(ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior

Solution 1:

The Baire Category Theorem implies that complete metric spaces are not first category (meager). That is, they are not a countable union of nowhere dense sets. A set $A$ is nowhere dense if $\text{Int}(\bar{A}) = \emptyset$.

Clearly $\mathbb{R}^n$ is a complete separable metric space. If $\mathbb{R}^n = \bigcup_k A_k$ where each $A_k$ is closed and has empty interior, then this contradicts the Baire Category Theorem.

So you just need that the Baire Category Theorem can be proved without the axiom of choice. This is true for Polish Space, i.e. complete separable metric spaces. Theorem 4 from this pdf from Berkeley has short proof of this result : http://math.berkeley.edu/~jdahl/104/104Baire.pdf

Solution 2:

This is essentially the Baire category theorem. Indeed in ZF it holds for separable complete metric spaces.

The argument is as follows:

Suppose that $(X,d)$ is a separable complete metric space, and $\{a_k\mid k\in\omega\}=D\subseteq X$ is a countable dense subset.

By contradiction assume that we can write $X=\bigcup F_n$ where $F_n$ are closed and with empty interior, we can further assume that $F_n\subseteq F_{n+1}$.

Define by recursion the following sequence:

1. $x_0 = a_k$ such that $k=\min\{n\mid a_n\notin F_0\}$;
2. $r_0 = \frac1{2^k}$ such that $d(F_0,x_0)>\frac1k$, since $x_0\notin F_0$ such $k$ exists.
3. $x_{n+1} = a_k$ such that $k=\min\{n\mid a_n\in B(x_n,r_n)\setminus F_n\}$;
4. $r_{n+1} = \frac1{2^k}$ such that $d(F_n,x_{n+1})>\frac1k$, the argument holds as before.

Note that $x_n$ is a Cauchy sequence, therefore it converges to a point $x$. If $x\in F_n$ for some $n$, first note that $d(x_k,F_n)\leq d(x_k,x)$, by the definition of a distance from a closed set.

If so, for some $k$ we have that $d(x,x_k)<r_n$, in particular $d(F_n,x_k)<r_n$. First we conclude that $n<k$, otherwise $d(F_n,x_n)>r_n$. Now we note that:

$$d(F_n,x_n)\leq d(x,x_n)\leq d(x,x_k)+d(x_k,x_n)\leq r_n+r_n=2r_n$$

It is not hard to see that $2r_n< d(F_n,x_n)$, which is a contradiction to the choice of $x_n$.

Added: Why is there $a_k$ in every step of the inductive definition?

Note that $D$ is dense therefore $\overline{D}=X$. In particular, if $F_n$ is closed and has a dense subset then $F=X$. Since we assume that the interior of $F_n$ is empty we have that $F_n\cap D$ is not dense in $X$, otherwise $F_n=X$ and has a non-empty interior.

We have that $D\setminus F_n$ is dense, since $D\cap F_n$ is not dense. In particular this means that in every open set there is some $a_k\in D\setminus F_n$, and thus the induction can be carried in full.