I am a bit confused because I don't fully understand when the axiom of choice can be used and when it cannot. Here is the definition of AC that interests me – it relies purely on sets (and importantly for me, it does not use the concept of functions):
For every family $S$ of non-empty disjoint sets, there exists a set $V$ (a so-called "selector"), which contains exactly one element from each of the sets belonging to the family $S$..
Formally:
$$\forall_{S} \Big\{\big[\forall_{X \in\, S} X \ne \varnothing\big]\! \land\! \big[\forall_{X, Y \in\, S}\, (X \ne Y \Rightarrow X \cap Y = \varnothing)\big] \Rightarrow \exists_V \forall_{X \in\, S}\, \exists_x \big(X \cap V = \{x\}\big)\Big\}$$
Notation:
$(a,b) = \{a, \{a,b\}\}$ – Kuratowski short ordered pair for elements
$(a;b)$ – Open interval in $\mathbb R$
(i.e., , denotes a pair, and ; denotes an interval)
$[a;b]$ – Closed interval in $\mathbb R$
$(a_i)_{i=1}^{n}=(a_1,a_2,…,a_n) = \{ (1,a_1), (2,a_2), (3,a_3),…, (n,a_n)\}$ , for $n>2$ – Finite sequence
$(a_i)_{i=1}^{\infty}=(a_1,a_2,…) = \{ (1,a_1), (2,a_2), (3,a_3),…\}$ , for $n>2$ – Infinite sequence
$\mathbb Z'$ is a certain subset of $\mathbb Z$, $\mathbb R'$ is a certain subset of $\mathbb R$
Description
Now let me cite the following examples of families of sets:
$$\begin{align}
A &= \{ \{a\}: a \in \mathbb{Z}' \} \\
B &= \{ \{a,b\}: a,b \in \mathbb{Z}' \} \\
C_n &= \{ \{a_1, a_2,…, a_n\}: a_i \in \mathbb{Z}' \} \\
C &= \{ \{a_1, a_2,…, a_n\}: a_i \in \mathbb{Z}', n\in\mathbb{N} \} \\
D &= \{ \{a_1, a_2,…\}: a_i \in \mathbb{Z}', \} \\
\\
E &= \{ (a,b): a,b \in \mathbb{Z}' \} \\
F_n &= \{ (a_1,a_2,…,a_n): a_i \in \mathbb{Z}' \} \\
F &= \{ (a_1,a_2,…,a_n): a_i \in \mathbb{Z}', n\in\mathbb{N} \} \\
G &= \{ (a_1,a_2,…): a_i \in \mathbb{Z}' \} \\
\\
H &= \{ \{a\} : a\in\mathbb{R}' \} \\
I &= \{ (a;b) : a,b\in\mathbb{R}', a<b \} \\
J &= \{ [a;b] : a,b\in\mathbb{R}', a<b \} \\
K &= \{ (a;a+1) : a\in\mathbb{Z}' \} \\
L &= \{ [a;a+1] : a\in\mathbb{Z}' \} \\
\end{align}$$
For simplicity, we assume that from all these sets we remove elements that have a common part with other elements of this set (so as to use the definition of AC referring to disjoint sets)
Based on each of the above-defined sets (which contain information about a specific subset of $\mathbb{Z}'$ or $\mathbb{R}'$), I would like to generate a set $X$ containing exactly one element from each subset of the given set, e.g. $X_A = \{ a: \exists b \in A, a\in b \} $, or exactly one element from the pair/sequence, e.g., the first element from each pair in $E$ i.e. $X_E = \{ a: \exists b \in E, a\in b \} $
Question
For the generation/creation of which sets $X_A,X_B,X_C,X_D,X_E,X_F,X_G,X_H,X_I,X_J, X_K,X_L$ will I need to use the axiom of choice and why?
Update:
What if $\mathbb {Z}'$ is special set where you cannot compare elements (and find minimal or center element) ? (we exclude cases H-L)
Best Answer
Assuming excluded middle:
None of $A$, $B$, $C_n$, $C$, require choice since all of their members have a (unique) minimal element.
$D$ doesn't require choice either, as you can always pick the minimal element in absolute value (and if both $a \in S$ and $-a \in S$, pick the positive one).
Similarly, none of $E$, $F_n$, $F$, $G$ require choice since all of their members have a first component.
Similarly, none of $H$, $I$, $J$, $K$, $L$ require choice since all of their members have a "middle element", in the sense of $S \mapsto \frac{\mathrm{inf}(S) + \mathrm{sup}(S)}{2}$.