Cardinality and Bijection of finite sets

[Strange]

1 minute ago, Edgard Neuman said:

But it is enough to prove the bijection ? I don't think so.

You have ignored the most important part of that post. The part that will prove your argument wrong.

[Edgard Neuman]

1 minute ago, Strange said:

You have ignored the most important part of that post. The part that will prove your argument wrong.

What did I ignore ? 

[studiot]

Just now, Edgard Neuman said:

What did I ignore ? 

My last comments.

[Strange]

8 minutes ago, Edgard Neuman said:

You said the set is unlimited.. so you must prove your "etc" part for the bijection to be complete.. go ahead..  (take the infinite time in the univers)

This is true by definition: every integer has a successor and so the series is unlimited.

[Edgard Neuman]

16 minutes ago, studiot said:

This is the sort of imprecise mathematics that worries me (and others it seems) and lead to my earlier post the last time I looked in.

You defined the cardinality of N to be

right at the outset.

 

So I asked you to explain what number system and arithmetic system are you using when you introduced   (ω−1)     since omega is not finite.

 

"since omega is not finite" 
You accuse me of being imprecise, and you use the adjective "finite" to describe a element. (but Ok that maybe a english way of speach) 

 assume you meant positioned at infinite. . And as I said before, you do your math based on construction you made to accommodate the bijection. I don't. 
You wanted a precise definition of ω

I already gave you one : 

ω is defined by :
(1) for any n ∈ ℕ, we have n<ω, 
(2) there is no n with n<ω and n not in  ℕ
- so ω-1 ∈ ℕ,  and ω don't ℕ

ω-1 ∈ ℕ  because of (2). 

You just can't suppose that
Card(ℕ)=ω
and ω-1 is not in ℕ

because then It would contradict (2)

For some reason, current mathematician decided to leave a "space" between ℕ and ω for numbers  ω-1, I don't. because then Card(ℕ) wouldn't be ω

Edited October 18, 2018 by Edgard Neuman

[Strange]

6 minutes ago, Edgard Neuman said:

What did I ignore ? 

This is painful.

You ignored this:

Quote

The only problem would be if there was a maximum integer n, then 2*n would not be in the set.

It's up to you to demonstrate that there is some integer n in that set, but not n+1...

It is up to YOU to prove that there is a maximum integer in N. 

[Edgard Neuman]

8 minutes ago, Strange said:

This is true by definition: every integer has a successor and so the series is unlimited.

it doesn't prove the bijection.. how does it ? 

6 minutes ago, Strange said:

This is painful.

You ignored this:

It is up to YOU to prove that there is a maximum integer in N. 

I've repeated all along that you can ETHER assume Card(ℕ)=ω or NOT ASSUME. 
In the first case, (ω-1) * 2 is not in ℕ => NO BIJECTION 
In the second case, you can't prove the bijection because it's a AND operation that is never finished and the last FALSE can always disprove it. => NO BIJECTION

Edited October 18, 2018 by Edgard Neuman

[Strange]

3 minutes ago, Edgard Neuman said:

it doesn't prove the bijection.. how does it ? 

It disproves your arguments about there being integers that are not in N (or whatever nonsense you are claiming).

4 minutes ago, Edgard Neuman said:

I've repeated all along that you can ETHER assume Card(ℕ)=ω or NOT ASSUME. 

This is not an assumption, it is the definition. We know Card(N) is infinite because it is larger than any n in N.

6 minutes ago, Edgard Neuman said:

In the first case, (ω-1) * 2 is not in ℕ

Correct. Because you can't do integer arithmetic on infinity.

6 minutes ago, Edgard Neuman said:

operation that is never finished

It is not, as you said before, a sequential operation. So this is a bogus argument.

[Carrock]

7 minutes ago, Edgard Neuman said:

13 minutes ago, Carrock said:

Maybe a process using unlimited sets would make the (lack of a) problem with infinite sets clearer.

You have the unlimited integer set 1,2,3,4...

2*1 is in that set as is 2*2, 2*3, 2*4 etc.

The only problem would be if there was a maximum integer n, then 2*n would not be in the set.

It's up to you to demonstrate that there is some integer n in that set, but not n+1...

But it is enough to prove the bijection ? I don't think so.


