Not $\sigma$-compact set without axiom of choice

Solution 1:

The axiom of choice is not needed to construct a subset of $\Bbb R$ which is not $\sigma$-compact (which I guess is ultimately what this question is about).

The irrational numbers form such a set, since any compact subset of $\Bbb{R\setminus Q}$ has an empty interior, and therefore any $\sigma$-compact set of irrational numbers is the countable union of closed nowhere dense sets, which means that it is meager. But Baire's theorem shows that the irrational numbers are not meager.

(To the initiated, Baire's theorem is provable in $\sf ZF$ when the space is separable, like $\Bbb{R\setminus Q}$ which has $\pi+\Bbb Q$ as a dense subset.)

Solution 2:

Note that - in $\mathbb{R}$ at least - a set is $\sigma$-compact only if (indeed, iff) it is the union of countably many closed sets. Such a set is called $F_\sigma$ (or $\Pi^0_2$). A fundamental theorem concerning the topology of $\mathbb{R}$ is the Baire category theorem, which states that the intersection of countably many dense open sets is dense. An easy corollary is that no countable dense set can be the intersection of countably many open sets (exercise!), and as a consequence the complement of a countable dense set cannot be $F_\sigma$.

So in particular, the set of irrationals is not $\sigma$-compact.