Prove that the following statements are equivalent:
- If $A$ is a non empty set, there exists a function $f:\mathcal{P}(A) \setminus \{\emptyset\} \rightarrow A$ such that $f(X) \in X$ for every $X \in \text{dom}(f)$.
- Every non empty family $\mathfrak{F}$ of sets such that any two distinct sets of the family are disjoint has a set $X$ such that $|X \cap F|=1$ for every $F \in \mathfrak{F}$.
Now, this is how I've done $1 \Rightarrow 2$:
We notice that for every non empty family $\mathfrak{F}$ of disjoint sets we have that: $$\mathfrak{F} \subseteq \mathcal{P} \left(\bigsqcup _{F \in \mathfrak{F}} F \right) \setminus \{\emptyset\}$$ If we set $A=\bigsqcup _{F \in \mathfrak{F}} F$, we have (for hypotesis) a function $f$ such that $f(Y) \in Y$ for every $Y \subseteq A$. So, in particular, we have that $f(F) \in F$ for every $F \in \mathfrak{F}$, and also that $f(F) \notin F'$ for every $F' \in \mathfrak{F}$ such that $F' \neq F$. So, if we consider the set $X=\{f(F) \mid F \in \mathfrak{F}\}$, we obtain the thesis.
For $2 \Rightarrow 1$ I really don't know what to do. Any suggestions?