A G-delta-sigma that is not F-sigma? A subset of $\mathbb{R}^n$ is

*

*$G_\delta$ if it is the intersection
of countably many open sets

*$F_\sigma$ if it is the union of    countably many closed sets

*$G_{\delta\sigma}$ if it is the union
of countably many $G_\delta$'s

*...

This process gives rise to the Borel hierarchy.
Since all closed sets are $G_\delta$, all $F_\sigma$ are $G_{\delta\sigma}$. What is an explicit example of a $G_{\delta\sigma}$ that is not $F_\sigma$?
 A: Any dense $G_{\delta}$ with empty interior is of II Baire category, and cannot be $F_\sigma$ by the Baire theorem (and of course it is in particular a $G_{\delta\sigma}$).
A: Here are some examples from recursion theory which are boldface $\mathbf{\Pi^0_3}$ (the first is $\Pi^0_3(\emptyset')$ and the others are lightface $\Pi^0_3$) but not boldface $\mathbf{\Sigma^0_3}$:
The collection of weakly-2-random reals;
The collection of Schnorr random reals;
The collection of computably random reals.
References:


*

*The Arithmetical Complexity of Dimension and Randomness. 
John M. Hitchcock, Jack H. Lutz, and Sebastiaan A. Terwijn.
ACM Transactions on Computational Logic, 2007.

*Descriptive set theoretical complexity of randomness notions. Liang Yu. To appear.
A: You want a $\Sigma^0_3$ set which is not $\Pi^0_3$.   The canonical answer is a "universal $\Sigma^0_3$ set".  You can find these concepts in books on Descriptive Set Theory (such as the one by Moschovakis or the one by Kechris).
A specific example:  Let $N_{10}$ be the set of normal numbers (each digit appears with the right asymptotic frequency in the decimal expansion).   Then $N_{10}$ is a
$\Pi^0_3$ set which as complicated as $\Pi^0_3$ sets get (in particular: every $\Pi^0_3$
set is a continuous preimage of it).   In particular, it is not $\Sigma^0_3$.  So the complement of $N_{10}$ is what you want.
References:

*

*Haseo Ki and Tom Linton, "Normal numbers and subsets of $\mathbb N$ with given densities", Fundamenta Math (1994). MR1273694


*Goldstern, "Complexity of uniform distribution", Mathematica Slovaca (1994). MR1338422
A: There is a class of (natural?) examples here.
A: I just stumbled upon this old question, and thought I would add a simple and natural example, which is $G_{\delta \sigma}$ but neither $G_{\delta}$ nor $F_{\sigma}$.
Consider $f\colon \mathbb{C}\to\mathbb{C}; z\mapsto e^z$, and its iterates
$$f^n(z) = \underbrace{f(f(\dots f(z)\dots))}_{\text{$n$ times}}=e^{e^{\cdot^{\cdot^{\cdot^{e^z}}}}}.$$
Consider the set
$$ X := \{z\in\mathbb{C}\colon f^n(z)\not\to\infty\} $$
and its subset
$$ X_0 := \{z\in\mathbb{C}\colon \{f^n(z)\colon n\geq 0\} \text{ is dense in $\mathbb{C}$}\}.$$
It follows from the definitions that $X_0$ is a $G_{\delta}$ and $X$ is a $G_{\delta \sigma}$. It is well-known that $X_0$ is dense in $\mathbb{C}$, as is the complement $I(f) = \mathbb{C}\setminus X$. (For an elementary proof, see The exponential map is chaotic, Amer. Math. Monthly 2017, arxiv:1408.1129.)
So $X_0$ contains a dense $G_{\delta}$, and can therefore not be $F_{\sigma}$, as noted by Pietro Majer.
That $X$ is not $G_{\delta}$ follows from the fact that $I(f)$ is not $F_{\sigma}$. (Escaping sets are not sigma-compact, arxiv:2006.16946.)
(More generally, for any transcendental entire function, the set of non-escaping points is a $G_{\delta \sigma}$ but neither $G_{\delta}$ nor $F_{\sigma}$.)
