I have to prove the following:
In $ZF^-$ the axiom of choice implies:
For every set X there exist $Y \subseteq \bigcup X$ such that:
I'll try to solve it using the Zorn's Lemma (that are equivalent to AC) but I can't see the way to solve it.
Thanks for the attention!
On
Indeed, this calls for Zorn. I think the idea should be straightforward, but maybe we should really spell it out:
Let $$\mathcal Y=\{\,Y\in\mathcal P(\bigcup X)\mid\forall Z\in X\colon \forall U,V\in Z\cap Y\colon U=V\,\}$$ Then $\mathcal Y$ is partially ordered by inclusion. If $\mathcal C\subseteq\mathcal Y$ is a chain, then $C=\bigcup \mathcal C$ is an upper bound for $\mathcal C$. Indeed, clearly $C\subseteq \bigcup X$ and $Y\subseteq C$ for all $Y\in \mathcal C$; and if $Z\in X$ and $U\in Z\cap C$, then $U\in Z\cap Y_U$ for some $Y_U\in \mathcal C$. Similarly, if $V\in Z\cap C$, then $U\in Z\cap Y_V$ for some $Y_V\in \mathcal C$. As $\mathcal C$ is a chain we have wlog $Y_U\subseteq Y_V$, hence $U,V\in Z\cap Y_V$ and so $U=V$. Zorn's lemma then says that $\mathcal Y$ has a maximal element.
Define a partial order of all subsets of $\bigcup X$ which meet every element of $X$ at most in a single point, and order those by inclusion.
Suppose that $\{Y_i\mid i\in I\}$ is a chain in this partial order, then $Y=\bigcup\{Y_i\mid i\in I\}$ is a subset of $\bigcup X$ (trivial) and if $x\in X$ then $x\in Y=\bigcup(Y_i\cap x)$. Suppose by contradiction that $|x\cap Y|>1$, then there is some $Y_i,Y_j$ such that $x\cap Y_i\neq x\cap Y_j$, but neither is empty.
However, since this is a chain, $Y_i\subseteq Y_j$ or $Y_j\subseteq Y_i$. This means that there is some $i\in I$ such that $x\cap Y_i$ has more than a single point. This is impossible because $Y_i$'s are assumed to be taken from our partial order. Therefore $x\cap Y$ has at most one point.
Therefore from Zorn's lemma there exists a maximal $Y$ in this partial order, which is what we were looking for.