Let $(S_i)_{i \in I}$ be a family of non-empty sets where $I$ is a Dedekind-finite set. Can we say without the Axiom of Choice that this family has a choice function? I know that for any finite family there is a choice function without the use of AC, but I do not know about the Dedekind-finite case.
Can we say that any Dedekind-finite family of sets has a choice function without AC
axiom-of-choiceset-theory
Related Solutions
You are working only with finite products, and this hold in general. In the Choice Function $\Rightarrow$ AoC direction you don't know if $D$ is nonempty, that is what you are trying to pare working only with finite products, and this hold in general. In the AoC $\Rightarrow$ Choce Function direction you need a choice function on $\mathcal{A},$ not in $D.$ Moreover, I think we need to clarify some definitions.
First one definition
Definition: Let $\mathcal{A}$ be a nonempty collection os sets. A surjective function $f$ from some set $J$ to $\mathcal{A}$ is called an indexing function for $\mathcal{A}.$ $J$ is called the index set. The collection $\mathcal{A},$ together with $f,$ is called an indexed family of sets.
I'll use your definition of axiom of choice.
Axiom of choice: For any indexed family $\{ A_\alpha \}_{\alpha \in J}$ of nonempty sets, with $J \neq 0,$ the cartesian product $$ \prod_{\alpha \in J} A_\alpha$$ is not empty. Recall that the cartesian product is the set of all functions $$ \mathbf{x}:J \to \bigcup_{\alpha \in J}A_\alpha$$ such that $\mathbf{x}(\alpha) \in A_\alpha$ for all $\alpha \in J.$
Now,
Existence of a choice function: Given a collection $\mathcal{A}$ of nonempty sets, there exists a function $$ c: \mathcal{A} \to \bigcup_{A \in \mathcal{A}}A$$ such that $c(A)$ is an element of $A: c(A)\in A$ for each $A \in \mathcal{A}.$
Axiom of choice $\Rightarrow$ Existence of choice function.
We are assuming that for any indexed family $\{ A_\alpha \}_{\alpha \in J}$ of nonempty sets, with $J \neq 0,$ the cartesian product $ \prod_{\alpha \in J} A_\alpha$ is not empty. Let $\mathcal{A}$ be a collection of nonempty sets. We have to prove that there exists a function $ c: \mathcal{A} \to \bigcup_{A \in \mathcal{A}}A$ such that $c(A) \in A$ for each $A \in \mathcal{A}.$ We first index $\mathcal{A}$: let $J=\mathcal{A}$ and $f:\mathcal{A} \to \mathcal{A}$ given by $f(A)=A.$ Then $\{A\}_{A \in \mathcal{A}}$ is an indexed family of sets, so we can consider its cartesian product $\prod_{A \in \mathcal{A}}A.$ By hypothesis, this product is nonempty, so there exists a function $$ \mathbf{x}: \mathcal{A} \to \bigcup_{A \in \mathcal{A}}A$$ such that $\mathbf{x}(A) \in A$ for each $A \in \mathcal{A}.$ Then $c=\mathbf{x}$ is the function we were looking for.
Existence of choice function $\Rightarrow$ Axiom of choice
Now we are assuming that the existence of choice function, and let $\{A_\alpha \}_{\alpha \in J}$ be an indexed family of nonempty sets, with $J \neq 0.$ This means that there is a nonempty collection of sets $\mathcal{A}$ and there is an indexing (i.e. surjective) function $f:J \to \mathcal{A}$ such that $f(\alpha)=A_\alpha \in \mathcal{A}$ for each $\alpha \in J.$ By the existence of choice function, there is a function $c:\mathcal{A} \to \bigcup_{A \in \mathcal{A}}A$ such that $c(A) \in A$ for each $A \in \mathcal{A}.$ Thus the function $$ \mathbf{x}:= c \circ f : J \to \mathcal{A} = \bigcup_{\alpha \in J}A_\alpha$$ satisfies $\mathbf{x}(\alpha)=c(f(\alpha))=c(A_\alpha) \in A_\alpha$ for each $\alpha \in J,$ so the product $\prod_{\alpha \in J}A_\alpha$ is nonempty.
I don't see any precise meaning for "midway" in this context. I suspect that all that was meant is that (1) the axiom of choice implies the ultrafilter principle, (2) the ultrafilter principle implies the axiom of choice for finite sets, and (3) neither of the preceding implications is reversible.
Best Answer
No.
If $A$ is an amorphous set, then its power set is Dedekind-finite. Therefore there is no choice function which chooses from its non-empty subsets. Since if $\mathcal P(X)\setminus\{\varnothing\}$ has a choice function, then $X$ can be well-ordered.
So if $A$ is amorphous, then we have a Dedekind-finite family without a choice function.
This is not an equivalence. For example, in Cohen's first model, the family of co-finite subsets of $A$, the canonical Dedekind-finite set, is Dedekind-finite. And it does not admit a choice function, since $$A\text{ is D-infinite} \iff\{A\setminus X\mid X\text{ is finite}\}\text{ admits a choice function}.$$
So again, in Cohen's model we have a Dedekind-finite family of sets without a choice function.
In general, given any Dedekind-finite set $X$, we can arrange an extension of the universe where $X$ remains Dedekind-finite, and there is a family of sets indexed by $X$ which does not have a choice function.