (i.e. surjective and injective) and M is an infinite set, then
> {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}
> is not a bijection (surjectivity is missing).

You need to be mathematically thorough.
A bijection is not given by just mentioning two sets, you need a function. Only a function can be a bijection. What this might stem from is that, IF there is a bijection between two sets, than they have the same cardinality (both finite and infinite).

So, at first you need to tell us what exactly the function here is. What you seem to want to say is that the bottom two sets do not have the same cardinality because there IS a function between the two that is not surjective. But that is not enough.

The function {a,b} -> {a,b}, a->a, b->a is not surjective, even tho the two sets (or the one set) of course is of same cardinality and there exists a bijection between the two.

In your case, it is pretty much the same. No, the function x-> x/1 is not a bijection between your bottom two sets. But that does not mean that there is no bijection between these two sets. Namely, the function 1->M, x->(x+1)/1 for x>1, x in N is a bijection between the bottom two sets, which is why they have the same cardinality.

[zelos...@gmail.com]

>Wrong. Mathematics describes motion, velocity, acceleration, etc. I mean real mathematics.

Correct, because mathematics is about MATHEMATICAL objects not REAL objects.

What you mean is APPLICATIONS OF MATHEMATICS, but that is not mathematics itself! In mathematics time does not exist!

>The sequence of these sets is often denoted as the potentially infinite set.

A sequence is a set and it doesn't change. So show me one set that changes in mathematics.

>Then a new room has been created. Then sets are not invariable.

No room was created.

>And the number of persons has decreased? The number of occupied rooms has decreased by one if the number of rooms is the same.

None left the hotel, everyone just moved up a room.

>aleph_0 is not a number.

It is a number, it is a CARDINAL number

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

You are so uneducated

It describes the prime numbers and the algebraic numbers. Do you claim these sets are one and the same?

They have the same cardinality

>You do not show this but you believe it erroneously.

We do show it, it is you are so stupid you think that infinite sets should behave like finite ones.

>You believe even that the number of points on the diameter of a quark is same as the number of points in the whole universe.

We are not talking about physical objects here, we are talking about mathematical objects.

>Hardly possible to imagine a stronger perversion

You're the idiot that brings in physics when we talk mathematics that is independent of physics.

[WM]

Nikolai vdB schrieb am Donnerstag, 21. Oktober 2021 um 03:44:01 UTC+2:
> WM schrieb am Montag, 18. Oktober 2021 um 10:25:41 UTC+2:
> > If
> > {1, 2, 3, 4, 5, ...} <--> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
> > is a bijection (i.e. surjective and injective) and M is an infinite set, then
> > {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}
> > is not a bijection (surjectivity is missing).
> You need to be mathematically thorough.
> A bijection is not given by just mentioning two sets, you need a function.

The function is trivial. Cantor gives only one set. The indices {1, 2, 3, 4, 5, ...} are applied in their natural order.

>
> So, at first you need to tell us what exactly the function here is.

The indices {1, 2, 3, 4, 5, ...} are applied in their natural order to the elements of the RHS set in the given order .

> No, the function x-> x/1 is not a bijection between your bottom two sets.

Of course it is.

Regards, WM

[WM]

zelos...@gmail.com schrieb am Mittwoch, 20. Oktober 2021 um 12:53:01 UTC+2:

> >The new room was empty.

It canot be made empty without a guest occupying a room which was empty before.

>> Or has it been created in a set which does not change?
> The new room was MADE empty by having the person MOVE OUT and into the NEXT room along with EVERY OTHER INHABITANT!
>
> The amount of rooms have not changed, still aleph_0,

aleph_0 is not an amount and not a number. |N U {a} has one element more than |N but same aleph_0. Therefore either the number of rooms has changed or there was no bijection before, and probably also no bijection afterwards.

Regards, WM

[zelos...@gmail.com]

Stop thinking that infinite things behave like finite ones! THEY DO NOT!

[Greg Cunt]

On Thursday, October 21, 2021 at 8:55:31 PM UTC+2, WM wrote:

> aleph_0 is not [...] a number.

It is, YOU PSYCHOTIC ASSSHOLE FULL OF SHIT!!!

It's called a CARDIAL NUMBER.

See: https://en.wikipedia.org/wiki/Cardinal_number

[Greg Cunt]

On Wednesday, October 20, 2021 at 9:59:46 PM UTC+2, WM wrote:

> The set of prime gaps is potentially infinite.

You are mentally ill, man.

Hint: The set of prime gaps is infinite.

See: https://mathworld.wolfram.com/PrimeGaps.html

[Greg Cunt]

On Thursday, October 21, 2021 at 8:55:31 PM UTC+2, WM wrote:

> zelos...@gmail.com schrieb am Mittwoch, 20. Oktober 2021 um 12:53:01 UTC+2:
> > >

> > > The new room was empty. [WM]
> > >
> > The [first] room was MADE empty by having the person MOVE OUT and into the NEXT room along with EVERY OTHER INHABITANT! [Zelos]
> >
> It canot be made empty without <bla>

CAN'T YOU READ ANYMORE, YOU IMBECILE?!

Lost your last marble?

[Nikolai vdB]

The function is not trivial. You also seem to confuse things, if you think that the function x -> x/1 is a bijection between the bottom two sets, they are in bijection and have the same cardinality, which is what you wanted to disprove.
You need to put more care into mathematical thoroughness and reading.

To show that the bottom two sets are NOT in bijection, you need to show there exists NO bijective function between them. But the only thing you seem to want to show is that one certain function is not a bijection between them, which is not enough.

[WM]

Nikolai vdB schrieb am Freitag, 22. Oktober 2021 um 13:28:12 UTC+2:
> WM schrieb am Donnerstag, 21. Oktober 2021 um 20:49:52 UTC+2:

> The function is not trivial.

The many lines of text seem to confuse you. Here is the essence:

{1, 2, 3, 4, 5, ...} <--> {1/1, 2/1, 3/1, 4/1, 5/1, ...}

in the order from left to right in both sets is a bijection. No element must be missing and no element must be added unless the bijection is spoilt.

{1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}

in the order from left to right in both sets is not a bijection unless M is empty.

> To show that the bottom two sets are NOT in bijection, you need to show there exists NO bijective function between them.

No, that is a mistake.

> But the only thing you seem to want to show is that one certain function is not a bijection between them, which is not enough.

It is enough to show that if the left side is not changed but the right side is changed and if it is claimed that in both cases a bijection exists, that then long-distance effects have to be assumed, which is nonsense.

Regards WM

[Nikolai vdB]

You need to be mathematically thorough. If you talk of a bijection, you need to explicitly state it. I can see what you might see as the bijection here, but you still have to specify it. Elseways you miss thoroughness. In both cases.

