Cardinality of the set of binary-expressed real numbers

[pengkuan]

 

What rubbish.

 

The number 1/3 is an infinitely long number in decimal.

It is also an infinitely long number in binary.

 

Yet it is rational in both systems.ie 1/3 or 1/11

Ok. Non repeating no terminating digits give irrationals.

[uncool]

The number of binary number is 2ⁿ, when n=infinity, n=_^À0 , so the cardinality of binary number is 2^À0.

Which set of binary numbers? Finite-length only, or infinite length allowed as well?

 

And what do you think [math]2^{\aleph_0}[/math] means? Because if you mean the set of finite-length binary numbers is of size [math]2^{\aleph_0}[/math] in the usual sense, then you are wrong.

As this is the cardinality of the power set of the naturals, this set must be the real. But the set of rationals does not contain irrational. So, the power set of the naturals is not the reals.

 

I'm not proving. I explain.

You'd benefit from explaining fully, to the formal level. Edited December 9, 2015 by uncool

[pengkuan]

Which set of binary numbers? Finite-length only, or infinite length allowed as well?

 

And what do you think [math]2^{\aleph_0}[/math] means? Because if you mean the set of finite-length binary numbers is of size [math]2^{\aleph_0}[/math] in the usual sense, then you are wrong.

You'd benefit from explaining fully, to the formal level.

Length n is finite but goes to infinity. The cardinality of the power set of the naturals is computed this way.

 

I'm not able to explain formally because I'm not a mathematician and lack the formal language of mathematics.

[Strange]

I'm not able to explain formally because I'm not a mathematician and lack the formal language of mathematics.

 

And I think your informal approach (relying on vague undefined terms such as "binary numbers") is why you have managed to mislead yourself.

[pengkuan]

 

And I think your informal approach (relying on vague undefined terms such as "binary numbers") is why you have managed to mislead yourself.

Sure! The discussions bring me back to the right way.

[uncool]

Length n is finite but goes to infinity. The cardinality of the power set of the naturals is computed this way.

No, it is not.

 

The power set of the natural numbers includes infinite sets. It is the set of all subsets of the natural numbers, finite or infinite. There is a bijection between "binary numbers" of arbitrary (including infinite) length; for example, the set of the even numbers corresponds to 0.01010101... = 1/3.

 

An important thing to note: While [math]\aleph_0[/math] is a limit of the sequence [math]0, 1, 2, ...[/math], it is not true that [math]2^{\aleph_0}[/math] is the limit of the sequence [math]2^0, 2^1, 2^2, ...[/math]. That is not how exponentiation is defined for cardinals.

I'm not able to explain formally because I'm not a mathematician and lack the formal language of mathematics.

Then you should learn the formal language. It seriously helps, and allows you to more easily find errors. Edited December 9, 2015 by uncool

[pengkuan]

No, it is not.

 

The power set of the natural numbers includes infinite sets. It is the set of all subsets of the natural numbers, finite or infinite. There is a bijection between "binary numbers" of arbitrary (including infinite) length; for example, the set of the even numbers corresponds to 0.01010101... = 1/3.

 

An important thing to note: While [math]\aleph_0[/math] is a limit of the sequence [math]0, 1, 2, ...[/math], it is not true that [math]2^{\aleph_0}[/math] is the limit of the sequence [math]2^0, 2^1, 2^2, ...[/math]. That is not how exponentiation is defined for cardinals.

You are right. But if we append the finite binary numbers with infinitely many zero, we obtain a one-to-one correspondence with the power set of the natural numbers.

 

Take 0.000000000....................... This binary number is zero plus point plus a infinite sequence of zero. Then we pair up each zero to one natural number that is the rank of place of the zero. We create a subset of the naturals and simultaneously flip the zero to 1 if the corresponding natural number is in the subset of natural number. This binary number corresponds to this subset uniquely. Then, we have a one-to-one correspondence between each binary number and one subset of the naturals. As the cardinality of the power set of the naturals is , the cardinality of the binary number so constructed is the same, .

 

Then you should learn the formal language. It seriously helps, and allows you to more easily find errors.

 

I know, thanks.

[uncool]

You are right. But if we append the finite binary numbers with infinitely many zero, we obtain a one-to-one correspondence with the power set of the natural numbers.

No, we don't. Which "finite binary number with infinitely many zeros" corresponds to the subset of the even numbers?

Take 0.000000000....................... This binary number is zero plus point plus a infinite sequence of zero. Then we pair up each zero to one natural number that is the rank of place of the zero. We create a subset of the naturals and simultaneously flip the zero to 1 if the corresponding natural number is in the subset of natural number. This binary number corresponds to this subset uniquely. Then, we have a one-to-one correspondence between each binary number and one subset of the naturals. As the cardinality of the power set of the naturals is , the cardinality of the binary number so constructed is the same, .

You've shown a map in one direction, not the other. Another way to say it: you've constructed a map, but you haven't shown that it is surjective. Edited December 9, 2015 by uncool

[pengkuan]

No, we don't. Which "finite binary number with infinitely many zeros" corresponds to the subset of the even numbers?

The subset of the even numbers is {2,4,6,8............}. This corresponds to the binary number with the even number^th digits flipped to 1. That is

0.010101010101010101................................. The length is infinite.

 

Maybe you think that my set of binaries contains only finite length numbers. But it contains the infinite length numbers too.

 

You've shown a map in one direction, not the other. Another way to say it: you've constructed a map, but you haven't shown that it is surjective.

 

One-to-one is a bijection. The rationals are in a one-to-one with the naturals and is bijective.

[uncool]

The subset of the even numbers is {2,4,6,8............}. This corresponds to the binary number with the even number^th digits flipped to 1. That is

0.010101010101010101................................. The length is infinite.

 

Maybe you think that my set of binaries contains only finite length numbers. But it contains the infinite length numbers too.

That isn't what you said. You said you were taking finite-length numbers and "appending" numbers with infinitely many zeroes. You've repeatedly apparently switched between talking about binary numbers only of finite length and those of infinite length.

One-to-one is a bijection. The rationals are in a one-to-one with the naturals and is bijective.

If you're going to use "one-to-one" to mean bijection, then you haven't shown that your map is one-to-one. In order for a map to be a bijection, you have to show that it is also surjective, that is, that it "hits" every element of the output.

 

On another note, one-to-one in modern terminology does not mean bijective; it means injective. One-to-one correspondence is a phrase that once meant bijection; however, terminology has generally moved on.

 

As far as I understand it, you think you have a bijection between some countable set and some supposedly uncountable set. Is my understanding wrong? If not, what are your two sets and what is your bijection?

Edited December 9, 2015 by uncool

[pengkuan]

That isn't what you said. You said you were taking finite-length numbers and "appending" numbers with infinitely many zeroes. You've repeatedly apparently switched between talking about binary numbers only of finite length and those of infinite length.

 

In the beginning I wanted to explain that the binaries have length n=<1 that increases forever In another discussion, people remark me that my binary set is not of infinitely many digits. Indeed, they have finite digits that increases without end. So, they are in some sort always finite. So I have changed my terminology and say they have finite digits. But here you have another terminology. Hence the confusion.

 

I think the term of infinitely many zero that flip with the subsets of natural numbers is a good compromise. This scheme makes a bijection between the binaries and the power set of naturals.

That isn't what you said. You said you were taking finite-length numbers and "appending" numbers with infinitely many zeroes. You've repeatedly apparently switched between talking about binary numbers only of finite length and those of infinite length.

If you're going to use "one-to-one" to mean bijection, then you haven't shown that your map is one-to-one. In order for a map to be a bijection, you have to show that it is also surjective, that is, that it "hits" every element of the output.

 

On another note, one-to-one in modern terminology does not mean bijective; it means injective. One-to-one correspondence is a phrase that once meant bijection; however, terminology has generally moved on.

 

As far as I understand it, you think you have a bijection between some countable set and some supposedly uncountable set. Is my understanding wrong? If not, what are your two sets and what is your bijection?

 

For me the one-to-one is a bijection. So, my set of binaries has the same cardinality than the power set of the naturals. But this is in contradiction with the commonly accepted theory that states that the power set of the naturals is strictly bigger than the naturals and equals the reals.

 

What I'm showing here is not the countable set of binaries has a bijection with the reals, but has a bijection with the power set of the naturals. In this case, the power set of the naturals has the cardinality of the naturals, not that of the reals. So, the theory that the power set of the naturals is strictly bigger than the naturals and equals the reals is false.

 

Also, 2^À0 is not the cardinality of the reals, but equals _^À0

[studiot]

 

If you're going to use "one-to-one" to mean bijection, then you haven't shown that your map is one-to-one. In order for a map to be a bijection, you have to show that it is also surjective, that is, that it "hits" every element of the output.