You said the set is unlimited.. so you must prove your "etc" part for the bijection to be complete.. go ahead..  (take the infinite time in the univers)

If you're correct I'm sure you can define the properties of the largest finite number.

 

15 hours ago, Edgard Neuman said:

Card(ℕ) =ω
implies that
( a ∈ ℕ ) if and only if  ( a<ω )

I presume you checked every a before making that statement (and many others). Or is it only those who oppose you who must check every value?

[Edgard Neuman]

The second case is tricky so let me explain again..
 

4 minutes ago, Strange said:

It disproves your arguments about there being integers that are not in N (or whatever nonsense you are claiming).

so n is an integer and n*2 is an integer.. so you have n->n*2.. 
but if n*2 is an integer, to prove the bijection, n*2 must also have a counterpart.. right ? 
let's try n*4... but does n *4 have a counterpart ? lets calculate .. and so on 
CAN YOU PROVE ME that 
n*2*2*2*2*2*.... is an integer ? I don't think so. so the bijection is it proven for ALL integer ? 

Edited October 18, 2018 by Edgard Neuman

[Strange]

11 minutes ago, Edgard Neuman said:

In the first case, (ω-1) * 2 is not in ℕ => NO BIJECTION 

Neither (ω-1) * 2 nor (ω-1) are in N. So this is not relevant.

[Edgard Neuman]

11 minutes ago, Strange said:

Neither (ω-1) * 2 nor (ω-1) are in N. So this is not relevant.

Do you even read what I write ???  I constructed ω so ω-1 is in N. I don't care about you preconcieve notion of what you call ω. It's a ω I DEFINED SO, as someone asked me to explain. If you don't use my definition, why do you ask for it ?? 

I can copy and paste : 

ω is defined by :
(1) for any n ∈ ℕ, we have n<ω, 
(2) there is no n with n<ω and n not in  ℕ
- so ω-1 ∈ ℕ,  and ω don't ∈ ℕ

ω-1 ∈ ℕ  because of (2). 

You just can't suppose that
Card(ℕ)=ω
and ω-1 is not in ℕ

 

because then It would contradict (2)

SO YES ω-1 ∈ ℕ  because of (2). 

why the hell would you create Card(ℕ)=ω 
if ω-1 is not in ℕ ??? That would make no sense..

Card([0;4[) = 4; Card([0;4[) -1 = 3 ; 3 ∈ [0;4[
Card([0;5[) = 5; Card([0;5[) -1 = 4 ; 4 ∈ [0;5[
Card([0;6[) = 6; Card([0;6[) -1 = 5 ; 5 ∈ [0;6[
....
Card([0;n[) = n; Card([0;n[) -1 = n-1 ;n-1 ∈ [0;n[
...
and 
IF [0;ω[ = ℕ
Card([0;ω[) = ω; Card([0;ω[) -1 = ω-1 ;ω-1 ∈ [0;ω[
Card(ℕ) = ω; Card(ℕ) -1 = ω-1 ;ω-1 ∈ ℕ

 

 

Edited October 18, 2018 by Edgard Neuman

[uncool]

20 minutes ago, Edgard Neuman said:

I can copy and paste : 

ω is defined by :
(1) for any n ∈ ℕ, we have n<ω, 
(2) there is no n with n<ω and n not in  ℕ

You would have to prove there is such an ω. When considering the integers, there is no such ω; when working with extensions, you will have to define your operations (addition, subtraction, etc.)

If you are truly usinng cardinalities, then there is no subtraction. ω-1 does not make sense.

2 hours ago, Edgard Neuman said:

And again, if A contains every integer, B= {a*2 } contains elements that are NOT in A, therefor if A = N, those added numbers are NOT integers.
You can't "create them" at this imaginary point in your mind. And so your "couple" for the bijection doesn't exist.

I'm not "creating them". B does not contain elements that aren't in A. For any integer x, 2x is also an integer.

[Strange]

41 minutes ago, Edgard Neuman said:

Do you even read what I write ???  I constructed ω so ω-1 is in N.

That is not possible. You said that w = Card(N). Card(N)-1 is not in N.

 

41 minutes ago, Edgard Neuman said:

ω is defined by :
(1) for any n ∈ ℕ, we have n<ω, 
(2) there is no n with n<ω and n not in  ℕ
- so ω-1 ∈ ℕ,  and ω don't ∈ ℕ

ω-1 ∈ ℕ  because of (2). 

