Cardinality and Bijection of finite sets

[Edgard Neuman]

Ok I'll explain : 
ℕ= [0;ω[
ℕ= {0 ;1 ;2 ;3 ; ..... ; ω-2 ;ω-1} 

All your question about all "+" "*" and number construction are answered by that. 
and you have ℚ and ℝ in the first post. It is not a very complicated idea in fact.

By the way when I write 
Card(ℕ)=ω
It means exactly that there is ω elements in ℕ, not one more, not one less element.  It really is the "upper boundary" of the set. 

ok here is the weird part ..
To know if a number is in N Q or R isn't always trivial

For instance
ω+1 is not in ℕ
but 
(ω+1) -2  is in ℕ

in ℚ it's easy because if a and b are in ℕ, + - a/b is in ℚ

In ℝ it's tricky again..
for instance 

x= 0,0000.....000025 =  25/ (10^a)

It is in ℚ if and only if     10^a<ω,
so if a<log10(ω)

To be in ℝ, means there is a way to write it in the form of a sum of ω number from ℚ, that is not trivial
But if found that every number in ℝ
(when you compute it outside of the axiomatic of course) , are multiples of 1/A003418(ω-1)  



 

Edited October 19, 2018 by Edgard Neuman

[Strange]

54 minutes ago, Edgard Neuman said:

ℕ= {0 ;1 ;2 ;3 ; ..... ; ω-2 ;ω-1} 

So if you are talking about a finite set of numbers (not the set of natural numbers) then you some of your claims will be correct. But irrelevant to the set of natural numbers.

[Edgard Neuman]

3 minutes ago, Strange said:

So if you are talking about a finite set of numbers (not the set of natural numbers) then you some of your claims will be correct. But irrelevant to the set of natural numbers.

I understand why we don't understand each others :
When I write 

Second, if you define
Card(ℕ) =ω

I my mind, it's self explanatory than there is ω elements in ℕ
So no, it's not "your usual" ℕ

Edited October 19, 2018 by Edgard Neuman

[Strange]

1 hour ago, Edgard Neuman said:

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.

When you say things that are incorrect then it doesn't make any difference how many times people read it or how much they think about it. It is still wrong.

Going back to your first post, you claim to be talking about the set of natural numbers. But now you are clearly talking about some finite subset with an arbitrary upper bound which you call ω-1. 

I don't think I can go back and reread the whole thread replacing the assumption that N is the natural numbers with N being some arbitrary subset.

Edited October 19, 2018 by Strange

[Edgard Neuman]

12 minutes ago, Strange said:

When you say things that are incorrect then it doesn't make any difference how many times people read it or how much they think about it. It is still wrong.

Things are only "CORRECT" or "INCORRECT" relatively to an axiomatic. I a propose an axiomatic, I don't care what your consider ""CORRECT" the axiomatic DEFINES what is correct. 

Quote

Going back to your first post, you claim to be talking about the set of natural numbers. But now you are clearly talking about some finite subset with an arbitrary upper bound which you call ω-1. 

Yes that's true. That's because your "Card(ℕ) = Card(ℕ) + 1561651 " seems so absurd to me,

I'll go further .. if your Card(ℕ) is not a quantity, if operations with it doesn't work the same way, it is simply not related to numbers.. You just created a different set of numbers with their weird rules. You can invent numbers with the same weird properties of Card(ℕ), that are not defined by Card(ℕ)> any n.. 

for me 
Card([0;n[) = n
I don't even think it can be something else. I don't use your "strange" infinite things. In your view, everything become "different" at infinity. In my view Infinity is only the limit of finite, in every proposition.

Edited October 19, 2018 by Edgard Neuman

[Strange]

9 minutes ago, Edgard Neuman said:

I a propose an axiomatic, I don't care what your consider ""CORRECT" the axiomatic DEFINES what is correct. 

True. But as your axioms define something different from the natural numbers, any conclusions you produce only apply to the objects you have defined, not the natural numbers. So maybe you can't define bijection on the sets you define with your axioms. But that says nothing about bijection of the natural numbers.

