Let $A ≠ \varnothing$ and $f : \mathcal P(A) \setminus \{\varnothing\} → A$ a choice function for $\mathcal P(A) \setminus \{\varnothing\}$.
Prove that the following two conditions are equivalent:
- $∀B \,∀C \big( B ≠ \varnothing \,\&\, B ⊆ A \,\&\, C ≠ \varnothing \,\&\, C ⊆ A \implies f(B∪C) = f(\{f(B), f(C)\}) \big)$;
- there exists a well order $≤$ in $A$ such that for each non-empty $B$ in $A$ it holds that $f(B) = \min_≤ B$.
I have already posted this problem, but in this post I share with you my attempt to prove the problem in the other direction, namely:
My attempt for $(\rm ii) \!\implies\! (\rm i)$: Define $f(s)=\min_≤s$ for all $s⊆A$. Let $s'=f(s)=\min_≤s $. Then $s'$ is a minimal element of $S$ in regard to $≤$, so $≤$ is a well order, which implies that $≤$ is a linear order.
Therefore, for all $a,b∈A$ it holds: $f(\{a,b\}) ∈ \{a,b\} = a∪b$. But by definition, $f(\{a,b\})=\min_≤\{a,b\}$ which is $a$ if $a≤b$ and $b$ if $b≤a$.
Hence $f$ preserves the order and $f(\{f(a),f(b)\})=\min_≤\{f(a),f(b)\}$. WLOG let $f(\{f(a),f(b)\})=f(\{a\})=a$, hence $f(a)≤f(b)$.
Now $f(a∪b)=f(\{a,b\})=a$ since $f(a)≤f(b)$. Then we proved that $f(a∪b)=f(\{f(a),f(b)\})$ for $a≤b$.
Since I cannot say whether or not this is correct, please give me feedback.