 

uncool

On another note, one-to-one in modern terminology does not mean bijective; it means injective. One-to-one correspondence is a phrase that once meant bijection; however, terminology has generally moved on.

 

As far as I understand it, you think you have a bijection between some countable set and some supposedly uncountable set. Is my understanding wrong? If not, what are your two sets and what is your bijection?

 

 

This is not a bad summary

 

https://en.wikipedia.org/wiki/Bijection,_injection_and_surjection

 

But a few points.

 

The first set is called the domain (shown as X) and the second set the codomain (shown as Y)

 

Injections, surjections and bijections are mappings which show connections between members of different sets.

Towards the end of the article Wiki links these terms with monomorphisms, epimorphisms and isomorphisms respectively. This is correct, but it should be remembered that morphisms are about what happens to the relations between members of the same set when mapped to the second set.

So an isomorphism is more than just a bijection and so on.

 

The diagrams also give the impression that all the members of the domain are connected.

This is not the case.

Only in a bijection are the sets the same size.

Otherwise either the domain or codomain can contain members that are not linked and either may be 'bigger' than the other.

 

 

 

pengkuan

For me the one-to-one is a bijection.

 

uncool has mentioned that one to one correspondence means a bijection. This is not the same as one to one.

Edited December 10, 2015 by studiot

[uncool]

In this case, the power set of the naturals has the cardinality of the naturals, not that of the reals.

No one objects to the claim that the set of natural numbers is in bijection with the set of all finite-length binary numbers, or the claim that the powerset of natural numbers is in bijection with the set of all binary numbers with any length including infinite, or the claim that both of the latter are in bijection with the real numbers. The problem is that you seem to be going back and forth between which set of binary numbers you are using. The cardinality [math]2^{\aleph_0}[/math] refers to the latter; you seem to often use it to refer to the former.

 

So bringing it down to sets that we all know and can't confuse:

 

If you think the power set of the natural numbers has the cardinality of the naturals, you should be able to provide a bijection between the two. What is your bijection?

Edited December 10, 2015 by uncool

[pengkuan]

If you think the power set of the natural numbers has the cardinality of the naturals, you should be able to provide a bijection between the two. What is your bijection?

 

I will write a detailed proof of the bijection.

 

But why the one-to-one correspondence does not prove the bijection? To any subset of the natural numbers, there is one and only one binary number like 0.154894................ with infinite zero behind or with infinite sensible digits. This is a bijection. Isn't it?

 

My going back and forth is only due to the terminology I use to adapt with other people's objections. but the heart of my idea is the same.

 

This is not a bad summary

 

https://en.wikipedia.org/wiki/Bijection,_injection_and_surjection

 

But a few points.

 

The first set is called the domain (shown as X) and the second set the codomain (shown as Y)

 

Injections, surjections and bijections are mappings which show connections between members of different sets.

Towards the end of the article Wiki links these terms with monomorphisms, epimorphisms and isomorphisms respectively. This is correct, but it should be remembered that morphisms are about what happens to the relations between members of the same set when mapped to the second set.

So an isomorphism is more than just a bijection and so on.

 

The diagrams also give the impression that all the members of the domain are connected.

This is not the case.

Only in a bijection are the sets the same size.

Otherwise either the domain or codomain can contain members that are not linked and either may be 'bigger' than the other.

 

 

uncool has mentioned that one to one correspondence means a bijection. This is not the same as one to one.

 

I have to write something on bijection.

 

I said one-to-one is a bijection just to shorten one-to-one correspondence.

[studiot]

 

I have to write something on bijection.

 

I said one-to-one is a bijection just to shorten one-to-one correspondence.

 

 

Thank you for your response, but it does not tell me if my last post was of any use or help?

Yes many make this contraction and that sloppyness leads to communication difficulty and confusion.

 

I have a further development of it available if you wish.

[uncool]

I will write a detailed proof of the bijection.

Good. Remember, the bijection wanted is between the natural numbers and the powerset of the natural numbers.

But why the one-to-one correspondence does not prove the bijection?

Because you haven't provided the bijection between the sets people care about: the natural numbers and their powerset. All you've done is provide bijections between various things and sets of binary numbers, and equivocated on the definition of binary numbers. As far as I can tell, your argument boils down to saying that the natural numbers are in bijection with the set of finite-length binary numbers, and their powerset i in bijection with the set of all binary numbers, and because both sets are in bijection with a set of binary numbers, they must have the same cardinality. Which is clearly wrong, when stated explicitly like that.

