Let $R$ be an equivalence relation on a set $X$. $T\subseteq X$ is called a transverse of $R$ if $T$ intersects every equivalence class of $R$ in exactly one point. This is basically a representative system.
Axiom of Choice (AC): Every indexed family $(X_i)_{i \in I}$ of sets has a choice function. That means there's a function $c:I\to\bigcup\{X_i:i\in I\}$ such that $c(i)\in X_i$ for every $i\in I$.
I want to prove in ZF that (AC) is equivalent to the statement that every equivalence relation on a set has a transverse.
My ideas:
"$\Rightarrow$":
We assume that the axiom of choice holds. Now let $R$ be an equivalence relation on a set $X$.Now we have that $X/R$ (the set of equivalence classes) is a partition of $X$. Now I want to put $X/R=(\overline{x}_i)_{i\in I}$ where $\overline{x}$ is the equivalence class of $x\in X$ and then $T:=\{c(i)\mid i\in I\}$. Then $T$ should be a transverse of $R$. But the part where I put $X/R=(\overline{x}_i)_{i\in I}$ is weird for me... So we know that $\overline{x}$ is a set for every $\overline{x}\in X/R$ since $X$ is a set. But where do I get the set $I$ from? I would have to put $I$ as a representative System or something, but that would not make sense since I am currently trying to show its existence...
"$\Leftarrow$":
Let $(X_i)_{i\in I}$ be an indexed family. I am now trying to construct a function $c:I\to \bigcup\{X_i:i\in I\}$ such that $c(i)\in X_i$ for every $i\in I$. Let $X:=\bigcup\{X_i:i\in I\}$. Since $X_i$ is a set for every $i\in I$ we have that $X$ is a set. By assumption every equivalence relation on $X$ has a transverse. Now define $R\subseteq X\times X$ such that $$xRy:\Leftrightarrow x\in X_i\land y\in X_i$$. This is obviously an equivalence relation, hence it has a transverse $T$ by assumption. Now I should be able to set $c(i)=x$, if $x\in T\cap X_i$, because we have $|T\cap X_i|=1$ for every $i\in I$ since $T$ is a transverse of $R$ and $X/R=\{X_i:i\in I\}$. I'm quite confident on this direction.
Thank you for your help. Kind regards, Max.
Give $X,R,$ we define the quotient as:
$$X/R:=\{ J\subseteq X\mid \exists x\in J\forall y(y\in J\iff (x,y)\in R)\}$$
Then we can define $X_{J}=J$ for each $J\in \mathcal I=X/R.$ Then a choice function is a traversal.
On the other hand, given $\{X_i\}_{i\in I}$ non-empty, we can define $Y_i= \{i\}\times X_i.$ Show the $Y_i$ are disjoint and non-empty.
We then define $$X=\bigcup_{i\in I} X_i,\\Y=\bigcup_{i\in I} Y_i\subseteq I\times X.$$
Then we define an equivalence relation $R$ on $Y$ by saying, for $(i,x),(j,y)\in Y,$ $(i,x)R (j,y)$ iff $i=j.$
Show $R$ is an equivalence relation on $Y,$ with equivalence classes $Y_i.$
Then a traverse for $R$ is a set $T\subseteq Y\subseteq I\times X$ with certain properties. Show this subset satisfies the definition of a function $I\to X,$ and then show that this function works as a choice function for the $X_i.$
At heart, the general axiom of choice is equivalent to the case where the $X_i$ are disjoint. By taken the general case, we can convert it to a disjoint case, and the disjoint case can then be converted to a question about the traverse of the union with the disjoint sets as the equivalence classes.
It turns out, amusingly, that the process of ensuring disjointness makes the traverse literally equal to a choice function.