Let A ≠ ∅ and f : P(A) \ {∅} → A is a choice function for P(A) \ {∅}.
Prove that the following two conditions are equivalent:
∀B∀C(B ≠ ∅ & B ⊆ A & C ≠ ∅ & C ⊆ A ⇒ f(B∪C) = f({f(B), f(C)}));
exists such a well order ≤ in A, that for each non empty B in A it holds that f(B) = min ≤ B
My attempt:
From f : P(A) \ {∅} → A ⇒ Dom(f)=P(A)\ {∅}.
Hence Let F={f|f : P(A) \ {∅} → A (∀a∈Dom(f))[f(a)∈a]}. <F,⊆F > is partial ordered set in which every chain has an upper bound.
Indeed, let F'⊆F is a chain. Then for every f∈F',f⊆∪F'.
Since F' is a chain of functions, then ∪F' also is a function and Dom(∪F')=∪(Dom(f)|f∈F'}⊆P(A)\ {∅} and for every a∈Dom(∪F'),∪F'(a)∈a. Then (∪F')∈F and as such it is an upperbound for the chain F'.
And now ∪(B,C)∈F'⊂f. So f((B∪C))=f({f(B),f(C)}).
But I do not know how to go further. Any help is welcome.
I cannot make any sense of your argument. If you’re trying to prove that the first statement implies the second, you cannot look at a whole family of choice functions: you’re given one specific choice function, and you have to use it to construct a well-ordering of $A$. And if you’re trying to prove the other direction, you must simply define $f(S)=\min_\le S$ for each non-empty $S\subseteq A$ and show that this $f$ has the desired property. That direction is quite trivial, and I’ll leave it to you.
For the other direction, assume that such a choice function $f$ exists. Define a relation $\le$ on $A$ by setting $a\le b$ iff $f(\{a,b\})=a$. We first prove that $\le$ is a partial order on $A$. Reflexivity and antisymmetry are easy.
For each $a\in A$ it is certainly true that $f(\{a,a\})=f(\{a\})=a$, so $a\le a$, and $\le$ is reflexive. And if $a,b\in A$, $a\le b$, and $b\le a$, then $a=f(\{a,b\})=b$, so $\le$ is antisymmetric.
Proving transitivity takes a bit more work.
If $a,b,c\in A$, $a\le b$, and $b\le c$, then $f(\{a,b\})=a$ and $f(\{b,c\})=b$. Thus,
$$\begin{align*} f(\{a,b,c\})&=f\big(\{a,b\}\cup\{b,c\}\big)\\ &=f\big(\big\{f(\{a,b\}),f(\{b,c\})\big\}\big)\\ &=f(\{a,b\})\\ &=a\,. \end{align*}$$
On the other hand,
$$\begin{align*} f(\{a,b,c\})&=f\big(\{a,c\}\cup\{b,c\}\big)\\ &=f\big(\big\{f(\{a,c\}),f(\{b,c\})\big\}\big)\\ &=f\big(\big\{f(\{a,c\}),b\big\}\big)\,, \end{align*}$$
so $f\big(\big\{f(\{a,c\}),b\big\}\big)=a$, and therefore $f(\{a,c\})=a$ and hence $a\le c$.
Clearly $\le$ is linear: for any $a,b\in A$, $f(\{a,b\})\in\{a,b\}$, so either $a\le b$, or $b\le a$. It only remains to show that $\le$ is a well-order. Let $\varnothing\ne S\subseteq A$, and for convenience let $s_0=f(S)$. Suppose that $s_0\ne\min_\le S$; then there is an $s\in S$ such that $s<s_0$, i.e., $f(\{s,s_0\})=s$. This is impossible if $S=\{s,s_0\}$ (why?), so $S\setminus\{s,s_0\}\ne\varnothing$. Let $R=S\setminus\{s,s_0\}$; then
$$\begin{align*} f(\{s,f(R)\})&=f\big(\big\{f(\{s,s_0\}),f(R)\big\}\big)\\ &=f(\{s,s_0\}\cup R)\\ &=f(S)\\ &=s_0\,, \end{align*}$$
which is impossible, since $s\ne s_0\ne f(R)$. Thus, $s_0\le s$ for each $s\in S$, and hence $s_0=\min_\le S$. Every non-empty subset of $A$ therefore has a $\le$-minimum element, and $\le$ well-orders $A$.