Because if you do, you will see that your argument makes no sense. You would also have to specify what "long distance effects" means here or anything like that. Bottom line is: If you cannot even specify with exact mathematical phrases what you are talking about, it is all useless.

A bijection is a function, without giving the function, what you do is meaningless. And of course, the lower two sets have a function between them that maps every element on the left side to the right side (including the M). There is a bijection between them, just like in the upper case.

[WM]

Nikolai vdB schrieb am Freitag, 22. Oktober 2021 um 18:09:20 UTC+2:
> WM schrieb am Freitag, 22. Oktober 2021 um 17:52:25 UTC+2:

> > > But the only thing you seem to want to show is that one certain function is not a bijection between them, which is not enough.

I show that Cantor's mapping which over 150 years has been accepted as a bijection is not a bijection.

> You need to be mathematically thorough. If you talk of a bijection, you need to explicitly state it.

Look, Cantor stated this 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, ... Every learned reader understands it.

> Because if you do, you will see that your argument makes no sense.

You have not yet understood it. But that is not an argument.

> You would also have to specify what "long distance effects" means here or anything like that.

It means that when the right side is changed, the left side changes without cause.

> A bijection is a function, without giving the function, what you do is meaningless. And of course, the lower two sets have a function between them that maps every element on the left side to the right side (including the M).

TSuch a belief shows lack of reason or lack of understanding mathematics, in particular bijections. If

{1, 2, 3, 4, 5, ...} >--> {1/1, 2/1, 3/1, 4/1, 5/1, ...}

with the trivial mapping is a bijection, then this means that no element at the LHS or at the RHS may be added or deleted without spoiling the bijection. Therefore

{1, 2, 3, 4, 5, ...} >--> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}

is not a bijection and your statement is wrong.

Regards, WM

[WM]

zelos...@gmail.com schrieb am Freitag, 22. Oktober 2021 um 06:46:09 UTC+2:
> torsdag 21 oktober 2021 kl. 20:55:31 UTC+2 skrev WM:
> |N U {a} has one element more than |N but same aleph_0. Therefore either the number of rooms has changed or there was no bijection before, and probably also no bijection afterwards.
> >

> Stop thinking that infinite things behave like finite ones! THEY DO NOT!

Bijections are defined by surjectivity and injectivity. That is invariable for all kinds of sets. If you claim other properties then don't call these mappings bijections.

Regards, WM

[WM]

Greg Cunt schrieb am Freitag, 22. Oktober 2021 um 09:37:56 UTC+2:
> On Wednesday, October 20, 2021 at 9:59:46 PM UTC+2, WM wrote:
>
> > The set of prime gaps is potentially infinite.
>

> Hint: The set of prime gaps is infinite.

So you try to cheat.

Regards, WM

[Python]

Crank Wolfgang Muckenheim, aka WM wrote:
> zelos...@gmail.com schrieb am Freitag, 22. Oktober 2021 um 06:46:09 UTC+2:
>> torsdag 21 oktober 2021 kl. 20:55:31 UTC+2 skrev WM:
>> |N U {a} has one element more than |N but same aleph_0. Therefore either the number of rooms has changed or there was no bijection before, and probably also no bijection afterwards.
>>>
>> Stop thinking that infinite things behave like finite ones! THEY DO NOT!
>
> Bijections are defined by surjectivity and injectivity.
> That is invariable for all kinds of sets.

Sure, then you have to consider for what both means precisely instead of
throwing silly bunches of nonsensical gibberish, Crank Wolfgang
Muckenheim, from Hochschule Augsburg.

Acting so is called "doing math", doing otherwise, as you do, is called
"being a fool".

[Nikolai vdB]

You do not show anythings about bijections, because, for starters, there is no single map given. You need mathematical thoroughness before you can do math.

[WM]

Python schrieb am Samstag, 23. Oktober 2021 um 01:33:26 UTC+2:
> Crank Wolfgang Muckenheim, aka WM wrote:
> > zelos...@gmail.com schrieb am Freitag, 22. Oktober 2021 um 06:46:09 UTC+2:
> >> torsdag 21 oktober 2021 kl. 20:55:31 UTC+2 skrev WM:
> >> |N U {a} has one element more than |N but same aleph_0. Therefore either the number of rooms has changed or there was no bijection before, and probably also no bijection afterwards.
> >>>
> >> Stop thinking that infinite things behave like finite ones! THEY DO NOT!
> >
> > Bijections are defined by surjectivity and injectivity.
> > That is invariable for all kinds of sets.
> Sure, then you have to consider for what both means precisely

It is well known that matheologians try to change meanings if their nonsense becomes too obvious. The meaning is and remains, here defined by a simple example, in order to show the difference netween mathematics and matheology:
If {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...} with f(n) = n/1 is a bijection, then
{1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...,0} with f(n) = n/1 is not a bijection.

Regards, WM

[WM]

Nikolai vdB schrieb am Samstag, 23. Oktober 2021 um 03:10:58 UTC+2:

> WM schrieb am Freitag, 22. Oktober 2021 um 21:49:39 UTC+2:
If
> > {1, 2, 3, 4, 5, ...} >--> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
> > with the trivial mapping is a bijection, then this means that no element at the LHS or at the RHS may be added or deleted without spoiling the bijection. Therefore
> > {1, 2, 3, 4, 5, ...} >--> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}
> > is not a bijection and your statement is wrong.

> You do not show anythings about bijections, because, for starters, there is no single map given.

I analyze only a single mapping, namely that given by Cantor, slightly simplified.

Aside: If there is a bijection between two sets, then every injective mapping is surjective and vice versa. But that is probably too advanced to be discussed here.

Regards, WM

[Greg Cunt]

On Saturday, October 23, 2021 at 7:35:11 PM UTC+2, WM wrote:

> If f: {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...} with f(n) = n/1 is a bijection, then
> f: {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., 0} with f(n) = n/1 is not a bijection.

Actually, this can be simplified to

f: {1, 2, 3, 4, 5, ...} --> {0, 1/1, 2/1, 3/1, 4/1, 5/1, ...} with f(n) = n/1 for all n e {1, 2, 3, 4, 5, ...} is not a bijection.

Indeed! So what?

[Serg io]

you are the Math liar.

[Serg io]

On 10/20/2021 2:58 PM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 20. Oktober 2021 um 12:53:01 UTC+2:
>>> It is. But it is not how matheologians can think.
>> It isn't, if you do you are too stupid to do mathematics and should butt out.
>>> No, only potentially infinite set can. But they have no cardinality.
>> False, ALL sets have cardinality and sets can be subsets of each other and be same cardinality
>>
>> N and 2N are such. 2N is a subset of N, but same cardinality. My god you are retarded.
>>> But the set of numbers describing them over the time is potentially infinite.
>> Time does not exist in mathematics.
>
> Wrong. Mathematics describes motion, velocity, acceleration, etc. I mean real mathematics.

nope, that is Physics.

