Cardinality of Real Numbers Paradox?

[ndflyers]

Here is a question to think about.

 

We assume that any element of a set with cardinality A, when represented by symbols, must have less than or equal to A symbols, which I think is a fair assumption.

 

The real numbers have cardinality C. So any number in the set of reals, when represented by, whatever, 0's and 1's if you wish, can have at most C of these objects to represent it. The cardinality of the power set of the reals, is 2^C>C. Now, we can imagine a number who has 2^C symbols to represent it. This number is technically not in the reals, since it must be of a set with cardinality greater than the reals. But this number is what we would intuitively think of as a real number.

 

So what do you think of the implications of this? Or do you find an error in my reasoning?

[md65536]

The cardinality of the reals is infinite.

You're right in that infinity is not in the reals.

Intuitively thinking of this number (infinity) as a real number is the error.

 

 

[ndflyers]

Intuitively thinking of this number (infinity) as a real number is the error.

 

The element of the Real Numbers, doesn't have to be (and can't be) infinity. I was thinking more of an irrational number, for instance e. An irrational number has an infinite decimal expansion. If each of these symbols of it's decimal expansion is put into a set, then this set, one would think should not be able to have a cardinality greater than the real numbers. But we could conceive of a number which, when the symbols of it's decimal expansion are put into a set, that set has a cardinality which is greater than the real numbers.

[md65536]

The element of the Real Numbers, doesn't have to be (and can't be) infinity. I was thinking more of an irrational number, for instance e. An irrational number has an infinite decimal expansion. If each of these symbols of it's decimal expansion is put into a set, then this set, one would think should not be able to have a cardinality greater than the real numbers. But we could conceive of a number which, when the symbols of it's decimal expansion are put into a set, that set has a cardinality which is greater than the real numbers.

You mean like "infinity plus one"?

 

 

Yes, I misread what you wrote. I think the mistake is in treating infinity as a normal number. But the number you're imagining would be a real number. It just wouldn't require "more than infinity" symbols to describe it.

Edited October 15, 2011 by md65536

[ndflyers]

Thank you for the replies.

 

I think the problem can be summarized as follows:

A cardinal number greater than that of the real numbers can be shown to exist.

If a set with this cardinality could be constructed, then clearly we could construct this paradoxical number I spoke of earlier.

But there is no reason to assume a set with this cardinality can be constructed just because we can show its existence.

 

Thus, the paradoxical number can be shown to exist if the set with cardinality greater than that of the Real Numbers can be constructed.

 

My understanding is that the axioms which would allow for the construction are somewhat open questions, or are not generally excepted in the mathematics community.

[md65536]

Thank you for the replies.

 

I think the problem can be summarized as follows:

A cardinal number greater than that of the real numbers can be shown to exist.

If a set with this cardinality could be constructed, then clearly we could construct this paradoxical number I spoke of earlier.

But there is no reason to assume a set with this cardinality can be constructed just because we can show its existence.

 

Thus, the paradoxical number can be shown to exist if the set with cardinality greater than that of the Real Numbers can be constructed.

 

My understanding is that the axioms which would allow for the construction are somewhat open questions, or are not generally excepted in the mathematics community.

I'm out of my league, but I think there are further problems.

 

See http://en.wikipedia.org/wiki/Aleph_number

The cardinality of the set of symbols representing digits in your proposed irrational number would be countably infinite and so would be aleph-naught.

The cardinality of the reals is 2^aleph-naught, and the set is uncountable. It sounds like the "continuum hypothesis" implies that this really is a valid form of "infinity plus one".

 

I suspect that the ability to construct a set is similar to the ability to count it, so you could construct a set of symbols representing the digits of an irrational number -- or at least you could construct every given symbol in that set, which I guess isn't the same as constructing the complete set. Either way you couldn't do the same for the set of reals.

 

I would say that setting C to infinity and then manipulating it arithmetically and expecting relations to hold (like 2^C > C) is an error.

 

Anyway, your statements above might still be logically true in that "If we can do something impossible, then we can do something else impossible."

[DrRocket]

Here is a question to think about.

 

We assume that any element of a set with cardinality A, when represented by symbols, must have less than or equal to A symbols, which I think is a fair assumption.

 

This makes no sense.

 

 

 

The real numbers have cardinality C. So any number in the set of reals, when represented by, whatever, 0's and 1's if you wish, can have at most C of these objects to represent it. The cardinality of the power set of the reals, is 2^C>C. Now, we can imagine a number who has 2^C symbols to represent it. This number is technically not in the reals, since it must be of a set with cardinality greater than the reals. But this number is what we would intuitively think of as a real number.

 

So what do you think of the implications of this? Or do you find an error in my reasoning?

 

Your reasoning is nothing but errors.

 

Any real number can be represented as an infinite decimal -- so with at most countably many ([math]\aleph _0[/math] symbols. But the cardinality of the real numbers is strictly greater than [math] \aleph _0[/math]

 

I have no idea what "anumber who has 2^C symbols to represent it" would be, andneither do you. Whatever it might possibly be, it most certainly is not a real number.

 

The cardinality of the reals is 2^aleph-naught, and the set is uncountable. It sounds like the "continuum hypothesis" implies that this really is a valid form of "infinity plus one".

 

[math] \aleph _0 [/math] is the cardinality of the natural numbers, and any set of cardinality [math] \aleph _0 [/math] is called "countably infinite". If a set has cardinality [math]K[/math] then the "power set", the set of all subsets of that set has cardinality [math] 2^K[/math] -- this is easily seen to be true if K is finite and is the definition of [math]2^K[/math] if [math]K[/math] is infinite.