You just can't suppose that
Card(ℕ)=ω
and ω-1 is not in ℕ

because then It would contradict (2)

It doesn't contradict (2) because ω-1 = ω (if it is defined at all)

The problem with your argument is that you say there is an integer (ω-1) in N such that adding 1 is no longer in N. Firstly, this contradicts the definition of integers (every integer has a successor) but it also says there is a largest integer. Please tell us what the value of this largest integer is. And then why you can't just add 1 to it.

 

Edited October 18, 2018 by Strange

[studiot]

56 minutes ago, Edgard Neuman said:

I already gave you one : 

ω is defined by :
(1) for any n ∈ ℕ, we have n<ω, 
(2) there is no n with n<ω and n not in  ℕ
- so ω-1 ∈ ℕ,  and ω don't ℕ

ω-1 ∈ ℕ  because of (2). 

And I already asked about that sequence since it's logic is false.

(2) quite clearly states (although in poor English) that omega is not a member of N.

Therefore arithmetic between omega and any member of N is not defined in the normal Peano system, since Peano arithmetic is closed under addition.

So ω-1 is undefined in your system at the outset of your 'proof'

So I asked you what system you are using.

Edited October 18, 2018 by studiot

[Strange]

@Edgard Neuman Several people here think you are wrong. You are disagreeing with some of the greatest mathematicians who have ever lived. Have you ever stopped for a moment to consider that the problem might be with your understanding?

[Edgard Neuman]

1 hour ago, Strange said:

@Edgard Neuman Several people here think you are wrong. You are disagreeing with some of the greatest mathematicians who have ever lived. Have you ever stopped for a moment to consider that the problem might be with your understanding?

That's not how maths works.. 
 

1 hour ago, studiot said:

And I already asked about that sequence since it's logic is false.

(2) quite clearly states (although in poor English) that omega is not a member of N.

Therefore arithmetic between omega and any member of N is not defined in the normal Peano system, since Peano arithmetic is closed under addition.

So ω-1 is undefined in your system at the outset of your 'proof'

So I asked you what system you are using.

So then ω is obviously not Card(N) which contradict (1)
To be > any n isn't enough to be the cardinal of a set.
The cardinal of {1;2;3;4} isn't 12.

1 hour ago, studiot said:

Therefore arithmetic between omega and any member of N is not defined in the normal Peano system, since Peano arithmetic is closed under addition.

(-1) is not an addition.. so I can simply assert (by axiom) that ω-1 is in N

 

1 hour ago, studiot said:

(2) quite clearly states (although in poor English) that omega is not a member of N.

I can't wait to see what your French looks like

1 hour ago, studiot said:

(2) quite clearly states (although in poor English) that omega is not a member of N.

 

Therefore arithmetic between omega and any member of N is not defined in the normal Peano system, since Peano arithmetic is closed under addition.

So ω-1 is undefined in your system at the outset of your 'proof'

So I asked you what system you are using.

So ω + 1 or ω-1 doesn't make sense..  or is it  =ω ? 
let's assume  ω + 1 = ω
So if I take
S = 1+1+1+1+... 
and I take the first 1  and I leave the rest.. the R 
So we have S= 1 +R 
because R= 1+1+1 + ....   and ω = ω +1 
R= 1+1+1+ ( 1 + .... ) 
so R = S = R+0
So  0 = 1

We come to a contradiction, right ? So if you really do "serious" math,  it prove that ω + 1 != ω

How sorry that's right you can't subtract infinite value from an equation, I get that. 
So R =R +1 = R +1 +1 = R + 1+1+1 .... = 2*R  
So R= 2 * R  but you can't divide right.. 
So R = 2* R= R *2 *2 = R * 2 *2 *2 *2 * .... = R * (2 ^R ).. right ? 
So Cantor was wrong with his diagonal argument..  right ? 

