3
$\begingroup$

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 $G_{\delta\sigma}$?

$\endgroup$
2
  • $\begingroup$ You mean, other than the empty set, of course. $\endgroup$
    – Asaf Karagila
    Commented Aug 24, 2019 at 22:02
  • $\begingroup$ Yes, in the question I want a non-empty open subset of $X$, $\endgroup$ Commented Aug 24, 2019 at 22:15

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $\mathbb R$." $\endgroup$
    – LSpice
    Commented Aug 25, 2019 at 15:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .