Countable choice for countable subfamily

Question

Axiom of Countable Choice (CC) states that for every countable family $\left\{A_i\right\}_{i=1}^\infty$ of nonempty sets there exists choice function $f \colon \mathbb{N} \to \bigcup_{i=1}^\infty A_i$ such that $f(i) \in A_i$ for all $i \in \mathbb{N}$.

Consider the following axiom strongly related to the above axiom:

For every countably infinite family $\left\{A_i\right\}_{i=1}^\infty$ of nonempty sets there exists countably infinite subfamily $\left\{A_{i_k}\right\}_{k=1}^\infty$ such that $\left\{A_{i_k}\right\}_{k=1}^\infty$ has choice function i.e. there exists $f \colon \mathbb{N} \to \bigcup_{k=1}^\infty A_{i_k}$ such that $f(k) \in A_{i_k}$ for all $k \in \mathbb{N}$.

Let's call it Axiom of Countable Choice for Countable Subfamily (CCCS).

Of course CC implies CCCS. But I wonder if CCCS is strongly weaker than CC or equivalent to CC.

I can't imagine being the first person to come up with the idea of CCCS, so probably there is literature which consider this kind of choice axiom or maybe it's unnaturally to consider such an axiom. I'm amateur in that sort of set theory. What do you think? Maybe someone knows proper literature?

Answer 1

They are equivalent. Given a countable family $B_i$ of nonempty sets, let $A_i=\prod_{j=1}^iB_j$. That these are nonempty is provable in ZF (we only use finite products). Now for any choice function $f$ with $f(k)\in A_{i_k}$, since $i_k\geq k$, projecting $f(k)$ onto the $k$-th coordinate of $A_{i_k}=\prod_{j=1}^{i_k}B_j$ gives a choice function for the $B_i$.

I don't know a reference for this result, but I have seen similar arguments when establishing various equivalents of countable choice for subsets of $\mathbb R$, so I can't take full credit for the idea.