>>137
Technically the set of all subsets of a set A, the so called powerset: P(A), has greater cardinality than the set A, to be more precise
|P(A)| = 2|A|. The following does hold true:
|N| = |Z| = |Q| = \aleph_0 - or naturals, integers and rationals are countably infinite.
Through Cantor's diagonalization argument, it can be shown that reals have a greater cardinality than naturals, called the cardinality of the continuum
c, and assuming the Axiom of Choice, it's
\aleph_1. An interesting thing about the powersets of infinite sets is that they will always give you the next cardinal, so the powerset of naturals has the same cardinality as that of reals (assuming said axiom).
However, I don't think that's what you are talking about at all. I think most people here are merely talking about the cardinality of well-known sets such as naturals and reals, while it doesn't seem to me clear what you're talking about, although if I had to give it a try, I'll say this:
there are functions f from N to N (f:N->N). There are a countable infinity of such enumerable functions which are computable, although they are not all total functions, some are partial (to say it in a more clear language: some are undefined on some inputs, or they never halt). The actual set f:N->N of total functions is not enumerable (including non-computable): this is non-trivial, but it can be proven by diagonalization, the same technique used to prove the uncountability/unenumerability of the reals.