For the first question, let us consider the following statement:
$x\in\mathbb R$ and $x\ge 0$. It is consistent with this statement that:
- $x=0$,
- $x=1$,
- $x>4301$,
- $x\in (2345235,45237911+\frac{1}{2345235})$
This list can go on indefinitely. Of course if $x=0$ then none of the other options are possible. However if we say that $x>4301$ then the fourth option is still possible.
The same is here. If all sets are measurable then it contradicts the axiom of choice; however the fact that some set is unmeasurable does not imply the axiom of choice since it is possible to contradict the axiom of choice in other ways. It is perfectly possible that the universe of set theory behave "as if it has the axiom of choice" up to some rank which is so much beyond the real numbers that everything you can think of about real numbers is as though the axiom of choice holds; however in the large universe itself there are sets which you cannot well order. Things do not end after the continuum.
That been said, of course the two statements "$\mathbb R$ is countable union of countable sets and "There are non-measurable sets" are incompatible: if $\Bbb R$ is a countable union of countable sets, then there is no meaningful way in which we can have a measure which is both $\sigma$-additive and gives intervals a measure equals to their length; whereas stating that there exists a set which is non-measurable we implicitly state that there is a meaningful way that we can actually measure sets of reals. However this is the meaning of it is consistent relatively to ZF. It means that each of those can exist with the rest of the axioms of ZF without adding contradictions (as we do not know that ZF itself is contradiction-free to begin with.)
As for the second question, of course each set is countable and thus has a bijection with $\mathbb N$. From this the union of finitely many countable sets is also countable.
However in order to say that the union of countably many countable sets is countable one must fix a bijection of each set with $\mathbb N$. This is exactly where the axiom of choice comes into play.
There are models in which a countable union of pairs is not only not countable, but in fact has no countable subset whatsoever!
Assuming the axiom of countable choice we can do the following:
Let $\{A_i\mid i\in\mathbb N\}$ be a countable family of disjoint countable sets. For each $i$ let $F_i$ be the set of injections of $A_i$ into $\mathbb N$. Since we can choose from a countable family, let $f_i\in F_i$.
Now define $f\colon\bigcup A_i\to\mathbb N\times\mathbb N$ defined by: $f(a)= f_i(a)$, this is well defined as there is a unique $i$ such that $a\in A_i$. From Cantor's pairing function we know that $\mathbb N\times\mathbb N$ is countable, and so we are done.
The natural numbers are well-ordered without the axiom of choice. In fact they still serve as the definition for countability without the axiom of choice assumed. Therefore the basic things we know about the natural numbers hold regardless to the axiom of choice. In particular $\mathbb{N\times N}$ is still countable, and $\{A\subseteq\mathbb N\mid A\text{ is finite}\}$ is also countable.
However when the axiom of choice is negated some other weird things could happen in the power set of the natural numbers, and in other infinite sets:
- There could be a set which is infinite, but has no countably infinite subset.
- It could be that there are no free ultrafilters on the natural numbers.
- It could be that the power set of the natural numbers cannot be well-ordered (it can still be linearly ordered, though).
- Countable unions of general countable sets need not be countable.
- It is possible that there is no linear basis for $\mathbb R$ over $\mathbb Q$.
Let us focus on the first one for a moment, such sets are known as infinite Dedekind-finite sets. Their existence contradicts the axiom of countable choice, so if we assume that we can prove that every infinite set has a countably infinite subset. The fourth one has the same properties, it negates the axiom of countable choice.
Both the second, third and fifth points, however, are compatible with countable choice.
As for why we accept the axiom of choice, historically we did not accept it. People found its consequence strange (regardless to the fact they have used it intuitively). After it was proved that assuming the axiom of choice does not add inconsistencies to ZF, people began using the axiom of choice more and see its wonderful applications. It simply made things easier.
For more reading:
- Advantage of accepting the axiom of choice
- Motivating implications of the axiom of choice?
- Advantage of accepting non-measurable sets
- Why is the axiom of choice separated from the other axioms?
Best Answer
It is a bit hard to answer your first question, since a countable union of countable sets does not even need to be well-orderable. For example, we could have an uncountable set that is the countable union of sets of size $2$. These sets are called Russell cardinals. You may find the following paper interesting:
The paper discusses in a fairly self-contained manner the variety of Russell cardinals, and some of their basic properties. As described in the background section of the paper, the name "Russell cardinal" comes from an example due to Russell trying to explain the need for the axiom of choice: If we are given infinitely many pairs of shoes, and asked to pick one of each, we can easily do it since, say, we can pick the left shoes. But if we are given an infinite set of pairs of socks, then it is not clear how to make the choices. Russell's original example, amusingly enough, used boots rather than socks, which kind of makes not much sense, and it is later authors that switched to socks when describing the example.
Anyway, there are many Russell cardinals (at least as many as real numbers!), if there is one at all, and that is just countable unions of pairs.
If we look at countable unions of countable sets, we can have even more possibilities:
This is not to say that there are no limitations. For example, a countable union of countable well-ordered sets has size at most $\omega_1$. It is consistent (by a significant result of Gitik) that any well-ordered cardinal has cofinality $\omega$. This means that $\omega_2$, for example, could be written as a countable union of smaller sets, so as a countable union of countable unions of countable sets. However, it is not itself a countable union of countable sets.