TITLE: Transfinite Induction and the Axiom of Choice
QUESTION [25 upvotes]: My question is essentially this: Why does the principle of transfinite induction not suffice to show the axiom of choice when the sets to be chosen from are indexed by a well ordered set?
I have read that one can prove the axiom of finite choice from simple induction. You induct on the size of the system of sets you are choosing from and pick an element from each set. I understand this. However, my grasp of the details is sketchy.
1) Why does standard induction alone not suffice to show the axiom of choice for systems of countable sets? Doesn't induction show the truth of the statement for all natural numbers, and therefore for any system of sets that can be indexed by the natural numbers (countable sets)? I know this to be false, but I do not know why.
2) Why can't the above "proof" that induction implies the AoC for countable sets not be repaired by using transfinite induction? Isn't this the purpose of transfinite induction, to allow one to induct on sets of infinite size? Shouldn't transfinite induction suffice to prove the axiom of choice for any system of sets indexed by a well-ordered set? 
I am reading Jech right now, but my knowledge of ordinals and transfinite induction is very, very poor, so I would greatly prefer answers with a great amount of explanation and hand-holding.

REPLY [12 votes]: 1) Why does standard induction alone not suffice to show the axiom of choice for systems of countable sets? Doesn't induction show the truth of the statement for all natural numbers, and therefore for any system of sets that can be indexed by the natural numbers (countable sets)? I know this to be false, but I do not know why.

As Jason DeVito points out in the comments to his answer, just because something holds for all natural numbers $n$ doesn't mean that it holds for the set $\mathbb{N}$ of natural numbers itself: For a trivial example, every natural number $n$ is finite, yet $\mathbb{N}$ itself is infinite.

2) Why can't the above "proof" that induction implies the AoC for countable sets not be repaired by using transfinite induction? Isn't this the purpose of transfinite induction, to allow one to induct on sets of infinite size? Shouldn't transfinite induction suffice to prove the axiom of choice for any system of sets indexed by a well-ordered set?

A proof by ordinary induction has two parts: a base case and a successor step.  A proof by transfinite induction has three parts:  a base case, a successor step, and a limit step.  The limit step says that for an infinite stage (technically, an ordinal number) $\lambda$, if the property $P$ holds at every stage $\alpha < \lambda$, then it holds at stage $\lambda$.  For many properties $P$, the base case and successor step hold but the limit step fails.  To revisit the trivial example from above, let's try to prove that every set is finite.  Let $P(\alpha)$ be the statement "every set of size $\alpha$ is finite," and let us try to prove that $P(\alpha)$ holds for all $\alpha$ by transfinite induction. In this example the base case and successor step hold trivially, but the induction will fail at the very first limit step.
For the less trivial example of the axiom of choice, let $P(\alpha)$ be the statement "every sequence of sets of length $\alpha$ has a choice function."  As suggested in the question, the base case and induction step are trivial.  But the limit step fails again:  Let  $(A_i : i \in \mathbb{N})$ be an infinite sequence of sets and suppose by the induction hypothesis that for every finite $n$, the proper initial segment $(A_i : i < n)$ of the sequence has a choice function.
The natural attempt to define a choice function $f$ for $(A_i : i \in \mathbb{N})$ would be to take choice functions $f_n$ for $(A_i : i < n)$ and combine them in some way.  But first we would have to choose a choice function $f_n$ for each finite $n$, which is just as hard as the original problem!  So transfinite induction doesn't help us at all.