>
>>> It belongs to the potentially infinite set.
>> There exists no such sets in mathematics.
>>
>>> Wrong.
>>
>> Correct, sets do not change. They are static. Show me 1 set that "changes". And no "set of known primes" is shorthand for "set of primes known
>> at the date and time of now" which is constant as the next time we say it, it is a different set.
>
> The sequence of these sets is often denoted as the potentially infinite set.

wrong. no one "denotes". potentially infinite is bogus word for = finite

[Greg Cunt]

On Saturday, October 23, 2021 at 7:40:10 PM UTC+2, WM wrote:

> [f:] {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
> I analyze only a single mapping, namely that given by Cantor [...]

Nonsense. f clearly is NOT the "mapping given by Cantor", you psychotic asshole.

I guess, you mean the function f defined by f(n) = n/1 (for all n e IN) here.

> If there is a bijection between two sets, then every injective mapping is surjective and <bla bla>

Idiotic nonsense.

Hint: There's a bijection from IN onto IN, namely id_IN. But the injective mapping f: IN --> IN defined with f(n) = n + 1 for all n e IN is NOT surjective (since 1 is not in the image of f).

See: https://en.wikipedia.org/wiki/Dedekind-infinite_set

[Greg Cunt]

On Wednesday, October 20, 2021 at 9:58:27 PM UTC+2, WM wrote:

> The sequence of these sets is often denoted as the potentially infinite set.

Can you prove your claim by some quotes?

Hint: You are nuts, man.

[FromTheRafters]

WM explained :

Why not? Do you require right totalness in Muckymath?

Zero doesn't exist in your naturals.

[Greg Cunt]

On Saturday, October 23, 2021 at 8:58:07 PM UTC+2, FromTheRafters wrote:
> WM explained :
> >
> > [f:] {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., 0} with f(n) = n/1 is not a bijection
> >
> Why not?

Because there's no x in {1, 2, 3, 4, 5, ...} such that f(x) = 0.

[Nikolai vdB]

There is no mapping in your statement. Also, why do you keep mentioning cantor? This is not about cantor, but about your lack of thoroughness.

And your bottom statement is not true. If there is a bijection between two sets, still a map can be injective but not surjective.
For example, surely the the natural numbers are in surjection with themselves, but the function:

N -> N, n -> n+1 is still injective, but not surjective because there is no natural number mapped to 1. But... that is probably too advanced for you?

[FromTheRafters]

Greg Cunt formulated on Saturday :

So endsegment {5, 6, 7, 8, ...} is not in bijection with endsegment {1,
2, 3, 4, 5, ...} ?

[Serg io]

[WM]

This should teach you, that also f(n) = n/1 - 1 is not a bijection. But my example is in sharper contrast to the familiar errors.

Regards, WM

[Greg Cunt]

On Sunday, October 24, 2021 at 10:26:02 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Samstag, 23. Oktober 2021 um 19:46:00 UTC+2:
> > On Saturday, October 23, 2021 at 7:35:11 PM UTC+2, WM wrote:
> >
> > > If f: {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...} with f(n) = n/1 is a bijection, then
> > > f: {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., 0} with f(n) = n/1 is not a bijection.
> >
> > Actually, this can be simplified to
> >
> > f: {1, 2, 3, 4, 5, ...} --> {0, 1/1, 2/1, 3/1, 4/1, 5/1, ...} with f(n) = n/1 for all n e {1, 2, 3, 4, 5, ...} is not a bijection.
> >
> > Indeed! So what?
> >

> This should <bla>

On the other hand

g: {1, 2, 3, 4, 5, ...} --> {0, 1/1, 2/1, 3/1, 4/1, 5/1, ...} with g(1) = 0 and g(n) = (n-1)/1 for all n e {2, 3, 4, 5, ...} *is* a bijection.

[Nikolai vdB]

Why do you avoid saying anything to the disproval of your claim that every injective function between two sets that are in bijection to each other is also surjective? It is not true and I have disproven it. You could at least admit that.

[Gus Gassmann]

Some reasons. 1. It would invalidate essentially his life's work, so he ignores it. 2. Even if he looked at it, do you honestly believe he'd understand your proof. 3. Deny, deny, deny.

[Nikolai vdB]

I think he might. Elseways he would not choose to ignore it, if he didn't know I was right.

[zelos...@gmail.com]

nope, that is not enough. When it comes to cardinal arithmetic and such that is NOT ENOUGH!

You have to show that there is NO bijection between them. Providing 2 different bijections proves NOTHING for you!

[zelos...@gmail.com]

In finite sets all injections are surjections when they have the same cardinality.

In infinite sets this is not the case!

[WM]

Nikolai vdB schrieb am Montag, 25. Oktober 2021 um 03:32:57 UTC+2:
> WM schrieb am Sonntag, 24. Oktober 2021 um 22:26:02 UTC+2:

> > This should teach you, that also f(n) = n/1 - 1 is not a bijection. But my example is in sharper contrast to the familiar errors.
> >

> Why do you avoid saying anything to the disproval of your claim that every injective function between two sets that are in bijection to each other is also surjective? It is not true and I have disproven it. You could at least admit that.

I wanted to postpone this question until you will have understood the glaring mistake of Cantor's enumeration of the fractions. Then it will be easier to understand because there it is much more impressive than in the present case. Alas ...

I know that all of you are firmly believing that ℕ → ℕ, n → n+1 is an injective mapping. I did so too for along time, see my "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015), and it works, like Hilbert's hotel, for the potentially infinite collection ℕ considered there. But it is not true and does not work for the actually infinite set ℕ.

