If you are talking about Cantor's diagonalization, where the real number(R) p differs by a decimal digit from every real rational number(Q) n, and thus has no real number partner, the my answer is that p can not exist(in Q), or is an imaginary number. The reason is that since there is an countable infinite number of real rational number n's, you will never come to a conclusion on what p must be in Q. In other words, it will take an uncountable infinite amount of n numbers(which does not exist, as Q is only countable) for p to be created, or put another way, it will take an uncountable infinite amount of time, calculations, attempts, or whatever, in order to create p. So you will always get closer to creating p without actually creating it. But yet it can be constructed by other means.
Do your believe that for every set A there exists a set, called the power set of A, P(A), that contains every subset of the set A? (The axiom of power sets)?
No, I don't. There can be two sets that are not mathematically related in any way. We can still count them subjectively, as if by yanking two unrelated numbers out of a barrel of infinite numbers, and saying "This number here, for this number there" and perform this excersise infinitely.
No, you cannot count all the real numbers(R), you cannot create a one-to-one mapping between N and R, as there are not enough natural numbers, by cantors diagonal argument. If you give me a any mapping between N and R, that you
say is surjective, I can always constructively find a number in R that you miss with your surjection, and your surjection is not surjective.
Cardinality(roughly equvivalent of size) of set are defined by the existence of injections and bijections(and surjections, but that require AC) between them. As we can construct a bijection(f(x) = 2x) between the natural numbers(N) and the even natural numbers(2,4,6,8,...), we say that the even natural numbers have the same cardinality as all of the natural numbers.
On the other hand any injection from the natural numbers can't be surjective, and can't therefor be a bijection, we then say that R have a greater cardinality then N. But we can show that the set of all subsets of N have the same cardinality as R.
(You should really go read some elementary set theory, your posts about them seems that you are unable to comprehend what the fuck is going on. This is in no way criticism, set theory is at some times really counter intuitive, and most have, like you, difficulty of understanding it.)