 

It can be proved that [math]2^k>K[/math] for any cardinal number [math]K[/math]. If [math]c[/math] denotes the cardinality of the real numbers then [math]2^{\aleph _0} = c [/math]. The continuum hypothesis asserts the non-existence of any cardinal [math]K[/math] such that [math] \aleph _0 < K < 2^{\aleph _0}[/math].

 

It has been proved that the continuum hypothesis is independent of the usual Zermelo Fraenkel axioms of set theory.

Edited October 17, 2011 by DrRocket

[ndflyers]

Any real number can be represented as an infinite decimal -- so with at most countably many

 

Can this be proven? And if so, can you give me an outline of this, or a general direction to look into?

 

Your reasoning is nothing but errors.

 

From what you've written, you have only pointed out one error in my reasoning...and you haven't qualified(given proof or sources to back up) this statement yet.

 

And yes, we don't know what a number which has 2^C symbols to represent it looks like, but we also don't know what a set with 2^(Aleph-naught) elements look like, so I don't see why this is relevant.

[DJBruce]

Can this be proven? And if so, can you give me an outline of this, or a general direction to look into?

 

Prove that .999...=1 and then generalize your argument.

[ndflyers]

More specifically, I meant can it be proven that the decimal expansion of a real number has COUNTABLY many symbols to represent it. A proof of this would make my original post make "no sense".

[md65536]

More specifically, I meant can it be proven that the decimal expansion of a real number has COUNTABLY many symbols to represent it. A proof of this would make my original post make "no sense".

Still out of my league but for fun:

 

Yes. By the very nature of a "decimal expansion"... just map the n'th decimal place to the n'th natural number, and you have a 1-1 mapping between the number of digits and the natural numbers.

 

If you have an infinite number of digits on both sides of the decimal (not sure if that makes sense) then alternate between them when counting. The even naturals will map to the digits on one side of the decimal, and the odds will map to digits on the other side.

 

 

[ndflyers]

Yes. By the very nature of a "decimal expansion"... just map the n'th decimal place to the n'th natural number, and you have a 1-1 mapping between the number of digits and the natural numbers.

 

Yes, you're right. You can just map the nth decimal place to the nth natural number and thus the decimal expansion must be countable.

 

Thanks for clearing this up.

[DrRocket]

Can this be proven? And if so, can you give me an outline of this, or a general direction to look into?

 

The proof is the trivial fact that any real number can be represented as an infinite decimal. You ought to be able to find this in any elementary text on real analysis.

 

For a nice accessibe account of the theory of cardinal numbers read Naive Set Theory by Paul Halmos. A reading shopuld correct your many misconceptions.

 

 

 

From what you've written, you have only pointed out one error in my reasoning...and you haven't qualified(given proof or sources to back up) this statement yet.

 

Wrong. Go read the referent book by Halmos. The fact that you are all wet is pretty obvious to anyone who understands simple set theory. You have made several misstatements regarding well-known elementary facts from the theory of cardinal numbers.

 

And yes, we don't know what a number which has 2^C symbols to represent it looks like, but we also don't know what a set with 2^(Aleph-naught) elements look like, so I don't see why this is relevant.

 

Wrong again. There is no such "number". Read my post to see that the cardinality of the real numbers is [math]2^{\aleph _0}[/math]. If you are unfamiliar with the real numbers then you really need to study some remedial mathematics.

 

More specifically, I meant can it be proven that the decimal expansion of a real number has COUNTABLY many symbols to represent it. A proof of this would make my original post make "no sense".

 

This is both true and well-known to even a freshman calculus student. And most certainly your original post makes no sense.

Edited October 18, 2011 by DrRocket

[ndflyers]

Wrong again. There is no such "number". Read my post to see that the cardinality of the real numbers is . If you are unfamiliar with the real numbers then you really need to study some remedial mathematics.

 

Yes, my misunderstanding was why there can be no such number. The reason why this number can't exist really has nothing to do with the cardinality of the real numbers.

 

You have made several misstatements regarding well-known elementary facts from the theory of cardinal numbers.

 

Such as?

[DrRocket]

 

 

Such as?

 

Read what you have written, perhaps after reading the books suggested. If you cannot find several examples quickly then further discussion would be futile.

[ndflyers]

Perhaps there are a few things I have said that did not distinguish whether or not I was speaking about cardinality or a number, but other than that, I can only see things that other people have said...

 

For instance you say that

We assume that any element of a set with cardinality A, when represented by symbols, must have less than or equal to A symbols, which I think is a fair assumption.

makes no sense. This is clearly true for the finite case. The set ({0},{{0}}), can be represented by two or less symbols and the cardinality of the set is 2. In the infinite case, I think this would be independent of ZFC.

[Cap'n Refsmmat]

This is both true and well-known to even a freshman calculus student.

I think you vastly overestimate the capabilities of an average freshman calculus student, and hostility is certainly not going to bring anyone up to your desired standard of mathematical aptitude. If you believe replying to be an exercise in futility, please don't bother.

[DrRocket]

I think you vastly overestimate the capabilities of an average freshman calculus student, and hostility is certainly not going to bring anyone up to your desired standard of mathematical aptitude. If you believe replying to be an exercise in futility, please don't bother.

 

I have taught enough calculus classes to be aware of the capabilities of the average calculus student. Noting that a decimal expansion has countably many instances of the set of symbols 0,1,2,3,4,5,6,7,8,9 is well within those capabilities. That is what "decimal places" are all about.

 

There is no overestimation involved. In point of fact any decent high school algebra student can handle this easily.

Edited October 18, 2011 by DrRocket

[Cap'n Refsmmat]

Then I commend you on the exceptional abilities of your calculus students. In the future, however, please hold your hostility. SFN is a place for people to learn, not to have their abilities compared poorly with the average high school algebra student. The latter merely discourages people from attempting the former.