Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
I need to prove that the set $\{A^n : n \in \mathbb{N} \}$ using the ZFC axioms (without Replacement).
My (rough) plan would be to construct some set containing "more" sets than necessary, then use the comprehension axiom to remove the undesirable sets. Something like: for some $Y$, $\{ X \in Y : \exists n \in \mathbb{N}, X = A^n\}$.
I know how to prove that any one of these sets (e.g., $A^2$), exists, but I have no idea how to prove that the set containing all of these powers exists.
If you define $A^n$ as the set of functions from the ordinal $n$ into $A$, then the answer is positive.
Note that a function $n\to A$ is a subset of $n\times A$, and hence also a subset of $\omega\times A$. Thus, every function $n\to A$ is an element of $\mathcal P(\omega\times A)$, so any family of such functions (in particular, every family of all such functions for a fixed $n$) is an element of $\mathcal P(\mathcal P(\omega\times A))$.
