# Finite family of infinite sets / A.C.

Let $\{A_i\mid i\in n\}$ be a finite family of infinite sets. ( That is, $A_i$ is infinite for every $i\in n$ and $n\in \mathbb{N}$)

Here, we can choose representative $a_i$ from each $A_i$ and construct $\{a_i\mid i\in n\}$.

This process really doesn't use Axiom of Choice?

How do I write this process down formally (in mathematical language)

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@superM I don't get it.. Are you saying that it is 'well-defined?' I'm talking about 'choosing arbitrary element from each infinite set, five times' – Katlus Jul 26 '12 at 9:56
– Michael Greinecker Jul 26 '12 at 10:15

$\newcommand{\Zobr}[3]{#1 \colon #2 \to #3}$ Claim: Let $\{A_i; i\in n\}$ be a system of non-empty sets, $n\in\omega$. Then there is a choice function $\Zobr fn{\bigcup A_i}$.

Proof. Induction on $n$.

$1^\circ$ For $n=0$ put $f=\emptyset$.

For $n=1$: Since $A_0$ is non-empty, we know that $(\exists a\in A_0)$. For every such $a$ the function $f=\{(0,a)\}$ has required property.

We have used existential instantiation here.

$2^\circ$ Suppose that there is a choice function $\Zobr fn{\bigcup A_i}$. We want to extend it to a new function $g$ defined on $n+1=n\cup\{n\}$ in a such way that $g(n)\in A_n$. Again we know that there exists $a\in A_n$ and we put $g=f\cup\{(n,a)\}$. (Again, this is application of existential instantiation.)

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