[Edgard Neuman]

10 minutes ago, Strange said:

True. But as your axioms define something different from the natural numbers, any conclusions you produce only apply to the objects you have defined, not the natural numbers. So maybe you can't define bijection on the sets you define with your axioms. But that says nothing about bijection of the natural numbers.

yes, but what makes your numbers better than mines ?  (are they "more natural" ?) everywhere I look, in reality, in computers, infinities are strictly COMPARABLES, and epsilon quantities are not disappearing..  
There is always more numbers in a set, that there is even numbers in it. ALWAYS. You want to make it somewhat equal at infinity, I don't. In my univers the "infinite" quantity is really a measurable quantity. 
And in your philosophy there is the problem that the quantity of number you use at the right side of n->n*2 is always different that the quantity you use at the left side.. that's a problem for me. 

Edited October 19, 2018 by Edgard Neuman

[Strange]

Just now, Edgard Neuman said:

yes, but what makes your numbers better than mines ? 

Just that they are the ones we use all the time. It isn't clear that the rules of arithmetic work with your numbers (at least without some modification).

[studiot]

1 hour ago, uncool said:

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.

Exactly. +1

The point of symbolism is for the author to communicate with his readers.

If the author uses new or different symbols he will fail to communicate if he doesn't let his readers know what those symbols mean.

Edited October 19, 2018 by studiot

[uncool]

1 hour ago, Edgard Neuman said:

ℕ= {0 ;1 ;2 ;3 ; ..... ; ω-2 ;ω-1} 

This is not in any way how the usual natural numbers work. I'd recommend using a different name. N' or EN or something like that.

So you've defined these new "pseudo-natural numbers". OK. Now what do you plan to do with them?

Note that anything you prove with these "pseudo-natural numbers" does not affect statements about actual natural numbers.

 

Edited October 19, 2018 by uncool

[Edgard Neuman]

1 hour ago, uncool said:

This is not in any way how the usual natural numbers work. I'd recommend using a different name. N' or EN or something like that.

So you've defined these new "pseudo-natural numbers". OK. Now what do you plan to do with them?

Note that anything you prove with these "pseudo-natural numbers" does not affect statements about actual natural numbers.

 

Ok ok i suggest ℕ_ω

because you can actually use a number as  ω

the + if fairly simple

(a +b) ∈ ℕ_ω

_{if and only if a+b<ω}

_{if not, then it is not in.. ω is not in so it's not different}

_{in peano :
S(S(.... S(0)))} ∈ ℕ_{ω
UNTIL you reach
S(ω-1) which is not in} ℕ_{ω (and so all its successor)}
 

_{so a = S(S... (S(0))))
so b = S(S... (S(0))))}
a+b=_{S(S... (S(S(S... (S(0)))))))) 
is or is not in} ℕ_{ω according the number of S<ω}

It's simpler to just consider that numbers works the same in this axiomatic excepts they are limited by ω

Then you create new numbers in ℚ_ω by adding " /  " defined by
a/b = c <=> a= b *c
so 1/3 for instance is in ℚ_ω but not in ℕ_ω

And because infinities are comparable (but not in ℕ_ω of course₎ , when you write "...." you have to specify the number of terms in use ....[ ] 

for instance  if 

S= 1+1+1+1 + ....[ω]   
P= 1+1+1+1 + ....[ω+1] 

P-S = 1
if 
Q= 1+2+3+4 + ....[ω]   
R= 1+2+3+4 + ....[ω+1] 
R-Q = ω +1 

Q= (ω ² + ω) / 2  
R= (ω ² + ω) / 2   + ω +1

 1 + 3 + 5 + 7  + 9  + … [ω]  = ω² 

or

1– 2 + 3 – 4  + 5 ….[ω] =

 1 + 3 + 5 + 7 …[ω/2]
– 2 – 4 – 6 – 8 …[ω/2]

= 1 + 1  + 1 +  1 …[ω/2]
+ 0 + 2 +  4  + 6  …[ω/2]
– 2 – 4 – 6  – 8 …[ω/2]

=  ω / 2  + 0 - 2 * (ω / 2)
= - ω / 2

you can also group them by 2
1– 2 + 3 – 4  + 5 ….[ω] 
= (1-2)+ (3 -4) + (5-6) ….[ω/2] 
= -1 -1 -1 -1 ….[ω/2] 
= - ω / 2

 

Or another interesting one :
c = 1 + r^1 + r ^ 2 + … [ω]

=(1 – r^ω ) / (1 – r)

 

 

Edited October 19, 2018 by Edgard Neuman

[studiot]

23 minutes ago, Edgard Neuman said:

if and only if a+b<ω

So the set of all sums (a+b)>ω is not finite.

Of what use is an arithmetic that omits such a large number of possible sums?

