# Subsets of reals which are both $F_{\sigma\delta}$ and $G_{\delta\sigma}$

Let $$X\subseteq \mathbb R$$ such that

• $$X$$ is an $$F_{\sigma\delta}$$-set (in $$\mathbb R$$); and
• $$X$$ is a $$G_{\delta\sigma}$$-set.

It is not necessarily true that $$X$$ must be $$F_\sigma$$ or $$G_\delta$$. A counterexample is $$[\mathbb Q \cap (-\infty,0)]\cup [\mathbb P\cap (0,\infty)]$$, where $$\mathbb Q$$ and $$\mathbb P$$ are the rationals and irrationals, respectively.

Question. Is there necessarily an open subset of $$X$$ which is $$F_\sigma$$ or $$G_\delta$$ in $$\mathbb R$$?

Has there been a study of zero-dimensional spaces which are both $$F_{\sigma\delta}$$ and $$G_{\delta\sigma}$$?

• You mean, other than the empty set, of course. – Asaf Karagila Aug 24 '19 at 22:02
• Yes, in the question I want a non-empty open subset of $X$, – D.S. Lipham Aug 24 '19 at 22:15

No, let $$X$$ be the set of those irrationals in $$x\in (0,1)$$ with binary expansion $$x=0.x_1x_2\dots$$ such that if we define $$x^{\text{even}}, x^{\text{odd}}$$ by $$x^{\text{even}}=0.x_2x_4x_6\dots$$ $$x^{\text{odd}}=0.x_1x_3x_5\dots$$ then exactly one of $$x^{\text{even}}$$, $$x^{\text{odd}}$$ is irrational.
• Since yes-or-no questions sometimes are edited in ways that make yes-or-no answers hard to understand: this answer means "no, there need not be a non-empty open subset of $X$ that is $F_\sigma$ or $G_\delta$ in $\mathbb R$." – LSpice Aug 25 '19 at 15:24