I need help with one of my Set Theory homework tasks. This problem is in (ZFC) ( With Axiom of Choice). Can someone give me some idea with this proof. I thought that I can construct transversal and then remove from it some elements, but I am not really sure how to do it.
Transversal of one set $A$ is set $Y$ with property $(\forall x \in A)(x \cap Y \neq \emptyset).$
Proof that, if all elements of $A$ are finite and nonempty sets, there is set $Y$ that is transversal for $A$ and
$\forall Z((Z \subseteq Y \land(\forall x \in A)(x \cap Z \neq \emptyset)) \implies Z=Y)$.
Let $P$ be the set of all transversals of $A$, ordered by reverse containment (so $T\leq T'$ means $T\supseteq T'$). We seek a maximal element of $P$, i.e. a transversal of $A$ which is mimimal under containment).
So we apply Zorn's Lemma to $P$. Note first that $P$ is nonempty: letting $f$ be a choice function for $A$, $\{f(x)\mid x\in A\}$ is a transversal of $A$. So now we need to know that every chain in $P$ has an upper bound in $P$, i.e. that given any collection of transversals linearly ordered by inclusion, there is some tranversal contained in all of them.
So suppose $(T_i)_{i\in I}$ is a chain of transversals of $A$, and let $T = \bigcap_{i\in I} T_i$. If we show that $T$ is a transversal of $A$ we're done, since $T$ is an upper bound for the chain.
Let me pause and talk about the intuitive idea here. You might worry that since the $T_i$ get smaller and smaller, you might throw away more and more elements of some $x\in A$, until the intersection of $T$ with $x$ is empty. But since $x$ is finite, this can't happen: at some point, you'd have to throw away the last element of $x$, and at that point some $T_k$ would already fail to be a transversal. Ok, on with the formal proof.
Suppose for contradiction that $T$ is not a transversal of $A$. Then there is some $x\in A$ such that $x\cap T = \emptyset$. Since $x$ is finite, we can write $x = \{y_1,\dots,y_n\}$, and if $x\cap T = \emptyset$, then $y_j \notin \bigcap_{i\in I} T_i$ for all $1\leq j \leq n$. So there is some $i_j\in I$ such that $y_j\notin T_{i_j}$ for all $1\leq j \leq n$. Let $T_k = \max(T_{i_1},\dots,T_{i_n})$ in the order on $P$, i.e. $T_k \subseteq T_{i_j}$ for all $j$. Then $y_j\notin T_k$ for all $1\leq j\leq n$, and so already $x\cap T_k = \emptyset$. But this is a contradiction with the fact that $T_k$ is a transversal of $A$.