Cardinality and Bijection of finite sets

[Edgard Neuman]

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)

Numbers in ℚ are spreaded like a fractal and are symmetrical by the transformation x=> 1/x

Then I build each x in ℝ as a sum of ω numbers from ℚ :

ℝ={any x=Sum(a[n]/b[n]; n=1..ω}
with -ω<a[n]<ω et 0<b[n]<ω
and -ω<x<ω
(That sum can actually be > ω)

I find that :
Card(ℝ)=(A003418(ω-1)*ω*2)-1
and guess what : 

Number in ℝ are spreaded perfectly evenly in ]-ω;ω[ with a step of 1/A003418(ω-1) 
(that is not obvious at all given the spreading of numbers in ℚ)

I know it's extraordinary claims..  very different from what is currently the mainstream math..
but can you prove me wrong ? 
 

 

[uncool]

9 minutes ago, Edgard Neuman said:

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

A bijection isn't (generally) a step-by-step thing. It can be built step-by-step, but the function simply is.

For example: the identity map on the natural numbers, that takes each number to itself. This can be built step-by-step, sending n to n, but it simply exists: given any n, it returns n. It is clearly a bijection, just by checking definitions.

[Edgard Neuman]

43 minutes ago, uncool said:

A bijection isn't (generally) a step-by-step thing. It can be built step-by-step, but the function simply is.

For example: the identity map on the natural numbers, that takes each number to itself. This can be built step-by-step, sending n to n, but it simply exists: given any n, it returns n. It is clearly a bijection, just by checking definitions.

I understand that, but if you start with all the elements from ℕ as an hypothesis, you can't then use numbers out of that set. 
You can use n-> n (fortunately, I wouldn't contradict that!),

but you can't have n -> n*2  or n-> n+1  because if you assume that  you have ALL the elements from ℕ in the left set,
some elements from the right set are always bigger, so they are not in the ℕ you started with (unless you didn't have ALL, which is a contradiction with the hypothesis, and in that case, the bijection wouldn't be decided)

I came up with the construction of the bijection because I wanted to avoid using ω

(sorry for the multiple editing)
 

Edited October 18, 2018 by Edgard Neuman

[wtf]

6 hours ago, Edgard Neuman said:

but you can't have n -> n*2  or n-> n+1  because if you assume that  you have ALL the elements from ℕ in the left set,
some elements from the right set are always bigger, so they are not in the ℕ you started with (unless you didn't have ALL, which is a contradiction with the hypothesis, and in that case, the bijection wouldn't be decided)
 

I don't understand this point at all. 

Let's look at a simpler case. We have two sets A and B.

A = {1, 2, 3, 4, 5} and B = (2, 4, 6, 8, 10}.

It seems perfectly clear that we can biject the elements of A to those of B, despite the fact that the numbers in B are larger than some of the ones in A. And it's true that some of the elements of B are not in A. Why does this trouble you?

[studiot]

Perhaps it would help if Edgard told us what he understands Bijection to mean

That is what definition of bijection are you using.

So when you match some apparant inconsistency against this definition, where exactly do they conflict?

[Edgard Neuman]

1 hour ago, wtf said:

I don't understand this point at all. 

Let's look at a simpler case. We have two sets A and B.

A = {1, 2, 3, 4, 5} and B = (2, 4, 6, 8, 10}.

It seems perfectly clear that we can biject the elements of A to those of B, despite the fact that the numbers in B are larger than some of the ones in A. And it's true that some of the elements of B are not in A. Why does this trouble you?

You would admit that, in your example,

B can't be a part of A (because some elements in B are always bigger than the biggest in A)

So it works for finit sets, but not with ℕ

Because if 

A is ℕ, and
B can't be a part of A

some elements in B are not in ℕ (so it's not a bijection)

you have something like : 

1->2
2->4

n->n*2

... it's Ok.. but : 

card(ℕ)-1 -> (card(ℕ)-1)*2

card(ℕ)-1 is in ℕ 
but (card(ℕ)-1)*2 >=  card(ℕ)
so it's NOT in ℕ 
You have an element in A that have no counterpart in B if B is a set of elements from ℕ

you have
A = {0 ;1;2;3;4;...}
B = {0 ;2;4;6;8;...}
in that case, you can't have 

A include B and Card(A) = Card(B)

if Card(A) = Card(B) some elements of B are not in A

you can't add new numbers to ℕ when needed

And, in the other hand, if you don't define Card(ℕ)=ω, you can't prove that you have all the couples of elements from A and B (that is needed to say the bijection exists). 
Building a bijection is not a mathematical induction.. the fact that can always get new elements from A and B doesn't imply that the bijection exists or doesn't, because to do so, you have to reach the end.. (it's in fact, the opposite)

You say that n->n*2 always exists for any n, but how can you prove that ? the fact that n -> n*2 doesn't imply that (n+1)->(n+1)*2 exists at all.. 