[Edgard Neuman]

34 minutes ago, studiot said:

So the set of all sums (a+b)>ω is not finite.

(a+b)>ω is not finite
yes.. 

(ω -1 ) +1  is not in ℕ_ω

I consider that if ω  is the measurable "infinite", all that's below is finite, all that is above or equal  is infinite

34 minutes ago, studiot said:

Of what use is an arithmetic that omits such a large number of possible sums?

I found the usage of a measurable ω very useful.. it automatically include limits of sums as comparable numbers

The fact that ω is not in ℕ_ω  doesn't mean you can't use It.. 

ω  + ω  = ω *2

_{It's "infinite" numbers, but there, the sums works exactly like finite numbers>}
 

_{And for cardinals, it really preserve quantities.. Card(even numbers) =} ω /2  
so even with infinite sets,
Card(A union B) = Card(A)+Card(B) - Card(A inter B)

_Card(ℕ_{ω) = Card(evens) + Card(odds) =} ω

you can build a bijection between evens and odds (n->n+1)
but not between numbers and evens  (or odds)
In that case "bijection" really mean equality of cards and vice-versa

_{What use is a number that is equal to its double ??  (your Card(N))}

_{It's kind of symmetrical to 0.. it's the other end of N ... you have 0  1 2 .. ω-2 ω-1}

_{I tried to add an other axiom, but I'm really not very sure it works  :}

ω is divisible by any number n in ℕ_ω  
only if n<sqrt(ω )

Edited October 19, 2018 by Edgard Neuman

[Edgard Neuman]

1 hour ago, Edgard Neuman said:

ω is divisible by any number n in ℕ_ω  
only if n<sqrt(ω )

I think probably this would better be :

ω is divisible by any number n in ℕ_ω  
only if n!<ω 
i don't know

Edited October 19, 2018 by Edgard Neuman

[studiot]

1 hour ago, Edgard Neuman said:

(a+b)>ω is not finite
yes.

Thank you all though it would be better to say that

There exists an (a+b)>ω since some (a+b)<ω

 

1 hour ago, Edgard Neuman said:

(ω -1 ) +1  is not in ℕ_ω

well no, (ω -1 ) is not defined.

and you are moving on to multiplication of non finite quantities, before finishing with finite ones.

 

It is not good enough to suggest that

1 hour ago, Edgard Neuman said:

the + if fairly simple

(a +b) ∈ ℕ_ω

_{if and only if a+b<ω}

_{if not, then it is not in.. ω is not in so it's not different}

The point everyone is trying to make to you (yes everyone, not just me) is that

the conventional statement c = a + b , a, b in N guarantees that there exists a unique c, and that c is also in N

This is an integral part and parcel of the properties of conventional binary operation called addition and cannot be removed any more than any other part of it.

Yours does not do this for some c as we have  just agreed.

[Strange]

20 minutes ago, studiot said:

the conventional statement c = a + b , a, b in N guarantees that there exists a unique c, and that c is also in N

I think you can achieve this, even with Edgar's finite set of integers, using modulo arithmetic. 

However, he doesn't seem to want to do that, just saying that c is undefined which doesn't really make sense to me.

[Edgard Neuman]

45 minutes ago, studiot said:

well no, (ω -1 ) is not defined.

 

you use Peano.
BUT at some time you have a number ω, it is not in ℕ_ω
the one before is ω-1. It is defined and in ℕ_ω 

Do you really have a problem with the concept of "unknown" value ? like n ..
is n-1 defined ? how do you write n-1 in Peano ? 

Some number are in ℕ_ω some are not by axiomatic
ℕ_ω is simply a set


[0;1;2;;3;4[
That is a set. Each element is constructible and "unique"

a+b>ω  is NOT in ℕ_ω, 

but is as unique as it is in Peano
 

Edited October 19, 2018 by Edgard Neuman

[Strange]

Maybe you should ask the mods to change the title to "Cardinality and Bijection of FINITE sets" so it describes what you are actually talking about.

 

p.s. I have suggested this to the mods

Edited October 19, 2018 by Strange

[Edgard Neuman]

19 minutes ago, Strange said:

I think you can achieve this, even with Edgar's finite set of integers, using modulo arithmetic. 

