## Cardinality of the set of binary-expressed real numbers – page 3 – mathematics – science forums usd to ruble

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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 pln usd exchange rate. 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, .

Take 0.000000000………………….. This binary number is zero plus point plus a infinite sequence of zero stock outperform definition. 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.

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.

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?

Towards the end of the article Wiki links these terms with monomorphisms, epimorphisms and isomorphisms respectively put and call option agreement. 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.

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.

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 futures market cnbc. What is your bijection? Edited December 10, 2015 by uncool

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?

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?

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.

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.

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 xau usd forecast today. Once again: people are objecting to your claim of a bijection between the natural numbers and their powerset.

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.

1)Thanks binary form music. 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.

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?

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?

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 2 n 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) british pound historical exchange rate. 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 convert cny to usd. 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.

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 2 n 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 gold price chart. 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 dollar pak rupee exchange rate. 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.