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 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.

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. Such as?

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.

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".

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

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.

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.

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?