However, he doesn't seem to want to do that, just saying that c is undefined which doesn't really make sense to me.

hum no. a+b IS NOT in ω
and it is not  a+b modulo  ω that would be a totally different axiomatic

Just now, Strange said:

Maybe you should ask the mods to change the title to "Cardinality and Bijection of FINITE sets" so it describes what you are actually talking about.

OK i would agree with that, except it's FINITE with the boundary unkwown 

[Strange]

2 minutes ago, Edgard Neuman said:

hum no. a+b IS NOT in ω
and it is not  a+b modulo  ω that would be a totally different axiomatic

That's what I said. (Actually, what you mean is a+b is not in N - you can't even get your own notation right!  )

So you have defined a set of numbers where the usual rules of arithmetic don't apply. I'm not sure what the point is. As has been pointed out several times, nothing that you prove about these numbers has any relevance to the natural numbers used in the real world (ie outside this thread).

Edited October 19, 2018 by Strange

[studiot]

11 minutes ago, Edgard Neuman said:

Do you really have a problem with the concept of "unknown" value ?

 

No the problem I have is that it feels more like I am negotiating Brexit with the intransigent EU than having a discussion about Mathematics.

Success in Mathematics is about following the rules exactly (whilst hoping those rules ahve been properly set up in the first place)

Not only are you not following them, you have been hiding that fact.

[Edgard Neuman]

9 minutes ago, Strange said:

That's what I said.

So you have defined a set of numbers where the usual rules of arithmetic don't apply. I'm not sure what the point is. As has been pointed out several times, nothing that you prove about these numbers has any relevance to the natural numbers used in the real world (ie outside this thread).

except everything become much simpler. and conform to our intuition about quantities.. I gave you several example of the usage
a measurable infinite (you want to say it's "finite" if you really have trouble understanding that), but ω is really used to simulate an infinite quantity that is still mesurable .. sorry for you on that..

I had no idea it would be so hard for people to use another axiomatic.. 

In fact I should have explain the philosophy of the idea before all.. I've done it on my blog.. the idea that Peano stop and some numbers are not allowed is very easy to understand for me, because I 'm among other things a software developper :
when you do math with 32bits integer you really know how math and arithmetics with limited range  works..  
 

Edited October 19, 2018 by Edgard Neuman

[Strange]

3 minutes ago, Edgard Neuman said:

except everything become much simpler.

Simpler but less useful.

3 minutes ago, Edgard Neuman said:

and conform to our intuition about quantities..

Your intuition, maybe.

I have no problems with an intuitive understanding of the natural numbers.

4 minutes ago, Edgard Neuman said:

I had no idea it would be so hard for people to use another axiomatic.. 

It wouldn't have been, if you hadn't been so unclear (verging on dishonest) about what you meant.

 

[Edgard Neuman]

42 minutes ago, Strange said:

Simpler but less useful.

So tell me :
if 
A= 1+2+3+4 + ...
B=  r^1+ r ^2 + r^ 3 + r^4 .... 

is A > B or A < B when r = 1.5 ?

and what is the difference equal to  ?

 

A-B = (ω ² + ω) / 2   - (1 – 1.5^ω ) / (1 – 1.5)
= ω ²/2 + ω/ 2    + 2*(1-  1.5^ω)
= ω ²/2 + ω/ 2    + 2-  2* 1.5^ω

ok it's still complicated but it's much closer to be solved

I suppose that 1.5^ω > ω² 

Anyway If you really don't see how using a measurable value for infinite is usefull, i'm sorry I can't do more for you

And I really curious to see where your infinite has been usefull.. It's simply "not related" to integer numbers. It's just "bigger".. How is it usefull in anyway ? 
You have to use alternative like "limits" and you can even sum them or substract them. Yeah very usefull.
You can't even solve 

x + Card(N) = Card(N)

x is a number ? 

YEAH VERY USEFULL 

inf/inf = a number ! or maybe not.. WOW usefull ! 


No our intuition..

Cut a segment in 2 part . How many points does each part have ? is it more or less that the original segment ? When you cut a cake, how many particle of cake do you have ? Can you infinitely cut a cake and  feed the world ? 

Edited October 19, 2018 by Edgard Neuman

[swansont]

29 minutes ago, Strange said:

Maybe you should ask the mods to change the title to "Cardinality and Bijection of FINITE sets" so it describes what you are actually talking about.

 

p.s. I have suggested this to the mods

 

29 minutes ago, Edgard Neuman said:

OK i would agree with that, except it's FINITE with the boundary unkwown 

!

Moderator Note

Done.