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 $F_\sigma$.


1 Answer 1


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 $$ \underline{B_0} \cup (A_0\setminus B_0) \cup \underline{(B_1\setminus A_0)} \cup (A_1 \setminus B_1) \cup \underline{(B_2\setminus A_1)} \cup\cdots $$ Reduction of a union of F_sigma sets 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.


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