Does there exist a nowhere dense set in $\mathbb R$ which is not $F_\sigma$

Question

It doesn't seem apparent to me why a nowhere dense set should necessarily be the countable union of closed sets. However, I seem to be having trouble constructing a counterexample.

Answer 1

A counting argument shows the existence of such sets: let $C$ be the middle third Cantor set inside $[0,1]$. This is closed (compact even) and has empty interior, so $C$ is a nowhere dense subset of $\mathbb{R}$. This means that every subset of $C$ is also a nowhere dense subset of $\mathbb{R}$. $C$ has $2^{\aleph_0}$ points (as many as the reals) and so $2^{2^{\aleph_0}}$ many subsets, all of them nowhere dense in $\mathbb{R}$.

It is well known that the number of $F_\sigma$ sets of $\mathbb{R}$, or even the number of Borel sets of $\mathbb{R}$ is $2^{\aleph_0}$ which is much smaller than $2^{2^{\aleph_0}}$, showing that the overwhelming majority of the nowhere dense subsets of $\mathbb{R}$ is not $F_\sigma$.

To construct a concrete example of this: we again go to the Cantor set $C$. Take any countable dense subset $D$ of $C$. Then $C\setminus D$ cannot be $F_\sigma$ because otherwise $C\setminus D= \cup_n F_n$, $F_n$ closed in $C$ (and thus in $\mathbb{R}$). $F_n$ missing $D$ means that all $F_n$ are nowhere dense in $C$ (!) as well. But then $\{\{x\}: x \in D\} \cup \{F_n: n \in \mathbb{N}\}$ are all nowhere dense in $C$ and their union being $C$ contradicts the Baire theorem (for the subspace $C$ which is compact), so this shows that $C \setminus D$ is a nowhere dense in $\mathbb{R}$ (being a subset of $C$) that is not an $F_\sigma$ in $\mathbb{R}$.