Understanding a world without the axiom of choice (AOC)

Question

It is known that there exists a number $x$ in the set $\Bbb R^n$. AOC further assume that for example there exists an element $f=(f_i)_{i\in R}$ in $\Bbb R^\Bbb R$, such that each $f_i\in \Bbb R$. $\Bbb R$ is the set of real numbers.

But to me, how can AOC does not hold? I cannot image a world where $\Bbb R^\Bbb R$ does not have any element in it.

Could you please give me an example that $\Bbb R^\Bbb R$ does not have elements $f=(f_i)_{i\in R}$(, such that each $f_i\in \Bbb R$)? Why the example is important?

Answer 1

There are $2$ very simple mistakes being made in your reasoning; consider the following analogy:

Your statement: "Some cats are brown"

My statement: "All cats are brown".

My statement is clearly false, whilst yours is clearly true. To prove your statement, you only need to show me a brown cat; to prove mine, I would need to collect all the cats on earth, and check their fur colour one by one. So these are genuinely vastly different statements.

The axiom of choice is a statement of the second kind, so exhibiting an example is not sufficient to prove it.

The second mistake is that the axiom of choice is not just a statement about sets but one of collections of sets, (of course on the most basic level there is no difference between these two, since everything is a set, but it seems from your example that you are interpreting it as applicable to a set where we don't even care what the formal structure of the elements are.) The best explanation to gain a preliminary and intuitive understanding imo is Russell's shoe-sock example, see here

"one can define a function to select from an infinite number of pairs of shoes, for example by choosing the left shoe from each pair. Without the axiom of choice, one cannot assert that such a function exists for pairs of socks, because left and right socks are (presumably) indistinguishable."

There are plenty of posts on this site to take you further.

Answer 2

The axiom of choice (AC) says that for any collection $(X_i: i\in I)$ where each $X_i\ne \emptyset$, the product $\prod_{i\in I} X_i \ne \emptyset.$

The axiom of choice for sets of reals (AC($\mathbb R$)) says the same thing, but only for $X_i\subseteq \mathbb R.$

Neither of these is provable in ZF.

What is provable in ZF is that if there is an $x$ such that for all $i,$ $x\in X_i,$ then $\prod_{i\in I} X_i \ne \emptyset.$ To prove this, all we need to do is observe that the constant function $f(i) = x$ is in the product.

In particular, if all the factors in the product are the same nonempty set, as in your example, then there is clearly a common element and the product is nonempty, without any need for AC.

The reason why AC is different is because in the general case, there is no "rule" for determining which $x_i\in X_i$ to pick, like there is in the case where there's a common element ("just pick $x$"). If you can find a rule, then you don't need AC.

For instance, if you just want to prove the axiom of choice for finite sets of reals, you can do this in ZF, since any finite set of reals has a least element in the usual order of the reals, so you can just pick that. Similarly, for any set that is linearly orderable, the axiom of choice for collections of finite subsets of that set holds in ZF (but note in ZF it's not necessarily the case that every set is linearly orderable).

Russell's "shoes and socks" analogy is instructive.