What are the properties of the power set of the set of all integers?

Do you know what a power set is? If not, see the definition. Basically, you are dealing with sets instead of numbers. Each element of the domain and codomain is either a set of integers or the empty set. In other words, you are mapping (sets of integers and the empty set) to (sets of integers and the empty set).

If you haven’t looked at the wiki page on power sets, it’s worth checking out. There’s some good information there that should help your thinking. For instance, it explains that the power set of a countably infinite set (like the integers) is uncountably infinite.

If you have more specific questions, I can try to help you more. As it is, I’m not really sure if I’ve given you the kind of information you’re looking for. I also don’t know your current level of mathematics, so I’m not sure if I’ve explained too much or not explained enough in this answer. If you need me to elaborate on anything here, let me know.

For what it is worth, the magnitude of the power set of all integers is the same as the magnitude of the reals.

I was aware of Cantor’s ingenious proofs of his theorems using diagonal enumeration (rationals are countable) and decimal expansions of reals (reals are not). But until now I wasn’t aware of Cantor’s Theorem: For any set A, the set of all subsets of A has a strictly greater cardinality than A itself. Cantor cleverly constructs “non-selfish sets” to establish proof by contradiction. Wow! An unexpected bit of learning today – Fluther rules!

Power sets in general: X is a set, P(X) is the power set of X. The mapping x |-> {x} is an injection from X to P(X), so |X| <= |P(X)|. If there were a surjection f : X -> P(X), then the set {x|x not in f(x)} would be the image of some y in X so that y in f(y) iff y not in f(y). Thus, no such surjection exists, and |X| < |P(X)|.

@LostInParadise I think that depends on the axiom of choice.