Assume I have the following

(DIS) For every indexed family $\{A_i : i \in I \}$ there exists a family $\{B_i : i \in I \}$ of pairwise disjoint sets such that $B_i \subset A_i$ for all $i \in I$ and $\bigcup_{i \in I} B_i = \bigcup_{i \in I} A_i$.


(AC) For every family $x$ of non-empty, pairwise disjoint sets, there exists a set $z$ such that $|z \cap y | = 1$ for each $y \in x$.

I would like to show that $ZF \vdash (DIS) \rightarrow (AC)$ (that's another exercise in a book I'm currently reading).

So let's assume (DIS) and let $B_i$ be a family of non-empty pairwise disjoint sets. I don't see how to proceed. I am supposed to use (DIS) to construct $z$ such that $|z \cap B_i| = 1$ but my sets $B_i$ are already pairwise disjoint so what can I do? Thanks for your help.


Let $X$ be a set whose elements are non-empty and disjoint. Let $Y = \cup_{x \in X} x$. We want to show that there is a set $Z \subset Y$ such that $|Z \cap x| = 1$ forall $x \in X$.

Put $I = Y$, and look at the family $(A_y = \{ x \in X / y \in x \})_{y \in Y}$. Since the elements of $X$ are disjoint, each $A_y$ is a singleton (the unique $x$ containing $y$), and because every element of $X$ is non-empty, $\cup_{y \in Y} A_y = X.$

By the axiom (DIS) there is a family $(B_y)_{y \in Y}$ such that $B_y \subset A_y$, $\cup B_y = \cup A_y = X$ and the $B_y$ are pairwise disjoint. Now pick $Z = \{y \in Y / B_y \neq \emptyset \}$. Then $Z \cap x = \{y \in Y / y \in x \land B_y \neq \emptyset\} = \{y \in Y / B_y = \{x \}\} = \{y \in Y / x \in B_y\}$. Because the $B_y$ are disjoint, this set has at most one element, and because $x \in \cup B_y$, it has at least one element, so its cardinal is exactly 1.


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