=== Axiom of Countable Choice for Finite Sets implies Countable Union Condition for Finite Sets === Suppose that the Axiom of Countable Choice for Finite Sets holds. Let $\FF$ be a countable set of finite sets. {{WLOG}}, assume that $\O \notin \FF$. Then $\FF$ is either finite or countably infinite. If $\FF$ is finite, then $\bigcup \FF$ is finite by Finite Union of Finite Sets is Finite, and thus countable. Suppose instead that $\FF$ is countably infinite. Then there is a bijection $f: \N \to \FF$. For each $n \in \N$, let $\map e n$ be the set of all bijections from $\card {\map f n}$ to $\map f n$. {{Explain|What exactly does it mean to have a mapping from $\card {\map f n}$ to $\map f n$? Presumably $\card {\map f n}$ is the cardinality of $\map f n$. So is $\card {\map f n}$ being treated as the natural number $\card {\map f n}$ defined as, for example, the Von Neumann definition of the natural numbers? Clarification needed.}} By Cardinality of Set of Bijections $\map e n$ is finite for each $n \in \N$. $\Img e$, the image of $e$, is countable by Image of Countable Set under Mapping is Countable. Thus by the Axiom of Countable Choice for Finite Sets, $\Img e$ has a choice function $c$. Let $q = c \circ e$. Then for each $n \in \N$, $\map q n$ is a bijection from some finite ordinal to $\map f n$. Recursively define a mapping $g: \N \to \N \times \N$ thus: :$\map g 0 = \tuple {0, 0}$ :$\map g {n^+} = \begin{cases} \tuple {\map {g_1} n, \map {g_2} n^+} & : \map {g_2} n^+ \in \Dom {\map q {\map {g_1} n} } \\ \tuple {\map {g_1} n^+, 0} & : \text{otherwise} \end{cases}$ where $\map {g_1} n$ and $\map {g_2} n$ are the first and second coordinates of $\map g n$. Now define a mapping $h: \N \to \bigcup \FF$ by defining: :$\map h n = \map {\map q {\map {g_1} n} } {\map {g_2} n}$ {{explain|What is meant by the concatenation of $\map q {\map {g_1} n}$ with $\paren {\map {g_2} n}$ in this context?}} {{finish|Show that that $h$ is a surjection.}} {{qed|lemma}} === Countable Union Condition for Finite Sets implies Axiom of Countable Choice for Finite Sets === Suppose that the union of every countable set of finite sets is countable. Let $S$ be a countable set of non-empty finite sets. Then $\bigcup S$ is countable. Thus by Surjection from Natural Numbers iff Countable, there exists a surjection $f: \N \to \bigcup S$. Define a mapping $g: S \to \bigcup S$ thus: :$\map g x = \map f {\min \set {n \in \N: \map f n \in x} }$ This is a valid definition: For each $x \in S$, $x$ is non-empty, so it has an element $y$. Then by the definition of union, $y \in \bigcup S$. Since $f$ is a surjection, $\set {n \in \N: \map f n = y}$ is non-empty. Thus $\set {n \in \N: \map f n \in x}$ is non-empty. Thus by the Well-Ordering Principle, $\set {n \in \N: \map f n \in x}$ has a smallest element. {{improve|The above is rather long-winded.}} But by the definition of smallest element: :$\min \set {n \in \N: \map f n \in x} \in \set {n \in \N: \map f n \in x}$ so $\map g x \in x$. Thus $g$ is a choice function for $S$. As this holds for every countable set of finite sets, the Axiom of Countable Choice for Finite Sets holds. {{qed}} Category:Set Union h3kqwc9wumyqk59ikxvnydn3if3ckxh