Conditions that topologies must have if (only if) the condition "$G_\delta$ iff (open or closed)" holds? Consider the class of topological spaces $\langle X,\mathcal T\rangle$ such that the following are equivalent for $A\subseteq X$:


*

*$A$ is a $G_\delta$ set with respect to $\mathcal T$

*$A\in\mathcal T$ or $X\smallsetminus A\in\mathcal T$


Open sets, of course, are always $G_\delta$. So, equivalently, we are considering the topological spaces such that


*

*closed sets are $G_\delta$ and

*non-open $G_\delta$ sets are closed.


Clearly, not all spaces satisfy this equivalence. For example, with respect to the typical topology, $\Bbb R$ has (many!) $G_\delta$ subsets that are neither open nor closed. On the other hand, with respect to the order topology, the set $\omega_1\cup\{\omega_1\}$ has $\{\omega_1\}$ as a closed subset that is not $G_\delta$ (unless, of course, $\omega_1$ is a countable union of countable sets, as may happen in models of $\mathsf{ZF}$).
On the other hand, there are certainly spaces that do satisfy the equivalence, with the discrete and trivial topologies on any set giving us two (not-very-enlightening) examples.

I wonder, then, has the described class of topological spaces been studied in much depth? If so, I am curious about the following:


*

*Are there any non-trivial topologies that make such a space?

*Is there any common nomenclature for such spaces?

*Are there sets of conditions on a topology that imply (or are implied by) a space being part of the class of such spaces?

*Furthermore, the members of this class may clearly vary from model to model of $\mathsf{ZF},$ so is there any such set of conditions such that one implication or the other (or both) is equivalent to a Choice principle?



Edit: The examples so far (aside from indiscrete spaces on sets with at least two points) have the property that every subset of the underlying set is open or closed. Is it possible that all such space are indiscrete or have all subsets open or closed?
Another thing that is readily apparent (now that I'm a little more awake) is that $\langle X,\mathcal T\rangle$ has the desired property if and only if $$\sigma(\mathcal T)=\mathcal T\cup\{X\setminus U:U\in\mathcal T\},$$ where $\sigma(\mathcal T)$ is the (Borel) $\sigma$-algebra (on $X$) generated by $\mathcal T.$
 A: As a start: any space having exactly one limit point has your property.  In this case every set is either open or closed.  For call the limit point $x$.  Then every set $A$ not containing $x$ is open (indeed, $A$ is discrete) and any set containing $x$ is closed (its complement does not contain $x$ and thus is open).
An example of such a space is $X_0 = \{1/n : n \ge 1\} \cup \{0\}$, with its natural Euclidean topology.  Here 0 is the unique limit point.  It's homeomorphic to the ordinal $\omega + 1$.
Another observation: Suppose $X$ is a Hausdorff space with this property, and has a limit point $x$ which is $G_\delta$.  Then $X$ has no other limit points.  Proof: suppose $y$ is another limit point.  Choose disjoint open neighborhoods $U,V$ of $x,y$ respectively.  Let $A = \{x\} \cup V \setminus \{y\}$. Then:


*

*$A$ is $G_\delta$.  For by assumption there are open sets $U_n$ with $\bigcap_n U_n = \{x\}$.  Moreover, since we are in a Hausdorff space, $\{y\}$ is closed.  Thus $(U_n \cap U) \cup V \setminus \{y\}$ is open, and $\bigcap_n ((U_n \cap U) \cup V \setminus \{y\}) = A$.

*$A$ is not open.  If it were, then $A \cap U = \{x\}$ would also be open.  But $x$ is a limit point.

*$A$ is not closed.  If it were, then $A^c \cap V = \{y\}$ would be open.  But $y$ is a limit point.
As a corollary, if $X$ is a first countable Hausdorff space (so that all points are $G_\delta$), then $X$ has your property iff $X$ has at most one limit point.  In particular, this applies to any metric space.
A: Another example of such a space: let $\mathcal{F}$ be any ultrafilter on a set $X$. Define a topology $\mathcal{T} = \mathcal{F} \cup \{\emptyset\}$. One easily checks this is a topology, and it clearly has the required property (as for any subset $A \subseteq X$ we have $A \in \mathcal{F}$ or $X \setminus A \in \mathcal{F}$. If $\mathcal{F}$ is free, the space is $T_1$, but never Hausdorff, but connected.  
