# Problem about $F_\sigma$ and $G_\delta$ sets

Define a set to be bivalent iff it's both $$F_\sigma$$ and $$G_\delta$$.

Let $$X$$ be a $$G_\delta$$-space (i.e. all closed sets are $$G_\delta$$ sets).

Let $$G$$ and $$H$$ be disjoint $$G_\delta$$ sets.

Prove that there exists a bivalent set $$B$$ disjoint from $$H$$, such that $$G\subset B$$.

Note: if the exercise just asked for $$B$$ to be an $$F_\sigma$$, the proof would be immediate (just take the complement of $$H$$, which is $$F_\sigma$$.

Separation properties as the one stated above are studied in extenso in Descriptive Set Theory (DST). Let me fix some notation. If $$\Gamma$$ is class of subsets of topological spaces, $$\check\Gamma$$ stands for the complements of sets in $$\Gamma$$ and $$\Delta := \Gamma \cap \check\Gamma$$ stands for the ambiguous class.
One observation in DST states that under certain hypothesis, if every union $$A\cup B$$ of sets in $$\Gamma$$ can be reduced (written as a disjoint union of respectively smaller sets, also in $$\Gamma$$), then $$\check\Gamma$$ has the separation property: disjoint sets in $$\check\Gamma$$ can be separated by sets in $$\Delta$$. These results are stated in DST for metrizable spaces, but for $$\Gamma = F_\sigma$$ over $$G_\delta$$ spaces this works. Check Kechris' book for more on this.
Therefore, I'll show that a union of $$F_\sigma$$ sets can be reduced, and this easily implies your exercise. Fix two $$F_\sigma$$ sets written as increasing unions $$A = \textstyle\bigcup_h A_h \qquad B = \bigcup_n B_n$$ where $$A_h,B_n$$ are closed. The key observation is that
the difference of two closed sets is an $$F_\sigma$$ set.
Then we can write $$A\cup B$$ as $$\underline{B_0} \cup (A_0\setminus B_0) \cup \underline{(B_1\setminus A_0)} \cup (A_1 \setminus B_1) \cup \underline{(B_2\setminus A_1)} \cup\cdots$$ The union $$B^*$$ of the underlined sets (shaded in the picture above) is disjoint from the union $$A^*$$ of the rest, and both are $$F_\sigma$$ sets satisfying $$A^*\subseteq A\qquad B^* \subseteq B\qquad A^* \cup B^* = A\cup B.$$ By taking $$A:= X\setminus G$$ and $$B:=X\setminus H$$ above we are done.