Why? Because in actual infinity ℕ → ℕ, n → n is a bijection by symmetry consideration. That means every pair (n, n) is fixed although most pairs cannot be checked. They are dark, belonging to the ℵo natnumbers following upon every definable natnumber n. They must be dark because otherwise they could not become exhausted such that none remains. But in a bijection all natnumbers must participate. And this is the crucial point: If all natnumbers are in the bijection, then none can be added. There is simply no more than all. But we assume at the beginning ℕ → ℕ that all natnumbers are engaged. If this could be changed arbitrarily, then Cantor's theory would fail. Unfortunately this condition is rarely mentioned or remembered:

"Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen" [Cantor, p. 119]

"gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 238 ]

"Die sämtlichen Punkte l unsrer Menge L sind also in gegenseitig eindeutige und vollständige Beziehung zu sämtlichen Punkten f der Menge F gebracht," [Cantor, p. 241]

"Zwei wohlgeordnete Mengen M und N heissen von gleichem Typus oder auch von gleicher Anzahl, wenn sie sich gegenseitig eindeutig und vollständig unter beidseitiger Wahrung der Rangfolge ihrer Elemente auf einander beziehen, abbilden lassen;" [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

"Zwei bestimmte Mengen M und M1 nennen wir äquivalent (in Zeichen: M ~ M1), wenn es möglich ist, dieselben gesetzmäßig, gegenseitig eindeutig und vollständig, Element für Element, einander zuzuordnen." [Cantor, p. 412]

"doch gibt es immer viele, im allgemeinen sogar unzählig viele Zuordnungsgesetze, durch welche zwei äquivalente Mengen in gegenseitig eindeutige und vollständige Beziehung zueinander gebracht werden können." [Cantor, p. 413]

"eine solche gegenseitig eindeutige und vollständige Korrespondenz hergestellt [...] irgendeine gegenseitig eindeutige und vollständige Zuordnung der beiden Mengen [...] auch eine gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 415]

"Zwei n-fach geordnete Mengen M und N werden 'ähnlich' genannt, wenn es möglich ist, sie gegenseitig eindeutig und vollständig, Element für Element, einander so zuzuordnen," [Cantor, p. 424]

Why do you "interpret" vollständig as unvollständig? Because you use potential infinity. In actual infinity we have, for instance, all unit fractions 1/n in the interval (0, 1]. Not even a single one must be missing and not even a single one can be added.

Regards, WM

[WM]

zelos...@gmail.com schrieb am Montag, 25. Oktober 2021 um 06:28:00 UTC+2:

> In finite sets all injections are surjections when they have the same cardinality.
> In infinite sets this is not the case!

In actually infinite sets this it the case too, if mathematics is taken seriously and precisely. It is only the potential infinity of the definable numbers that feigns variability of the complete set.

Regards, WM

[WM]

Greg Cunt schrieb am Sonntag, 24. Oktober 2021 um 22:34:44 UTC+2:

> On the other hand
>
> g: {1, 2, 3, 4, 5, ...} --> {0, 1/1, 2/1, 3/1, 4/1, 5/1, ...} with g(1) = 0 and g(n) = (n-1)/1 for all n e {2, 3, 4, 5, ...} *is* a bijection.

So there is one more natnumber than in
f: {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...} with f(n) = n/1 ?
Or one unit fraction less?

And what happens when
h: {2, 4, 6, 8, 10, ... } --> {1/1, 2/1, 3/1, 4/1, 5/1, ...} with h(2n) = n/1
or
h': {10^0, 10^10, 10^100, 10^1000, 10^10000 } --> {1/1, 2/1, 3/1, 4/1, 5/1, ...} is taken ?

Consider the geometric part of mathematics. Is it possible that the points occupied by unit fractions in the interval (0, 1] are variable by one or even more?

In my opinion they are not variable. If all are existing independently of what we recognize - that need not be assumed, but if it is assumed - then there is not even one more or less possible.

Remember: Gott, der Herr hat sie gezählet, dass ihm auch nicht einer fehlet.

Regards, WM

[WM]

zelos...@gmail.com schrieb am Montag, 25. Oktober 2021 um 06:26:44 UTC+2:
> fredag 22 oktober 2021 kl. 17:52:25 UTC+2 skrev WM:
> > Nikolai vdB schrieb am Freitag, 22. Oktober 2021 um 13:28:12 UTC+2:
> > > WM schrieb am Donnerstag, 21. Oktober 2021 um 20:49:52 UTC+2:
> >
> > > The function is not trivial.
> > The many lines of text seem to confuse you. Here is the essence:
> > {1, 2, 3, 4, 5, ...} <--> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
> > in the order from left to right in both sets is a bijection. No element must be missing and no element must be added unless the bijection is spoilt.
> > {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}

> You have to show that there is NO bijection between them.

I have proved that Cantor's bijection, the enumeration of all positive fractions accepted for over 100 years fails. Of course every other bijection between naturals and all fractions would fail.

> Providing 2 different bijections proves NOTHING for you!

Above I provide only one bijection.

Regards, WM

[zelos...@gmail.com]

no it isn't.

f(x)=2x from N to N is injective but not surjective!

[zelos...@gmail.com]

you provide A bijection but have not proven it is THE ONLY one. and it isn't!

[FromTheRafters]

on 10/23/2021, FromTheRafters supposed :

What about {A, B, C, D} and {elephant, gargoyle, darknumber, unicorn} ?

Can there be a bijection without non-empty intersection?

What about

{0, 1, 2,...} and {|, ||, |||, ||||, |||||, ...}

[WM]

> you provide A bijection but have not proven it is THE ONLY one. and it isn't!

If there is a bijection, then every mapping between transposed sets is a bijection too.
I proved that Cantor's enumeration of the naturals is not a bijection. And I am able to prove the same for every mapping between infinite sets which cannot be bijected by symmetry consideration.

Regards, WM

[zelos...@gmail.com]

You proved nothing of the sort. You think you did but you didn't.

[Gus Gassmann]

On Monday, 25 October 2021 at 08:39:02 UTC-3, WM wrote:
> If there is a bijection, then every mapping between transposed sets is a bijection too.

Where the hell did you get that screwy idea from? Let A = {1, 2, 3} and B = {4, 5, 6}. Then f defined by f(n) = n+3 is a bijection from A to B. Now take g defined by g(n) = 4 for n in A. Are you actually delusional enough to think that this is a bijection? Conversely, do you actually think because g is *NOT* a bijection, f does not exist? I said a few weeks back that you have no clue what mappings are, and here is conclusive proof of that. Your brain is fried to the point that you should be placed in a padded cell for your own protection.

[Greg Cunt]

On Monday, October 25, 2021 at 11:08:51 AM UTC+2, WM wrote:
.
> I have proved that Cantor's bijection, the enumeration of all positive fractions accepted for over 100 years fails

Time for the Abel Prize, I'd say!

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

[Serg io]

gad.

[Serg io]

which proves both "potential" and "definable" are Fake Math

[Greg Cunt]

On Monday, October 25, 2021 at 1:39:02 PM UTC+2, WM wrote:

> I proved that Cantor's enumeration of the naturals is not a bijection.

Cool!

It's really time for the Abel Prize, I'd say!

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

[Gus Gassmann]

Well, the fact that he hasn't won the Abel Prize yet, or the Fields medal, even, is clear proof that the mathematical community is trying to silence him and ignore his inconvenient truths.

[Nikolai vdB]

No. You cannot have it both ways. Either you talk about actual modern mathematics and have to stick to the definitions if you want to prove there are inconsistencies, or you stick to whatever you make up about Cantor.

