Axiom of choice for subsets

Question

I'm trying to solve following problem. Let $A_{i,j}$ be subsets of a set $X$ for $i,j\in \mathbb{N}$. Show that $$\bigcap_{i=0}^\infty \bigcup_{j=0}^\infty A{i,j}=\bigcup_{(a_{i})}\bigcap_{i=0}^\infty A_{i,a_{i}},$$ where the second union is over all sequences $(a_i)^\infty_{i=0}$ of natural numbers. Do I need to use axiom of choice here?

Answer 1

No, you don't need choice here, because the space of possible $j$s is already well-ordered. When you need to construct an $(a_i)$ sequence that verifies that some element is in your resulting set, you can just choose the smallest $j$ that works in each position.

If, on the other hand, the $j$s were drawn from an arbitrary index set rather than $\mathbb N$, you would need countable choice, though.