You just have to use the logic that "everything infinite" = "everything else infinite" to quickly come to something nobody never came up with, and you genius mathematician didn't add an axiom like "you can't do that, because it's forbidden by the bijection" .. and over the years you add a lot of those rules .. like an infinity of them
Why do you think I am here ? 
I have a nice video from numberphile that says 1+2+3+4+5+ ....  = -1/12 .. but because ... = ... +1 ? Isn't it 
1+2+3+4+5+ ....  = -1/12  +1   ? I'm confused 

 

Edited October 18, 2018 by Edgard Neuman

[studiot]

33 minutes ago, Edgard Neuman said:

 

Quote

Why do you think I am here ? 

 

 

Well the mockery in the rest of your post is a shame because I only have a few minutes and then I will be otherwise occupied for a couple of hours.

 

It is axiomatic in Peano that for all a, b in N then (a+b) = c is in N

Those who are closer to this stuff will tell you which of the Peano axioms this is but I seem to remember it was one of the first.

You are not only trying to contradict more advanced results of arithemtic you are trying to contradict the basic axioms.

 

Think on that.

[Strange]

38 minutes ago, Edgard Neuman said:

So ω + 1 or ω-1 doesn't make sense..  or is it  =ω ? 
let's assume  ω + 1 = ω
So if I take
S = 1+1+1+1+... 
and I take the first 1  and I leave the rest.. the R 
So we have S= 1 +R 
because R= 1+1+1 + ....   and ω = ω +1 
R= 1+1+1+ ( 1 + .... ) 
so R = S = R+0
So  0 = 1

We come to a contradiction, right ?

Which is a common (informal) proof that you can't do arithmetic using infinity. So your ω + 1 or ω-1 doesn't make sense.

40 minutes ago, Edgard Neuman said:

I'm confused 

We can probably all agree on that. As I say, take some time to think about where the problem might be.

[Edgard Neuman]

 

that's a axiom I add. It's different that what you know, and It solve the problem. 

I define ω SO THAT Peano stops at ω.. so yes, In my axiomatic, (a+b) ISN'T always in N, that's exactly how I built it. 
What I use is a limited version of Peano.. that's how I calculate the Card(Q) and Card(R)..

It is in my FIRST POST, right there : 

18 hours ago, Edgard Neuman said:

I think I have proven that what's usually accepted as bijection and cardinality is all wrong, and nobody seems to believe me or care.. what should I do ? I know you don't easily reject old accepted ideas, but can you at least really think about it ? 

First : bijection between two infinite set are not true, they are UNDECIDABLE. When you build a bijection, it is true only when you don't get a new element from one of the sets or both, not just when you can still get elements from the sets. So with 2 infinite set, the bijection is NEVER proven true, because you can always get new elements from both. It just keep infinitely UNDECIDABLE.
So you can NEVER prove a bijection between for instance ℕ and ℚ (precisely because you can always get new elements from both)..
There's an infinite number of ways of ordering elements in ℚ (even in ℕ), that doesn't define the bijection as true or anything.  (that seems important to the people I spoke to for some reason)

Second, if you define
Card(ℕ) =ω
You can in fact easily find Card(ℚ) and Card(ℝ) : 

Card(ℕ) =ω
implies that
( a ∈ ℕ ) if and only if  ( a<ω ) 
so 
( a+b ∈ ℕ ) if and only if  ( a+b<ω ) 
and so 
( (a/b) ∈ ℚ ) if and only if   (-ω<a<ω )  and (b<ω)
So you can actually count the number of element in ℚ according to ω :

Card(ℚ+)=A018805 (ω-1) +1

Card(ℚ)=Card(ℚ+)*2-1 = A018805 (ω-1) *2 +1

Card(ℚ)=~ ω² * (3/Pi²)

 

You want to leave a infinite space between ω and the integer, because you want to respect Peano infinite. 

I try to solve all the bijection problem, by stating that the only proper way to work with infinite set that is always coherent, is to treat them as the limit case of finite sets.. because if all proposition are true together, you can have a first set of Peano elements there and a bigger (when it have some new elements built in, it's bigger) one somewhere else in the same universe (Especially at the two ends of the same bijection). That's the whole philosophy of it. You asked me how I build integer so I said the usual way : with Peano.. I never said that  my set of axiom is the whole set of axiom from Peano, uncontradicted somewhere. 
ω is just the variable I play with to access this limit. So in my axiomatic N = [0;ω[ and yes ω-1 ∈ ℕ

(I remember now that I had the same kind of arguments years ago, so that's why I wanted to come up with a way of comparing infinite set closer to what "intuition about quantities" tells us)

To come up with A018805  and A003418, I assumed ω is an integer and I computed the Card(Q) and Card(R) with this limited set [0;ω[..