To any subset of the natural numbers, there is one and only one binary number like 0.154894................ with infinite zero behind or with infinite sensible digits. This is a bijection. Isn't it?

That is a bijection. That isn't the thing people are objecting to. Once again: people are objecting to your claim of a bijection between the natural numbers and their powerset.

My going back and forth is only due to the terminology I use to adapt with other people's objections. but the heart of my idea is the same.

Then try to drop binary numbers, and deal with the natural numbers and their powerset, as those are unambiguous.

I have to write something on bijection.

 

I said one-to-one is a bijection just to shorten one-to-one correspondence.

That isn't how "one-to-one" is used in modern terminology; use the word bijection instead.

[pengkuan]

 

Thank you for your response, but it does not tell me if my last post was of any use or help?

Yes many make this contraction and that sloppyness leads to communication difficulty and confusion.

 

I have a further development of it available if you wish.

Thanks. I must read carefully the link you have given. Since I'm not familiar with the formal language of set theory and I cannot understand things like "domain or codomain". Your help is useful in the future article I will write on the bijection between the natural number and its power set.

Good. Remember, the bijection wanted is between the natural numbers and the powerset of the natural numbers.

Because you haven't provided the bijection between the sets people care about: the natural numbers and their powerset. All you've done is provide bijections between various things and sets of binary numbers, and equivocated on the definition of binary numbers. As far as I can tell, your argument boils down to saying that the natural numbers are in bijection with the set of finite-length binary numbers, and their powerset i in bijection with the set of all binary numbers, and because both sets are in bijection with a set of binary numbers, they must have the same cardinality. Which is clearly wrong, when stated explicitly like that.

That is a bijection. That isn't the thing people are objecting to. Once again: people are objecting to your claim of a bijection between the natural numbers and their powerset.

Then try to drop binary numbers, and deal with the natural numbers and their powerset, as those are unambiguous.

That isn't how "one-to-one" is used in modern terminology; use the word bijection instead.

Thanks for your advise. Effectively, bijection between binary numbers and the power set of the natural numbers is not the same thing than bijection between natural numbers and its power set. I will explain the latter bijection more carefully.

[studiot]

 

pengkuan

 

1)Thanks. I must read carefully the link you have given. Since I'm not familiar with the formal language of set theory and I cannot understand things like "domain or codomain". Your help is useful in the future article I will write on the bijection between the natural number and its power set.

 

 

2)Thanks for your advise. Effectively, bijection between binary numbers and the power set of the natural numbers is not the same thing than bijection between natural numbers and its power set. I will explain the latter bijection more carefully.

 

 

 

Domain and codomain are not really the language of set theory but my proposed discussion covers and explains this.

 

Bijections are special because they involve every member of both sets.

 

There is a great deal of mathematical terminology to absorb and much of it seems pointless pedantry to the beginner.

My whole idea is to ease this by setting the terms in context, showing why we need them as they are, and how they relate to other terms discussed in the same manner.

 

Doing this involves some effort on my part so I will only bother if you are really interested, not just being polite. and I am still having trouble determining this?

[pengkuan]

 

Domain and codomain are not really the language of set theory but my proposed discussion covers and explains this.

 

Bijections are special because they involve every member of both sets.

 

There is a great deal of mathematical terminology to absorb and much of it seems pointless pedantry to the beginner.

My whole idea is to ease this by setting the terms in context, showing why we need them as they are, and how they relate to other terms discussed in the same manner.

 

Doing this involves some effort on my part so I will only bother if you are really interested, not just being polite. and I am still having trouble determining this?

I have read the link of bijection.... I learned that domain is departure and codomain is arrival in some sense.

[Keen]

In fact, fractional binaries cannot write all the reals in [1,0[ because they cannot have infinitely many digits. They are only rationals.

Ok, I thought you were considering this type of numbers. I just wasn't sure.

 

Because infinitely many digits have no sense. They cannot be used to perform mathematics.

It actually does make perfect sense. You just consider the infinite sequence (u_n) of 0 and 1 and to this sequence you can associate a unique number which is the limit of the infinite series [math] \Sigma [/math] u_n 2^-n.