Edited October 18, 2018 by Edgard Neuman

[studiot]

8 hours ago, Edgard Neuman said:

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)

Well this is using non standard definitions for arithmetic and number systems so you need to start with definitions of you symbols and proofs of the properties of the entities you invoke to be able to 'proove' anything.

I also asked about your definition of bijection?

Why did you not take the opportunity to answer when you had more than 20 minutes logged on here after your last post?

[Edgard Neuman]

1 hour ago, studiot said:

Well this is using non standard definitions for arithmetic and number systems so you need to start with definitions of you symbols and proofs of the properties of the entities you invoke to be able to 'proove' anything.

I also asked about your definition of bijection?

Why did you not take the opportunity to answer when you had more than 20 minutes logged on here after your last post?

you can use Peano or the {{{{}}}} to create integers.
It could also be simplified by "elements" that are different from each other,  you don't even need to order them to define a bijection.
In any case, if one of hypothesis is you have "all" the elements, you can't then use the creation process.. creating a new element different than all you already have means you didn't had all in the first place.

my definition of bijection is the standard one.. I suppose
You create a bijection between to sets A and B when each element of A is paired with a unique element of B and vice-versa
(you can call it a "function" if you want, it's a set of couples from A * B)

in other words : 
exist(bijection) = exist(a[1];b[1]) AND exist(a[2];b[2]) AND exist(a[3];b[3]) .. 

with obviously all the a[] and b[] different elements  and no remaining elements from a or from b. 

If you give me an other one, without adding extra axiom to suit you demonstration, i will work with it.

[Of course if you arbitrarily add "except for infinite sets", or you add "when ordered a certain way" you just added an axiom to avoid the contradiction i'm talking about..
If you add special axiom to the definition for infinite sets, why not add anything else you want ? ]


"Why did you not take the opportunity to answer when you had more than 20 minutes logged on here after your last post?" 

(because I'm in France, in lunchtime, and in fact I don't think questioning my definition of the bijection is relevant)

1 hour ago, studiot said:

Well this is using non standard definitions for arithmetic and number systems so you need to start with definitions of you symbols and proofs of the properties of the entities you invoke to be able to 'prove' anything.

I start with assuming that Card(ℕ)=ω

ω here is like a imaginary boundary
- for any n ∈ ℕ, we have n<ω, 
- there is no n with n<ω and n not in  ℕ
- so ω-1 ∈ ℕ,  and ω don't ℕ

ℕ are integers using Peano or anything else that define the usual a+b nothing new here
so : 
ℕ=[0;ω[
(so of course)

ℚ={any q= a/b} with -ω<a<ω and 0<b<ω (a and b from ℕ)
with the ordinary identities.. we can add that PGCD(a;b)=1 to remove equal couples
a/b=c means a=b*c,  so it implies a/b= a*x/b*x... (a,b,c,x integers and without the problematic zeroes)

ℝ={any x=sum(q[n] with n=1..ω} (q[n] from ℚ)
with -ω<x<ω

I think I sometimes speak and come as an annoying person, I hope I didn't offend you in any way.  that should'nt have any impact on the "math" though. 

Edited October 18, 2018 by Edgard Neuman

[Edgard Neuman]

I don't see "speculations" here.. but ok

to obtain Card(Q) and Card(R), I simply counted the set according to values of ω, and it matches the sequences

you add a lot of axiom everywhere to avoid the annoying paradoxes created solely by those bijections 
like Card(N) = Card(Even numbers)  and Card(N) > Card(Even numbers) ..

so for instance if 
S = 1 +1 +1 +1 ... 
and you separate the  1 in two equal sets S1 and S2,    if you use your bijection, you get S=S1+S2 and S=S1=S2
so S=0.. yeah good job with your bijections !
 In theory if your math come to a contradiction, you don't admire it as a magical mystery, you don't add a specific axiom to remove the one proposition you can't handle. you simply look for the wrong hypothesis.. (the bijection, obviously)

The modern math is full of these absurdities to the point where a guy explain to you seriously that 1+2 + 3 +.... = - 1/12
And you call my idea "speculations".. I'm way better with my ideas, thanks

 

Edited October 18, 2018 by Edgard Neuman

[uncool]

13 hours ago, Edgard Neuman said:

I understand that, but if you start with all the elements from ℕ as an hypothesis, you can't then use numbers out of that set. 
You can use n-> n (fortunately, I wouldn't contradict that!),

but you can't have n -> n*2  or n-> n+1  because if you assume that  you have ALL the elements from ℕ in the left set,
some elements from the right set are always bigger, so they are not in the ℕ you started with (unless you didn't have ALL, which is a contradiction with the hypothesis, and in that case, the bijection wouldn't be decided)

I came up with the construction of the bijection because I wanted to avoid using ω

(sorry for the multiple editing)
 

Again: a bijection doesn't exist "step by step". It exists all at once. For every number, it is possible to double that number. This process is injective, and by definition of even numbers surjective to the even numbers. Therefore, it is a bijection.

[Edgard Neuman]

39 minutes ago, uncool said:

Again: a bijection doesn't exist "step by step". It exists all at once. For every number, it is possible to double that number. This process is injective, and by definition of even numbers surjective to the even numbers. Therefore, it is a bijection.

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.
At most, if you say your a*2  is in A, you then have to check that a*2 has is own element in B. and because this never end, you really DON'T know. 

So no, as strange as it sound, "For every number, it is NOT possible to double that number". You can't double infinite and except it to fit in itself. 
For the number you can reach, yes, but, not for ALL.. "all" means ALL, to the END of the set. You don't see it, because you mistake reaching  ALL for a recursive process.. But in reality, you recursive thinking NEVER really reach ALL. You do for one, for the other etc.. and you think it's ALL, but it's not. By definition, you can't reach ALL, because it's a infinite set.  The only thing you can do, is put a unit of measurement,  (ω) and see how your set behave according to  ω. (that's what I did).. .
In one sentence : if you assume N has a last number, this number doesn't have any counterpart in the bijection, and if you assume that  N doesn't have a last element, you bijection is never proven true (because bijection requires all couples to exists).

In fact, there is a very simple to prove it, but you will reject it because you trained to see it as something to "avoid" because of infinity instead of simply accepting it.. 
For every set if

A = B union C
and
B and C are not empty and does not intersect.
Card(A) = Card(B)  + Card(C)
and 
Card(A) > Card(B) and  Card(A) > Card(C)
So if N = {even numbers} union {odd numbers}
Card(N) > Card(even numbers). NOT EQUAL. 

If this is not a proof, what is it ? There you have a reasoning, that consider all numbers together, absolutely out of time and steps, that doesn't involve knowing or building something from the elements, and simply consider the properties of numbers (being even or odd). And for some reason, modern math decided to go the other way that is just creating a lot of problems for no  reason. 



It's like the Hilbert hotel.. the infinite hotel is full.. the paradox state that you can slide everybody to the next room. But really can you ? The hotel is full, so there is NO empty room at the end of the hotel. 

 

 

Edited October 18, 2018 by Edgard Neuman

[Strange]

39 minutes ago, Edgard Neuman said:

And again, if A contains every integer, B= {a*2 } contains elements that are NOT in A,

Huh? 

If A is the set ofall integers, then it contains all the integers multiplied by 2, therefore it contains B.

40 minutes ago, Edgard Neuman said:

therefor if A = N, those added numbers are NOT integers.

So you are saying that the result of multiplying an integer by 2 is not an integer?

41 minutes ago, Edgard Neuman said:

So no, as strange as it sound, "For every number, it is NOT possible to double that number".

Obviously nonsense.

45 minutes ago, Edgard Neuman said:

You don't see it, because you mistake reaching  ALL for a recursive process.. But in reality, you recursive thinking NEVER really reach ALL. You do for one, for the other etc.. and you think it's ALL, but it's not. By definition, you can't reach ALL, because it's a infinite set. 

You seem to be the one who thinks it is a recursive process. You don't have to "reach" all because it is not a recursive process. So if we make a statement about all integers then it is true for all integers.

47 minutes ago, Edgard Neuman said:

It's like the Hilbert hotel.. the infinite hotel is full.. the paradox state that you can slide everybody to the next room. But really can you ? The hotel is full, so there is NO empty room at the end of the hotel.

Yes you can. The hotel may be full but you can still add an infinite number of new guests. This is at the heart of your error.

1 hour ago, Edgard Neuman said:

I don't see "speculations" here..

Really?

[Edgard Neuman]

17 minutes ago, Strange said:

Huh? 

So you are saying that the result of multiplying an integer by 2 is not an integer?

