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A G-delta-sigma that is not F-sigma?

MathOverflow Asked by Julián Aguirre on August 20, 2020

A subset of $mathbb{R}^n$ is

  • $G_delta$ if it is the intersection
    of countably many open sets
  • $F_sigma$ if it is the union of countably many closed sets
  • $G_{deltasigma}$ if it is the union
    of countably many $G_delta$‘s

This process gives rise to the Borel hierarchy.

Since all closed sets are $G_delta$, all $F_sigma$ are $G_{deltasigma}$. What is an explicit example of a $G_{deltasigma}$ that is not $F_sigma$?

5 Answers

Any dense $G_{delta}$ with empty interior is of II Baire category, and cannot be $F_sigma$ by the Baire theorem (and of course it is in particular a $G_{deltasigma}$).

Correct answer by Pietro Majer on August 20, 2020

I just stumbled upon this old question, and thought I would add a simple and natural example, which is $G_{delta sigma}$ but neither $G_{delta}$ nor $F_{sigma}$.

Consider $fcolon mathbb{C}tomathbb{C}; zmapsto e^z$, and its iterates $$f^n(z) = underbrace{f(f(dots f(z)dots))}_{text{$n$ times}}=e^{e^{cdot^{cdot^{cdot^{e^z}}}}}.$$ Consider the set $$ X := {zinmathbb{C}colon f^n(z)nottoinfty} $$ and its subset $$ X_0 := {zinmathbb{C}colon {f^n(z)colon ngeq 0} text{ is dense in $mathbb{C}$}}.$$

It follows from the definitions that $X_0$ is a $G_{delta}$ and $X$ is a $G_{delta sigma}$. It is well-known that $X_0$ is dense in $mathbb{C}$, as is the complement $I(f) = mathbb{C}setminus X$. (For an elementary proof, see The exponential map is chaotic, Amer. Math. Monthly 2017, arxiv:1408.1129.)

So $X_0$ contains a dense $G_{delta}$, and can therefore not be $F_{sigma}$, as noted by Pietro Majer.

That $X$ is not $G_{delta}$ follows from the fact that $I(f)$ is not $F_{sigma}$. (Escaping sets are not sigma-compact, arxiv:2006.16946.)

(More generally, for any transcendental entire function, the set of non-escaping points is a $G_{delta sigma}$ but neither $G_{delta}$ nor $F_{sigma}$.)

Answered by Lasse Rempe-Gillen on August 20, 2020

There is a class of (natural?) examples here.

Answered by Ashutosh on August 20, 2020

Here are some examples from recursion theory which are boldface $mathbf{Pi^0_3}$ (the first is $Pi^0_3(emptyset')$ and the others are lightface $Pi^0_3$) but not boldface $mathbf{Sigma^0_3}$:

The collection of weakly-2-random reals;

The collection of Schnorr random reals;

The collection of computably random reals.

References:

  1. The Arithmetical Complexity of Dimension and Randomness. John M. Hitchcock, Jack H. Lutz, and Sebastiaan A. Terwijn. ACM Transactions on Computational Logic, 2007.

  2. Descriptive set theoretical complexity of randomness notions. Liang Yu. To appear.

Answered by 喻 良 on August 20, 2020

You want a $Sigma^0_3$ set which is not $Pi^0_3$. The canonical answer is a "universal $Sigma^0_3$ set". You can find these concepts in books on Descriptive Set Theory (such as the one by Moschovakis or the one by Kechris).

A specific example: Let $N_{10}$ be the set of normal numbers (each digit appears with the right asymptotic frequency in the decimal expansion). Then $N_{10}$ is a $Pi^0_3$ set which as complicated as $Pi^0_3$ sets get (in particular: every $Pi^0_3$ set is a continuous preimage of it). In particular, it is not $Sigma^0_3$. So the complement of $N_{10}$ is what you want.

References:

  1. Haseo Ki and Tom Linton, "Normal numbers and subsets of $mathbb N$ with given densities", Fundamenta Math (1994). MR1273694

  2. Goldstern, "Complexity of uniform distribution", Mathematica Slovaca (1994). MR1338422

Answered by Goldstern on August 20, 2020

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