Edited October 18, 2018 by Edgard Neuman

[studiot]

2 hours ago, Edgard Neuman said:

that's a axiom I add. It's different that what you know, and It solve the problem. 

I define ω SO THAT Peano stops at ω.. so yes, In my axiomatic, (a+b) ISN'T always in N, that's exactly how I built it. 
What I use is a limited version of Peano.. that's how I calculate the Card(Q) and Card(R)..

Did you say somewhere

3 hours ago, Edgard Neuman said:

That's not how maths works.. 

 

Sorry but you start with axioms in Maths.

 

You don't pick them up in Carrefour somewhere along the journey.

 

The point about this axiom is that you are drawing upon a whole structure of maths that rests upon it

Almost everything in arithmetic depends directly upon the property of being closed under addition.

Donc, ce n'est pas possible de conduire un entraîneur et des chevaux à travers elle.

 

Bonsoir

[wtf]

3 hours ago, Edgard Neuman said:

I have a nice video from numberphile that says 1+2+3+4+5+ ....  = -1/12 .. but because ... = ... +1 ? Isn't it 
1+2+3+4+5+ ....  = -1/12  +1   ? I'm confused 

 

I'm sure Numberphile explained that this is an INFORMAL way of looking at something much more subtle. You need to use the analytic continuation of the Riemann zeta fuction via a process called zeta function regularization, to show that zeta(-1) = -1/12. You can CASUALLY think of zeta(-1) as sort of looking like the formal expression 1 + 2 + 3 + ... but it actually isn't. That series diverges to infinity. 

It's sad that so many people have gotten confused by this video. Numberphile is usually reputable but they unleashed a lot of confusion online about this issue.

https://blogs.scientificamerican.com/roots-of-unity/does-123-really-equal-112/

https://plus.maths.org/content/infinity-or-just-112

Edited October 18, 2018 by wtf

[uncool]

4 hours ago, Edgard Neuman said:

that's a axiom I add. It's different that what you know, and It solve the problem. 

If you're adding an axiom, you will have to demonstrate that your new system is consistent (or at least, equiconsistent). 

[Edgard Neuman]

7 hours ago, uncool said:

If you're adding an axiom, you will have to demonstrate that your new system is consistent (or at least, equiconsistent). 

Well, there is something that I don't understand, each time I speak with people 
Why don't you THINK BY YOURSELF ? 
Everything is written is THE FIRST POST.

Can't you figure out that when I write  "when a+b>=ω it's not in N " It means exactly what is written  ?
I get really annoyed I have to explain how I "build numbers". So annoyed that I get confused myself in my own idea. 
At some point I said I used Peano or {{{}}} or even any system that generate new elements different than the precedents 
Because that don't matter here. It's NOT in the FIRST POST. 

What next ? You want me to explain the sign " +" the sign ">", the usage of the alphabet ? 

You should be able

- to REALLY read the hypothesis, the axioms, understand them and what they mean. 
(why does it take 20 post for you to figure out what  the axiom means ? )
- to think about everything they imply (EVERYTHING) BY THEMSELF, and certainly not by OTHER AXIOM YOU KNOW  
(why did you ever suppose that ω-1 is not in N ? Did I once said ω is a ordinal number ?? WHERE ? 
It's in the hypothesis ! Do I really have to explain each and every axiom I DON'T USE because somebody can't figure out that a letter is just a letter ? )

So no, "I don't have to demonstrate" it is "consistent".  (What does that even mean ? I think a axiom while contradict some other axiom ?)
You do that, according to the hypothesis, not you own math. 
And if you find something, you tell me. That's why I'm here. 
Stop asking question about thing that you should be able to figure out from the hypothesis. 
 

Edited October 19, 2018 by Edgard Neuman

[uncool]

4 minutes ago, Edgard Neuman said:

Can't you figure out that when I write  "when a+b>=ω it's not in N " It means exactly what is written  ?

I can't figure out what that means, no. You're clearly using "+" to mean something different from how it is usually used. To answer a later question in your post, yes, since you are using "+" in a nonstandard way, I do expect you to define it.