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Can someone please clear my doubts regarding a result concerning the axiom of choice:

Prove that (i) implies (ii) where:

(i) For every nonempty set whose elements are non empty sets there exists a choice function.

(ii) If $\{a_i\}$ is a family of nonempty sets indexed by a nonempty set $I$, then there exists a family $\{x_i\}$ with $i \in I$ such that $x_i \in a_i$ for each $i \in I$.

Here is the proof which is in my book: Let $A$ be a collection of disjoint sets. We have $A \subset \mathcal{P}(\cup_{i \in I}a_i)$. ($\mathcal{P}$ is the power set). By (i) there exists a choice function $f$ on $\mathcal{P}(\cup_{i \in I}a_i)$. Let $b$ be the image of $A$. Pick an element $a \in A, f(a) \in a \cap b$ since $f(a) \in a$. Let $y \in b$ where $y \neq f(a)$ thus we have $y = f(a')$ where $a'\neq a$, and thus $y \in a'$. Since $a$ and $a'$ are disjoint $y \notin a$. Thus the only element of $b \cap a$ is $f(a)$.

My doubts are:

  • Why do we need to prove this? Won't the choice function $f$ give us the family ${f(i)}$ anyway.

  • How is the existence of $A$ guaranteed?

  • The choice function $f$ has what domain and co-domain? If its co-domain is $\mathcal{P}(\cup_{i\in I} a_i)$ then how is that set indexed? Or is the indexing set $\mathcal{P}(\cup_{i\in I} a_i)$?

  • How does $b \cap a$ being $f(a)$ guarantee that (ii) holds.

Thanks a lot.

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The "proof" is completely opaque to me. For one thing there seems to be no construction of $\{x_i\}$. –  Marc van Leeuwen Sep 18 '12 at 14:09
    
I don't understand the proof properly myself. I will welcome an alternate proof. (I think b is the family $\{x_i\}$) –  Shahab Sep 18 '12 at 14:14
    
What is a choice function in the setup of (i) anyway? I presume that if $S$ is a set such that $S\ne\emptyset$ and $\emptyset\notin S$ then a choice function is a function $f\colon S\to \bigcup S$ such that $f(x)\in x$ holds for all $x\in S$. This has nothing to do with "collection of disjoint sets"; that's another formulation pf AC: If $S$ is a set of pariwise disjoint nonempty sets then there exists a set $C$ such that for each $x\in S$ the set $x\cap C$ is a singleton. –  Hagen von Eitzen Sep 18 '12 at 14:34
    
@Hagen von Eitzen: In the setup of (i) the choice function is precisely what you said, or as I understand if X is a nonempty set with nonempty elements the choice function is a member of the cartesian product (of those elements). Essentially my question 3 is whether $S=\mathcal{P}(\cup a_i)$. –  Shahab Sep 18 '12 at 14:40
    
$S=\mathcal P(\ldots)$ makes no sense as then $\emptyset\in S$. –  Hagen von Eitzen Sep 18 '12 at 14:42
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2 Answers

up vote 2 down vote accepted

Let $\{a_i\}_{i\in I}$ be a family of sets, indexed by the nonempty set $I$. Then by the Axiom Schema of Replacement, $S:=\{a_i\mid i\in I\}$ is a set. By assumption and $I\ne \emptyset$, the set $S$ is a nonempyty set of nonempty sets. By (i) there exists a choice function, i.e. a function $f\colon S\to \bigcup S$ such that $f(x)\in x$ for all $x \in S$. For $i\in I$ let $x_i=f(a_i)$. Then $\{x_i\}_{i\in I}$ is a family with the required properties.

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The statement says, essentially, that if you can choose then you can choose in a somewhat one-to-one fashion.

The assertion (ii) seems a bit weird for me, perhaps you wish to include the assumption that the family is a family of pairwise disjoint sets? The proof seems to assume that...

About the proof, you assume that $A$ is given to you. Note that every set $A$ is a subset of $\mathcal P(\bigcup A)$, this is not hard to check. Using the assumption that every collection of non-empty sets admits a choice function we simply use a choice function on $\mathcal P(\bigcup A)\setminus\{\varnothing\}$, and by the assumption that $A$ is a collection of pairwise disjoint sets we have that $\{f(a_i)\}$ is our beloved family of singletons.

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