What is the rank of the set of functions from $\omega$ to $\omega$?

Question

My stab was to say each member is a set of pairwise distinct ordered pairs of $\omega$ into $\omega$, so each function has rank $\omega + 1$ , therefore the entire set of functions has rank $\omega+ 2$

Does this work?

Answer 1

This is the right idea but slightly off. Each ordered pair of elements of $\omega$ has finite rank, so each function has rank $\omega$, not $\omega+1$ (it has no elements of rank $\omega$). So, the set of all such functions has rank $\omega+1$.

Answer 2

HINT. $rank(S)=\sup \{rank(T)+1:T\in S\}.$