If you have Card(N), you have Card(N) -1 that is an integer but (Card(N)-1)  * 2   is not. 
If you don't have Card(N), then there's no point in bijections anyway, because you can't compare or equal a quantity that doesn't exist. 

17 minutes ago, Strange said:

Yes you can. The hotel may be full but you can still add an infinite number of new guests. This is at the heart of your error.

And again, "adding a room" contradict the very idea that the hotel contains ALL the rooms. People can't really ADD numbers to sets. Sets are  equal to themselves only. A set union a new element, is a different set. Numbers are in or are not. You can create a set, by deciding a number has the property to be in the set.. (and then it's a DIFFERENT set after you added your new number) 
The Hilbert hotel can't be a "two state" hotel, with a "before" and a "after". You yourself state that bijection is not a stepped process. the Hotel is neither. 
The hotel is infinite and full. Any room you can imagine build, or create, IS ALREADY full.
 

Edited October 18, 2018 by Edgard Neuman

[Strange]

12 minutes ago, Edgard Neuman said:

If you have Card(N), you have Card(N) -1 that is an integer but (Card(N)-1)  * 2   is not. 

What is N in this notation?  

This statement doesn't seem to make much sense, perhaps if you define what N is, it might make some sense.

 

Edit: Is N supposed to the set of natural numbers? If so then the cardinality of N is infinity and so is Card(N)-1. And so is (Card(N)-1)  * 2

But you are right that NONE of the these are integers. But that doesn't seem relevant to anything you have said before.

You don't seem to understand the concept of infinity.

14 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..

The reason nobody believes you is because you don't know what you are talking about.

Quote

what should I do ?

Study mathematics?

Edited October 18, 2018 by Strange

[Edgard Neuman]

4 minutes ago, Strange said:

What is N in this notation?  

This statement doesn't seem to make much sense, perhaps if you define what N is, it might make some sense.

 

Edit: Is N supposed to the set of natural numbers? If so then the cardinality of N is infinity and so is Card(N)-1. And so is (Card(N)-1)  * 2

But you are right that NONE of the these are integers. But that doesn't seem relevant to anything you have said before.

You don't seem to understand the concept of infinity.

sorry it's not very easy to systematicaly copy an paste the "ℕ" character .. I thought it was obvious enough since we are talking about integer.. (the set  ℕ)

Edited October 18, 2018 by Edgard Neuman

[Strange]

Just now, Edgard Neuman said:

sorry it's not very easy to systematicaly copy an paste the "ℕ" .. I thought it was obvious enough since we are talking about integer.. (the set  ℕ)

OK. So Card(N)-1 and (Card(N)-1)*2 don't really mean anything because Card(N) is not an integer.

[Edgard Neuman]

hum...
Card([0;n[) = n
and 
Card(N)-1 just can't be = Card(N)
I know you created a class of specific "not number" with there all set of properties and operation, only because you accept the bijection. But I don't, and so I don't assume Card(N) behave diffently that any number and I don't see why I should

Edited October 18, 2018 by Edgard Neuman

[Strange]

2 minutes ago, Edgard Neuman said:

hum...
Card([0;n[) = n

So what. How is that relevant?

[Edgard Neuman]

Just now, Strange said:

So what. How is that relevant?

Card([0;n[) = n
Card([0;n-1[)  = n -1 =  Card([0;n[) -1
and
n -1 ∈ [0;n[
so 
Card([0;n[) -1  ∈ [0;n[
and because every n is in ℕ
Card(ℕ) -1  ∈ ℕ


 

[studiot]

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

15 hours ago, Edgard Neuman said:

Card(ℕ) =ω

right at the outset.

 

So I asked you to explain what number system and arithmetic system are you using when you introduced   [math]\left( {\omega  - 1} \right)[/math]    since omega is not finite.

 

[Carrock]

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...

[Strange]

1 minute ago, Edgard Neuman said:

and because every n is in ℕ
Card(ℕ) -1  ∈ ℕ

Every n is in N but Card(N) is not an n. So your argument fails.

Edited October 18, 2018 by Strange

[Edgard Neuman]

2 minutes ago, Strange said:

Every n is in N nut Card(N) is not an n. So your argument fails.

ok that's not what I meant.. I meant there is no m (in ℕ of course)  so m <n that is not in [0;n[

 

Edited October 18, 2018 by Edgard Neuman

[Strange]

1 minute ago, Edgard Neuman said:

ok that's not what I meant.. I meant there is no m (in ℕ of course)  so m <n that is not in [0;n[

 

OK. So answer Carrock's post above, then.

[Edgard Neuman]

5 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.

 

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)

Edited October 18, 2018 by Edgard Neuman