Fact is: Your "symmetry" is not necessary to prove a bijection, neither is talk about actual or potential infinity, a concept that cantor might have talked about, but that is just not part of actual modern mathematics.

IF you talk about BIJECTIONS, then you HAVE TO STICK TO THE DEFINITION OF THE WORD. Elseways, don't talk about it.

A function f:A -> B is injective if there is only one a in A for every b in B so that f(a)=b, or rather, if for a, a' in A it holds that f(a)=f(a'), then a=a' follows.
And that is true for the function I gave you. The definition of injection holds.
If you have other definitions entirely, then dont talk about it as if you'd use actual mathematical definitions.

Nobody cares about what you think cantor might have said because whatever he might have said is not relevant for the actual, real, mathematical definition of the word injective.

[WM]

Nikolai vdB schrieb am Montag, 25. Oktober 2021 um 17:37:30 UTC+2:
> WM schrieb am Montag, 25. Oktober 2021 um 10:32:45 UTC+2:
> > Nikolai vdB schrieb am Montag, 25. Oktober 2021 um 03:32:57 UTC+2:
> > > WM schrieb am Sonntag, 24. Oktober 2021 um 22:26:02 UTC+2:
> >
> > > > This should teach you, that also f(n) = n/1 - 1 is not a bijection. But my example is in sharper contrast to the familiar errors.
> > > >
> > > Why do you avoid saying anything to the disproval of your claim that every injective function between two sets that are in bijection to each other is also surjective? It is not true and I have disproven it. You could at least admit that.
> > I wanted to postpone this question until you will have understood the glaring mistake of Cantor's enumeration of the fractions. Then it will be easier to understand because there it is much more impressive than in the present case. Alas ...
> >
> > I know that all of you are firmly believing that ℕ → ℕ, n → n+1 is an injective mapping. I did so too for along time, see my "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015), and it works, like Hilbert's hotel, for the potentially infinite collection ℕ considered there. But it is not true and does not work for the actually infinite set ℕ.
> >
> > Why? Because in actual infinity ℕ → ℕ, n → n is a bijection by symmetry consideration. That means every pair (n, n) is fixed although most pairs cannot be checked. They are dark, belonging to the ℵo natnumbers following upon every definable natnumber n. They must be dark because otherwise they could not become exhausted such that none remains. But in a bijection all natnumbers must participate. And this is the crucial point: If all natnumbers are in the bijection, then none can be added. There is simply no more than all. But we assume at the beginning ℕ → ℕ that all natnumbers are engaged. If this could be changed arbitrarily, then Cantor's theory would fail. Unfortunately this condition is rarely mentioned or remembered:
> >
> > "Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen" [Cantor, p. 119]
> >
> > "gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 238 ]
> >
> > "Die sämtlichen Punkte l unsrer Menge L sind also in gegenseitig eindeutige und vollständige Beziehung zu sämtlichen Punkten f der Menge F gebracht," [Cantor, p. 241]
> >
> > "Zwei wohlgeordnete Mengen M und N heissen von gleichem Typus oder auch von gleicher Anzahl, wenn sie sich gegenseitig eindeutig und vollständig unter beidseitiger Wahrung der Rangfolge ihrer Elemente auf einander beziehen, abbilden lassen;" [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
> >
> > "Zwei bestimmte Mengen M und M1 nennen wir äquivalent (in Zeichen: M ~ M1), wenn es möglich ist, dieselben gesetzmäßig, gegenseitig eindeutig und vollständig, Element für Element, einander zuzuordnen." [Cantor, p. 412]
> >
> > "doch gibt es immer viele, im allgemeinen sogar unzählig viele Zuordnungsgesetze, durch welche zwei äquivalente Mengen in gegenseitig eindeutige und vollständige Beziehung zueinander gebracht werden können." [Cantor, p. 413]
> >
> > "eine solche gegenseitig eindeutige und vollständige Korrespondenz hergestellt [...] irgendeine gegenseitig eindeutige und vollständige Zuordnung der beiden Mengen [...] auch eine gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 415]
> >
> > "Zwei n-fach geordnete Mengen M und N werden 'ähnlich' genannt, wenn es möglich ist, sie gegenseitig eindeutig und vollständig, Element für Element, einander so zuzuordnen," [Cantor, p. 424]
> >
> > Why do you "interpret" vollständig as unvollständig? Because you use potential infinity. In actual infinity we have, for instance, all unit fractions 1/n in the interval (0, 1]. Not even a single one must be missing and not even a single one can be added.
> >

> No. You cannot have it both ways. Either you talk about actual modern mathematics and have to stick to the definitions if you want to prove there are inconsistencies, or you stick to whatever you make up about Cantor.

Modern mathematics is not based upon completeness of bijections?

>
> Fact is: Your "symmetry" is not necessary to prove a bijection,

It is valid and it turns out necessary. Do you think that it is invalid?

> neither is talk about actual or potential infinity, a concept that cantor might have talked about, but that is just not part of actual modern mathematics.

Modern mathematics has not changed the way Cantor did it, but they try to cheat their disciples by not distinguishing between actual and potential infinity.

>
> IF you talk about BIJECTIONS, then you HAVE TO STICK TO THE DEFINITION OF THE WORD. Elseways, don't talk about it.

I stick to mathematics. If

{1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...}

is a bijection, then
{1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}, with M=/= { }
is not a bijection.

> Nobody cares about what you think cantor might have said because whatever he might have said is not relevant for the actual, real, mathematical definition of the word injective.

What of the above quotes is no longer valid?

Regards, WM

[Greg Cunt]

On Tuesday, October 26, 2021 at 11:22:13 AM UTC+2, WM wrote:

> I stick to [mückenmath]. If

> {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
> is a bijection,

No. This is just nonsense.

> then <bla bla bla>

Whatever.

Want to try again, Mückenheim?

Hint:

> If [the function f:] {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...} is a bijection [from {1, 2, 3, 4, 5, ...} onto {1/1, 2/1, 3/1, 4/1, 5/1, ...}], then
> [the function g:] {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}, [with g(x) = f(x) for all x e {1, 2, 3, 4, 5, ...}]
> ​is not a bijection [from {1, 2, 3, 4, 5, ...} onto {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}; if M !e {1/1, 2/1, 3/1, 4/1, 5/1, ...}].

And no

> M =/= { }

doesn't ensure for that. Hint: Let M = 1/1, then the function g: {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}, with g(x) = f(x) for all x e {1, 2, 3, 4, 5, ...} *is* a bijection from {1, 2, 3, 4, 5, ...} onto {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}.

