Does this construction rely on the axiom of choice?

Question

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.

Answer 1

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.

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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.