Recent content by ben21

So I think we are able to construct aleph zero these functions.. P(N) contains as subset the set of N so.. We can build this function like this: F: P(N) -> N; F(N) = {case 1 - the power of N equals aleph zero: G(N); {case 2 - in other case (the power of N is less than aleph zer)...

OK, but.. amount of subset of P(N) is more than natural numbers.. it is continuum, not aleph zero... edit::: wrong... I'm a little bit confused.. We need to find a amount of functions f: P(B) -> n, where B is a proper subset of N and n in B...

by "P(N) has N amount of number except zero" are you meaning, that P(N) has \aleph one-element subsets excpet zero? f(N)? Do you mean one-element subsets by N as argument?

There are \aleph (aleph zero) numbers in N-{0}... So 2^\aleph equals \Im (continuum)... So it proves, that the power of P(N) equals \Im.. However, how to combine it with a choice function?

Homework Statement Find the power of set of all choice functions for P(N) - {0}. The Attempt at a Solution I really don't know how to start with that. Hope you will give some clues.