[Nikolai vdB]

No you do not stick to mathematics because you claim things like N -> N, n -> n+1 is not an injective function, even tho it certainly satisfies the definition of injective.
NONE of what Cantor says matters. Only what the current definitions are matters.

You struggle with very basic concepts, like you still do not write down what exactly your function is that may or may not be a bijection and why that would matter. Certainly, none of what you says satisfies any mathematical thoroughness.

If you want to show that N is not in bijection to Q, you have to show there is no single bijection between them. And you can't and you didn't.
The same goes for showing that N and N_0 are not in bijection. You cannot and you did not.

[WM]

Greg Cunt schrieb am Dienstag, 26. Oktober 2021 um 14:23:01 UTC+2:
> On Tuesday, October 26, 2021 at 11:22:13 AM UTC+2, WM wrote:
>

> > I stick to mathematics. If

> > {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
> > is a bijection,
> No. This is just nonsense.

It appears so to you because permutations of the sets could be not in bijection in your opinion. But that is nonsense. Every transposition of one of the sets will yield a bijection.

Cantor unconsciouly wrote the correct statement, destroying all his theory: "Die Frage, durch welche Umformungen einer wohlgeordneten Menge ihre Anzahl geändert wird, durch welche nicht, läßt sich einfach so beantworten, daß diejenigen und nur diejenigen Umformungen die Anzahl ungeändert lassen, welche sich zurückführen lassen auf eine endliche oder unendliche Menge von Transpositionen, d. h. von Vertauschungen je zweier Elemente."

Of course this must be true if all elements of the domain are existing and are connected to their partners of the range.

>
> > If [the function f:] {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...} is a bijection [from {1, 2, 3, 4, 5, ...} onto {1/1, 2/1, 3/1, 4/1, 5/1, ...}], then
> > [the function g:] {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}, [with g(x) = f(x) for all x e {1, 2, 3, 4, 5, ...}]
> > ​is not a bijection [from {1, 2, 3, 4, 5, ...} onto {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}; if M !e {1/1, 2/1, 3/1, 4/1, 5/1, ...}].
>

Hopefully you will understand that Cantor's famous enumeration of the positive fractions all the more must fail.

Regards, WM

[WM]

Nikolai vdB schrieb am Dienstag, 26. Oktober 2021 um 17:52:33 UTC+2:
> WM schrieb am Dienstag, 26. Oktober 2021 um 11:22:13 UTC+2:

> > Modern mathematics is not based upon completeness of bijections?

No answer.

> > >
> > > Fact is: Your "symmetry" is not necessary to prove a bijection,
> > It is valid and it turns out necessary. Do you think that it is invalid?

No answer.

> > > Nobody cares about what you think cantor might have said because whatever he might have said is not relevant for the actual, real, mathematical definition of the word injective.
> > What of the above quotes is no longer valid?

> No you do not stick to mathematics because you claim things like N -> N, n -> n+1 is not an injective function, even tho it certainly satisfies the definition of injective.

It is not a bijection because ℕ → ℕ is a bijection. That means completeness. No single elements can be added in domain or range without destroying the completeness.

> NONE of what Cantor says matters. Only what the current definitions are matters.

Dpoes it not matter that tey lead to false results?

>
> You struggle with very basic concepts, like you still do not write down what exactly your function is

For a bijection, this is irrelevant because the function cannot apply to dark numbers.

> If you want to show that N is not in bijection to Q, you have to show there is no single bijection between them. And you can't and you didn't.

I emphasized several times that I only show that one special mapping, which over decades had been believed to be a bijection, provably is not a bijection if

{1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...}

is a bijection, and

{1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}

is not a bijection.

That is the present topic! Can you agree to the last statement, where the pairs of the mappings are taken from left to right?

Regards, WM

[FromTheRafters]

WM formulated the question :

Maybe M is {1/2, 2/2, 3/2, 4/2, 5/2, ...} so the joining can be
interleaved from {1/1, 2/1, 3/1, 4/1, 5/1, ..., M} to {1/1, 1/2, 2/1,
2/2, 3/1, 3/2, 4/1, 4/2, 5/1, 5/2, ...} both sets were countable and so
is the resulting set.

[zelos...@gmail.com]

>It is not a bijection because ℕ → ℕ is a bijection. That means completeness. No single elements can be added in domain or range without destroying the completeness

Non one adds elements! Just because f(x)=x from N to N is a bijection and g(x)=x+1 is a bijection from N to N\{0} does not mean there is an element created!

[Greg Cunt]

On Wednesday, October 27, 2021 at 1:16:22 PM UTC+2, FromTheRafters wrote:

> Maybe M is {1/2, 2/2, 3/2, 4/2, 5/2, ...} so the joining can be
> interleaved from {1/1, 2/1, 3/1, 4/1, 5/1, ..., M} to {1/1, 1/2, 2/1,
> 2/2, 3/1, 3/2, 4/1, 4/2, 5/1, 5/2, ...}

NO, IT CAN'T --- HOLY SHIT!

I guess you meant that {1/1, 2/1, 3/1, 4/1, 5/1, ...} u M = {1/1, 1/2, 2/1, 2/2, 3/1, 3/2, 4/1, 4/2, 5/1, 5/2, ...}

> both sets

{1/1, 2/1, 3/1, 4/1, 5/1, ...} and M = {1/2, 2/2, 3/2, 4/2, 5/2, ...}

> were countable and so is the resulting set

namey the UNION of {1/1, 2/1, 3/1, 4/1, 5/1, ...} and M.

Holy cow!

Hint: {1/1, 2/1, 3/1, 4/1, 5/1, ..., M} \ {1/1, 2/1, 3/1, 4/1, 5/1, ...} = {M}, and {M} =/= {1/2, 2/2, 3/2, 4/2, 5/2, ...}.

Please stop talking/writing mückennonsense.

So, NO, {1/1, 2/1, 3/1, 4/1, 5/1, ..., M} does not "mean" {1/1, 2/1, 3/1, 4/1, 5/1, ..., 1/2, 2/2, 3/2, 4/2, 5/2, ...} in any reasonable way. At least not in the context of _proper_ SET THEORY.

[Greg Cunt]

On Wednesday, October 27, 2021 at 1:46:05 PM UTC+2, Greg Cunt wrote:
> On Wednesday, October 27, 2021 at 1:16:22 PM UTC+2, FromTheRafters wrote:

if

> > M is {1/2, 2/2, 3/2, 4/2, 5/2, ...}

then

> > {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}

is {1/1, 2/1, 3/1, 4/1, 5/1, ..., {1/2, 2/2, 3/2, 4/2, 5/2, ...}}

(by the rule of substitution of identities), which isn't

> > {1/1, 1/2, 2/1, 2/2, 3/1, 3/2, 4/1, 4/2, 5/1, 5/2, ...}

in the context of set theory.

