Let $X = \{0,1\}$. We define $f: \mathbb Z_+ \rightarrow X^\omega$ as $$f(n) = (\underbrace{0,…0}_{n-1},1,0…)$$ According to my understanding of the choice axiom, $f$ shouldn't rely on it. Let us now define another very similar construction. Let $A = \{\{x_i\}\}_{i \in \mathbb Z_+}$ be an indexed family, where $x_i = (\underbrace{0,…,0}_{i-1},1,0…) \in X^\omega$. Now there exists ($A$ is a collection of nonempty sets) a choice function $$c: A \rightarrow \bigcup_{i \in \mathbb Z_+} \{x_i\}$$ where $c(\{\{x_i\}\}) = x_i, \forall i \in \mathbb Z_+$. If we now define $g: \mathbb Z_+ \rightarrow A$ such that $g(i) = \{\{x_i\}\}, \forall i \in \mathbb Z_+$ does $g \circ c$ rely on the axiom of choice? Clearly $g \circ c$ and $f$ are almost the same, but I'm not sure if defining $A$ or $g$ constitutes as making an arbitrary amount of choices. Note I am just learning to use the axiom of choice so my intuition isn't quite there yet, hence the question.
Does this construction rely on the axiom of choice
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
When you are dealing with singletons there is no room for arbitrary choices. Therefore the axiom of choice is not necessary.
If we have a set of singletons $A$, we can easily describe a choice function: $F(\{a\})=a$. In other words, if $u$ is in $A$, then we know that there is a unique $a$ such that $u=\{a\}$, therefore map $u$ to this unique $a$.
Now, since you wrote $A$ already as the image of a function $g(i)=\{\{x_i\}\}$, there was really no involvement of the axiom of choice.
A modicum of intuition about choice.
The axiom of choice is needed when you do not have a uniform way to specify elements from your set. That means that you cannot identify some property that exactly one element from each set will satisfy.
When we say property, we allow it to depend on parameters, or be very complicated such as "if the set is finite and has $3$ elements, choose this and that; if the set has infinitely many elements ...", it just needs to provably select a single member from each member of the family.
One example of a "complicated property" can be given when the family is finite. Then we can prove by induction on the size of the family that there is a choice function. This means that we can instantiate a choice function, and define the property as being the chosen element.
But to your case here, where we're only dealing with sets that have exactly one element, then it's easy to uniformly define a single element from each singleton. Well, it's that element.
Another example is if we are given a set with three elements, $\{a,b,(a,b)\}$ (where $(a,b)$ is the ordered pair $a,b$), then I can either choose the ordered pair, or I can choose the left one, or I can choose the right one. Or I can mix them.