I am trying to understand the relationship between two forms of the axiom of choice:
If $T=\{X_0,X_1\cdots \}$ is a family of non-empty mutually disjoint finite sets, then $\cup T$ contains at least one subset having exactly one element in common with each element of $T$.
Let $X\ne\emptyset$ be countable and $\mathcal{F}(X)$ be the collection of all finite subsets of it. Then there is a function $f:\mathcal{F}(X)-\emptyset\to X$ such that $f(x)\in x\forall x\in \mathcal{F}(X)-\emptyset$.
Which one implies the other? What will be the formulation of 1. in terms of the choice function?
Thanks.
Follow up question: If in (2) the condition $X\ne\emptyset$ is countable is removed, i.e. $X$ is an arbitrary nonempty set then will it imply (1)?