Constructing a tuple that satisfies a certain relation without axiom of choice

Question

Suppose we have sets $C$ and $S$ and a relation $R \subset C \times S$. Suppose $\boldsymbol{c}\in C^n$. Suppose we know that for $0\le i < n$ with $i, n \in \mathbb{N}$ that we know that there exists $s\in S$ with $(\boldsymbol{c}_i, s) \in R$ where $\boldsymbol{c}_i$ is the $i^{th}$ element of $\boldsymbol{c}_i$. How can we (rigorously) prove that there exists $\boldsymbol{s}\in S^n$ with $(\boldsymbol{c}_i, \boldsymbol{s}_i)$ for each $0 \le i < n$?

There may be multiple $s \in S$ related to each $\boldsymbol{c}_i \in C$. So it feels like I would need some axiom of choice to pick out exactly one of these for each $\boldsymbol{c}_i$ to construct the tuple $\boldsymbol{s}$. But somehow this feels like overkill. How to do this without AC? I feel like I'm missing something.

If it matters I'm defining an $n$-tuple on a set $X$ to be a function from $\{0, \ldots, n-1\} \to X$ and $X^n$ is the collection of all such functions.

The proof of existence of such a $\boldsymbol{s} \in S^n$ need not be constructive..

Perhaps we require a proof by induction on $n$ that such a $\boldsymbol{s}$ exists. Clearly if $n=1$ and we identify $S^1$ with $S$ it is clear the base case holds. If there exists $\boldsymbol{s}\in S^n$ that satisfies the property for $\boldsymbol{c} \in C^n$ then you could append to that $\boldsymbol{s}$ to find a corresponding $\boldsymbol{s}'\in S^{n+1}$ that satisfies the property for $\boldsymbol{c}'\in C^{n+1}$.

Answer 1

Your inductive idea is exactly right, this is essentially the axiom of choice for finite sets i.e. for every finite family of non empty sets there exists a choice function. The axiom of choice for finite sets is provable in ZF and does not need AC, one can prove this by a simple inductive argument(which again is essentially the same as yours). This argument has been explored on the site before for example here: Do We Need the Axiom of Choice for Finite Sets?