TITLE: Intuition behind the Axiom of Choice
QUESTION [6 upvotes]: Why is it different to make one choice or many choices than to make infinite choices from a theoretical point of view in which indeed you are not going to do any?
How could that be different from making infinite additions for example $\sum_{i=0}^{+\infty}A_n$.
Thanks a lot.

REPLY [4 votes]: The axiom of choice can be seen as a generalization of the principle of induction. So, since the principle of induction is technically a lot simpler, let's ask why is there a need to accept the principle of induction. 
The principle of induction, for the purposes of this answer, says that for a property $P(n)$ about a natural number $n$, if $P(0)$ holds and if $P(n)$ holds, then $P(n+1)$ holds, then in fact $P(n)$ holds for all natural numbers $n$. 
This principle seems obvious enough (just like the axiom of choice (in at least one of its forms) seems obvious) so why the fuss about calling it a principle? Well, let's first agree that any proof must be a finite list of characters. Now, how does one argue to convince the skeptic about the validity of the principle of induction? One way is to say, well suppose you want to prove that $P(1)$ holds. Then here is a (finite!) proof: $P(0)$ is known. It is also known that $P(0)\implies P(1)$, thus Modus Ponens tells us that $P(1)$ holds. QED.
This of course is far from proving $\forall n\in \mathbb N \quad P(n)$. So we go on. Suppose you want to establish $P(2)$. Well, here is a (finite!) proof: $P(1)$ was already established (i.e., cut and paste prvious (finite!) proof here), and it is given that $P(1)\implies P(2)$. Thus, Modus Ponens again, gives us that $P(2)$ holds. QED. 
Usually one then concludes with the not so convincing argument "and so on" to then argue that we actually established $\forall n\in \mathbb N \quad P(n)$. Well, here is the problem then. We didn't actually prove that! What we did was give a hand-wavy argument that the two assertions 1) $P(0)$ holds and 2) $P(n)\implies P(n+1)$ holds, are sufficient to convince one that one has a recipe for proving $P(n)$ for all $n\in \mathbb N$. In other words, one seems to be convinced that for any given $n$, one can find a (finite!) proof that $P(n)$ holds. But, do we now have a single finite proof that $\forall n\in \mathbb N\quad P(n)$ ? Well, the answer would be yes if you accept the recipe for proofs as an actual proof. In other words, if you accept the principle of induction. 
So, accepting the principle of induction can be said to be the acceptance of a finite recipe of finite proofs for $P(n)$ (where the length of the proof of $P(n)$ depends on $n$ and will typically tend to infinity with $n$) as a single finite proof of all $P(n)$ in one go. It seems very reasonable to accept such a proof recipe as a proof, which is why the principle of induction is doubted by very few. 
Now, the principle of induction is equivalent to the existence of a least element in any non-empty subset of $\mathbb N$, namely to $\mathbb N$ being well-ordered. The axiom of choice, is equivalent to the existence of a well-ordering on any non-finite set. So the axiom of choice allows for more intricate recipes of proofs and is no longer so easily accepted.