Note that we aren't engaged in mereology or mückenmath here.

See: https://en.wikipedia.org/wiki/Mereology
and: https://en.wikipedia.org/wiki/Delusional_disorder

[Gus Gassmann]

On Wednesday, 27 October 2021 at 06:53:40 UTC-3, WM wrote:
[...]

> I emphasized several times that I only show that one special mapping, which over decades had been believed to be a bijection, provably is not a bijection if

> {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...} [*]
> is a bijection, and
> {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M} [**]
> is not a bijection.

What utter nonsense. First, yes, [*] can be written out to look like a proper bijection. (Do it!) But your implication is bullshit. [*] and [**] can both be bijections, since you are dealing with infinite sets. (But of course you have no clue about infinity, so it is no wonder that you fuck up at least once an hour.)

[Greg Cunt]

On Wednesday, October 27, 2021 at 3:21:45 PM UTC+2, Gus Gassmann wrote:
> On Wednesday, 27 October 2021 at 06:53:40 UTC-3, WM wrote:
> [...]

> > if
> > {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...} [*]
> > is a bijection, and
> > {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M} [**]
> > is not a bijection.
> >
> What utter nonsense.

Especially since neither "{1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...}" nor "{1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}" denote functions in mathematics. Actually, "A --> B" usually is read as "(from) A into B", hence "A --> B" doesn't denote a function whatever.

With other words, Mückenheim's utterings are just nonsense.

[FromTheRafters]

Greg Cunt explained :

Yes, adjoined by the union of the two infinite sets in the muckymath
notation. No, I was not talking about equality, only the size of the
resulting infinite set. I was talking about bijection in general not
specifically any numerically bijective functions or element/set
generators.

WM says Cantor was wrong to sugggest a larger infinity than the
countable one we know and love. If WM's M is countable, so is the
union, but he doesn't say whether or not M is countable because we are
actually discussing muckymath and he likes to discount zero and the
continuum if it helps to muddy the waters.

If you thought you were having a non-mückennonsense discussion with
him, I will leave you to it.

Hint: He won't learn.

[Serg io]

On 10/27/2021 4:38 AM, WM wrote:
> Greg Cunt schrieb am Dienstag, 26. Oktober 2021 um 14:23:01 UTC+2:
>> On Tuesday, October 26, 2021 at 11:22:13 AM UTC+2, WM wrote:
>>
>>> I stick to mathematics. If
>>> {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
>>> is a bijection,
>> No. This is just nonsense.
>
> It appears so to you because permutations of the sets could be not in bijection in your opinion. But that is nonsense. Every transposition of one of the sets will yield a bijection.
>
> Cantor unconsciouly wrote the correct statement, destroying all his theory: "Die Frage, durch welche Umformungen einer wohlgeordneten Menge ihre Anzahl geändert wird, durch welche nicht, läßt sich einfach so beantworten, daß diejenigen und nur diejenigen Umformungen die Anzahl ungeändert lassen, welche sich zurückführen lassen auf eine endliche oder unendliche Menge von Transpositionen, d. h. von Vertauschungen je zweier Elemente."

Got URL ? or are you just making stuff up again ?

[Nikolai vdB]

No. Where is your definition of "completeness" from? Certainly not from modern mathematics. Certainly, bijective is not defined by any kind of 'completeness', but by being surjective and injective.
The function i gave you satisfies the definition of injective, anything about 'completeness' you make up is not part of real mathematics.

If you want to talk about real mathematics, you need to use its definition, elseways don't talk about it.

The definition of injective is that for a1 =! a2 in A, for a function f: A->B, it holds that f(a1) =! f(a2). Not more, not less. And that holds, which is why the function is injective. Elseways show me your definition of what injective means.

You NEED mathematical thoroughness, like that you are not saying anything because you just make up things.

[WM]

zelos...@gmail.com schrieb am Mittwoch, 27. Oktober 2021 um 13:34:49 UTC+2:
> >It is not a bijection because ℕ → ℕ is a bijection. That means completeness. No single elements can be added in domain or range without destroying the completeness
> Non one adds elements!

The set {1/1, 2/1, 3/1, 4/1, 5/1, ...} is countable. The set {1/1, 2/1, 3/1, 4/1, 5/1, ..., x} contains the same and one added element. It is not countable.
If a set like

{1/1, 2/1, 3/1, 4/1, 5/1, ..., M}

is not countable then adding further elements cannot make it countable. Therefore the set {1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ..., M}
is all the more not countable.

Regards, WM

[WM]

Nikolai vdB schrieb am Donnerstag, 28. Oktober 2021 um 17:06:54 UTC+2:
> WM schrieb am Mittwoch, 27. Oktober 2021 um 11:53:40 UTC+2:

> > > > Modern mathematics is not based upon completeness of bijections?
> > No answer.
> > > > >
> > > > > Fact is: Your "symmetry" is not necessary to prove a bijection,
> > > > It is valid and it turns out necessary. Do you think that it is invalid?
> > No answer.

> > I emphasized several times that I only show that one special mapping, which over decades had been believed to be a bijection, provably is not a bijection if
> > {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
> > is a bijection, and
> > {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}
> > is not a bijection.
> >
> > That is the present topic! Can you agree to the last statement, where the pairs of the mappings are taken from left to right?

> No. Where is your definition of "completeness" from? Certainly not from modern mathematics. Certainly, bijective is not defined by any kind of 'completeness', but by being surjective and injective.

Surjective is only another word for complete.

> The function i gave you satisfies the definition of injective, anything about 'completeness' you make up is not part of real mathematics.

Surjectivity is required. Here

{1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...}

the function f(n) = n/1 is surjective.
Here

{1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}

it is not surjective if M stands for other fractions than n/1.
>
> mathematical thoroughness,

Will certainly support this argument:
If {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M} is not surjective, then
{1, 2, 3, 4, 5, ...} --> {1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ..., M} will not be surjective either.

Regards, WM

[Gus Gassmann]

On Friday, 29 October 2021 at 10:48:27 UTC-3, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 27. Oktober 2021 um 13:34:49 UTC+2:
> > >It is not a bijection because ℕ → ℕ is a bijection. That means completeness. No single elements can be added in domain or range without destroying the completeness
> > Non one adds elements!
> The set {1/1, 2/1, 3/1, 4/1, 5/1, ...} is countable. The set {1/1, 2/1, 3/1, 4/1, 5/1, ..., x} contains the same and one added element. It is not countable.

Oh, bullshit. You are such a *FUCKING WANKER*. Sets have no order. So the set {1/1, 2/1, 3/1, 4/1, 5/1, ..., x} is the same set as {x, 1/1, 2/1, 3/1, 4/1, 5/1, ..., }. Regardless of what x stands for (neglecting to explain that is of course one of your trademark omissions), this set has cardinality aleph_0 + 1 = aleph_0.