 

On the other hand your limit does not make much sense. You claim, that by making n going to the cardinal of aleph0, you obtain the result, that the cardinal of numbers with finitely many digits is aleph1. However you still do not explain in what sense you take your limit. It surely isn't the classical limit of sequences, since the limit of 2ⁿ in the classical sense is simply infinity, which does not say then anything on the cardinality of your set [math]B_F[/math], besides that it is infinite, but that's not really what you are trying to prove. Also I would like to add, that the notation of the powerset [math]2^{\mathbb{N}} [/math] is a mere notation and isn't anything particularly special, which makes your use of limit even weirder and if I am to be honest, I do not think there is a way to write your limit formally in a way that would be compatible with modern mathematics as it is a very well known result, that a set can't have the same cardinal as its powerset.

 

The proof of this statement is actually not that difficult and is reminiscent of Bertrand Russel's paradox, which I recommend you to check out. If there was a set A with the same cardinal as its powerset, there would be a one to one correspondence between A and [math] 2^A [/math], let's call it f. You can think of one to one correspondence as a label. To each member x of A, you associate a unique subset of A denoted f(x). One to one simply means, that each subset of A has a unique label. Now we consider a very particular subset of A, which has all the elements x, such that x is not a member of f(x). We call this subset B. Since we know that f is one to one, it means that there is y in A such that B=f(y) and now the question is whether y is in B or not. First let's suppose it is. Then by definition of B y is not in f(y). but f(y) is precisely B. If on the other hand it isn't in B, that means that y is in f(y) by definition of B, but f(y) is B, so y is in B, which is a contradiction. No matter how you put it you obtain something contradictory, which means that the set B does not exist and therefore neither the one to one correspondence f does.

[pengkuan]

Ok, I thought you were considering this type of numbers. I just wasn't sure.

 

It actually does make perfect sense. You just consider the infinite sequence (u_n) of 0 and 1 and to this sequence you can associate a unique number which is the limit of the infinite series [math] \Sigma [/math] u_n 2^-n.

 

On the other hand your limit does not make much sense. You claim, that by making n going to the cardinal of aleph0, you obtain the result, that the cardinal of numbers with finitely many digits is aleph1. However you still do not explain in what sense you take your limit. It surely isn't the classical limit of sequences, since the limit of 2ⁿ in the classical sense is simply infinity, which does not say then anything on the cardinality of your set [math]B_F[/math], besides that it is infinite, but that's not really what you are trying to prove. Also I would like to add, that the notation of the powerset [math]2^{\mathbb{N}} [/math] is a mere notation and isn't anything particularly special, which makes your use of limit even weirder and if I am to be honest, I do not think there is a way to write your limit formally in a way that would be compatible with modern mathematics as it is a very well known result, that a set can't have the same cardinal as its powerset.

 

The proof of this statement is actually not that difficult and is reminiscent of Bertrand Russel's paradox, which I recommend you to check out. If there was a set A with the same cardinal as its powerset, there would be a one to one correspondence between A and [math] 2^A [/math], let's call it f. You can think of one to one correspondence as a label. To each member x of A, you associate a unique subset of A denoted f(x). One to one simply means, that each subset of A has a unique label. Now we consider a very particular subset of A, which has all the elements x, such that x is not a member of f(x). We call this subset B. Since we know that f is one to one, it means that there is y in A such that B=f(y) and now the question is whether y is in B or not. First let's suppose it is. Then by definition of B y is not in f(y). but f(y) is precisely B. If on the other hand it isn't in B, that means that y is in f(y) by definition of B, but f(y) is B, so y is in B, which is a contradiction. No matter how you put it you obtain something contradictory, which means that the set B does not exist and therefore neither the one to one correspondence f does.

 

I agree that numbers with infinitely many digits have actually definite values, the limit of the sequences that you mentioned. This is commonly accepted. But in the discussion that I have about my article, I have found that there is some ambiguity that must be worked out. Essentially, as we cannot write infinitely many digits, a sequence does not equal its limit. So, I will discuss this point in detail later.

 

I agree also that my point about the power set of natural numbers is confusing, because I have not given a definite one-to-one correspondence between the two sets. But I'm writing it now and I will inform you when it will be done.

 

I was not aware that Russel's paradox was about power set in this way. It is a paradoxe when the set has finite members, but with infinite members, it is much less clear.

[uncool]

Pengkuan, have you figured out your bijection between the natural numbers and their powerset?

[pengkuan]

Pengkuan, have you figured out your bijection between the natural numbers and their powerset?

 

Yes. I'm writing it. Its a very big table.