I have run into the following statement in Rotman, Advance Modern Algebra
If $\{G_a\mid a\in A\}$ is a family of groups, then we may define a choice function $f:A\to\bigcup_{a\in A}G_a$ by $f(a)=1_a$, where $1_a$ is the identity element of $G_a$; we do not need the axiom of choice to define $f$. In contrast, if we merely "choose" some element $x_a\in G_a$, then the "function" $h:A\to\bigcup_{a\in A}G_a$ with $h(a)=x_a$ is not well-defined.
Intuitively, I have no problem with this statement: in the first case I am performing a kind of "determinate" or "natural" choice, because every group has an identity. In the second case, I'm really looking at the family of groups as a family of sets and hence I am really performing a choice.
However, I have been asked to show this "formally", i.e. in Set Theory, and my expertise of first-order logic is very poor.
First of all, I would like to show that the choice function $f:a\mapsto 1_a$ is well-defined without resorting to the Axiom of Choice. I know that $A$ is a set by assumption. $G:=\bigcup_{a\in A}G_a$ is a set by the Union Axiom of Set Theory and I can perform $A\times G$ which is a set. Now, I consider the formula
$$\phi:\quad \exists a(a\in A \wedge 1\in G_a\wedge u=(a,1))$$
which is an honest formula because built up from the atomic formulas $x\in y$ and $x=y$ and the constant $1$ (which exists in Group Theory) by means of the logic connectives and the quantifiers.
The choice function $f$ now should be
$$\{u \mid \phi(u)\},$$
which should mean exactly $f=\{(a,g)\in A\times G\mid a\in A \text{ and }g=1_a\in G_a\}$. Isn't it?
If that's correct, then my second question is: why I cannot use the same argument to prove that the general choice function is well-defined? Is it because in the latter case I should define $\phi$ as
$$\phi:\quad \exists a\exists g(a\in A \wedge g\in G_a \wedge u=(a,g))$$
and the critical point is $g\in G_a$? But then why exactly is it critical or in what does it differs from $1\in G_a$?
Many thanks in advance for any advice or help.
Best Answer
The issue here is that there are usually many ways (read: more than one) to make a set into a group. So you'd need to choose one, given a family of sets. But in the case you're dealing with, you are given these choices already. So you're simply decoding from this choice a different kind of choice.
Not to mention that the axiom of choice follows from the assumption "every non-empty set has a group structure". But that's besides the point.
The thing here is that a group has a distinguished element. So it is easy to choose that one. Similarly, if $\{A_i\mid i\in I\}$ is a family of sets, then $\{A_i\cup\{I\}\mid i\in I\}$ admits a choice function: simply choose $I$ from each one. On the other hand, sets, in general, do not have a distinguished element. So you cannot just pick one in a coherent and uniform way.