But even if you intended an order when writing {1/1, 2/1, 3/1, 4/1, 5/1, ..., x}, all it says is that the order is not a well-order. It does not in any way influence countability. A first-semester student of mathematics would be expected to know this. You evidently do not. You have no clue about mathematics, you demented old fool.

[FromTheRafters]

on 10/29/2021, WM supposed :

> zelos...@gmail.com schrieb am Mittwoch, 27. Oktober 2021 um 13:34:49 UTC+2:
>>> It is not a bijection because ℕ → ℕ is a bijection. That means
>>> completeness. No single elements can be added in domain or range without
>>> destroying the completeness
>> Non one adds elements!
> The set {1/1, 2/1, 3/1, 4/1, 5/1, ...} is countable. The set {1/1, 2/1, 3/1,
> 4/1, 5/1, ..., x} contains the same and one added element. It is not
> countable.

The naturals adjoined with zero makes your naturals starting at one not
countable by your new definition. One countably infinite set has no
zero and the other has -- oh no!! Somebody do something, we need dark
numbers quick, uncountably many in fact, any fewer will not suffice.

> If a set like {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}
> is not countable then adding further elements cannot make it countable.
> Therefore the set {1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1,
> 1/6, ..., M} is all the more not countable.

Incomprehensible means not comprehensible.

[Serg io]

On 10/29/2021 9:01 AM, WM wrote:
> Nikolai vdB schrieb am Donnerstag, 28. Oktober 2021 um 17:06:54 UTC+2:
>> WM schrieb am Mittwoch, 27. Oktober 2021 um 11:53:40 UTC+2:
>
>>>>> Modern mathematics is not based upon completeness of bijections?
>>> No answer.
>>>>>>
>>>>>> Fact is: Your "symmetry" is not necessary to prove a bijection,
>>>>> It is valid and it turns out necessary. Do you think that it is invalid?
>>> No answer.
>
>>> I emphasized several times that I only show that one special mapping, which over decades had been believed to be a bijection, provably is not a bijection if
>>> {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
>>> is a bijection, and
>>> {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}
>>> is not a bijection.
>>>
>>> That is the present topic! Can you agree to the last statement, where the pairs of the mappings are taken from left to right?
>
>> No. Where is your definition of "completeness" from? Certainly not from modern mathematics. Certainly, bijective is not defined by any kind of 'completeness', but by being surjective and injective.
>
> Surjective is only another word for complete.

Wrong, Entirely. As usual.

>
>> The function i gave you satisfies the definition of injective, anything about 'completeness' you make up is not part of real mathematics.
>
> Surjectivity is required. Here
> {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
> the function f(n) = n/1 is surjective.
> Here
> {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}
> it is not surjective if M stands for other fractions than n/1.

Wrong. M stands for Mistake.

>>
>> mathematical thoroughness,
>
> Will certainly support this argument:
> If {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M} is not surjective, then
> {1, 2, 3, 4, 5, ...} --> {1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ..., M} will not be surjective either.
>
> Regards, WM
>

Wrong

[FromTheRafters]

WM explained on 10/29/2021 :

> Nikolai vdB schrieb am Donnerstag, 28. Oktober 2021 um 17:06:54 UTC+2:
>> WM schrieb am Mittwoch, 27. Oktober 2021 um 11:53:40 UTC+2:
>
>>>>> Modern mathematics is not based upon completeness of bijections? No
>>>>> answer.
>>>>>>
>>>>>> Fact is: Your "symmetry" is not necessary to prove a bijection,
>>>>> It is valid and it turns out necessary. Do you think that it is invalid?
>>> No answer.
>
>>> I emphasized several times that I only show that one special mapping, which
>>> over decades had been believed to be a bijection, provably is not a
>>> bijection if {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...} is a
>>> bijection, and {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}
>>> is not a bijection.
>>>
>>> That is the present topic! Can you agree to the last statement, where the
>>> pairs of the mappings are taken from left to right?
>
>> No. Where is your definition of "completeness" from? Certainly not from
>> modern mathematics. Certainly, bijective is not defined by any kind of
>> 'completeness', but by being surjective and injective.
>
> Surjective is only another word for complete.

In muckymath. Maybe you should explain this.

>> The function i gave you satisfies the definition of injective, anything
>> about 'completeness' you make up is not part of real mathematics.
>
> Surjectivity is required. Here
> {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ...}
> the function f(n) = n/1 is surjective.
> Here
> {1, 2, 3, 4, 5, ...} --> {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}
> it is not surjective if M stands for other fractions than n/1.

Why not? Is there some muckyaxiom in play here? Is big M a non-empty
set of little m elements which you wish to randomly "merge" amongst the
n/1 sequence of elements?

How many of them?

[Serg io]

On 10/29/2021 10:31 AM, FromTheRafters wrote:
> on 10/29/2021, WM supposed :
>> zelos...@gmail.com schrieb am Mittwoch, 27. Oktober 2021 um 13:34:49 UTC+2:
>>>> It is not a bijection because ℕ → ℕ is a bijection. That means completeness. No single elements can be added in domain or range without destroying
>>>> the completeness
>>> Non one adds elements!
>> The set {1/1, 2/1, 3/1, 4/1, 5/1, ...} is countable. The set {1/1, 2/1, 3/1, 4/1, 5/1, ..., x} contains the same and one added element. It is not
>> countable.
>
> The naturals adjoined with zero makes your naturals starting at one not countable by your new definition. One countably infinite set has no zero and the
> other has -- oh no!! Somebody do something, we need dark numbers quick, uncountably many in fact, any fewer will not suffice.

Uncountably many Parashooting Dark Ants to the Rescue with Zero Replacement

>
>> If a set like {1/1, 2/1, 3/1, 4/1, 5/1, ..., M}
>> is not countable then adding further elements cannot make it countable. Therefore the set  {1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1,
>> 1/6, ..., M} is all the more not countable.
>
> Incomprehensible means not comprehensible.

WM's mislogic with M indicates misfiring neuron at the last Marble.

[Serg io]

correction


WM's Mislogic with M Means Misfiring Meuron at last Marble.

[Nikolai vdB]

Surjective is not another word for complete. Use real definitions. Complete usually means something else in mathematics.
And for injectivity, surjectivity is not required.
You claimed that the function i gave you, N -> N, n -> n+1 was not injective. But it is. I never claimed it was surjective, but you denied it was injective even tho the definition of injective is satisfied. YOu need mathematical thoroughness, like this, nothing you say is worth anything because it means nothing, you do not even use proper definitions.