Haar-positive closed subsets of Haar-positive analytic sets

Márton Elekes; Márk Poór; Zoltán Vidnyánszky

Home/Archive/arXiv

arXiv · 2003.06854

Haar-positive closed subsets of Haar-positive analytic sets

Márton Elekes·Márk Poór·Zoltán Vidnyánszky

View original source ↗Read source document ↗Export metadata ↓

Abstract

We show that every non-Haar-null analytic subset of $\mathbb{Z}^\omega$ contains a non-Haar-null closed subset. Moreover, we also prove that the codes of Haar-null analytic subsets, and, consequently, closed Haar-null sets in the Effros Borel space of $\mathbb{Z}^\omega$ form a $\mathbf{\Delta}^1_2$ set.

Explore related subjects

math.LO math.GR

Keep this discovery

BibTeX RIS

Márton Elekes, Márk Poór, Zoltán Vidnyánszky. 2020-03-15. Haar-positive closed subsets of Haar-positive analytic sets. https://arxiv.org/abs/2003.06854

Cite the original work for its findings. Save a collection to share your selection of sources.

KEEP EXPLORING

Related papers

arXiv2026

There is no maximal $K$-degree

The Kolmogorov complexity of a string characterize how complex it is to describe the string. If every prefix of a real $x$ is more complex to describe than every prefix (of the same length) of real $y$, then it is seen as $x$ is more complex to describe than $y$. It is wondered if there is a real $x$ so that no other reals are strictly more complex (to describe) than $x$. The behavior of Kolmogorov complexity functions generated by reals (namely $n\mapsto$ the minimal description length of the real) is quite chaos. Therefore, it is widely believed that there are many reals that are maximally complex to describe. For instance, it is conjectured that all random enough reals have maximal $K$-degree. In this paper, it is shown that there is no real with maximal $K$-degree. Actually, for almost all real $x$, we can uniformly computably find another real whose $K$-degree is strictly above $x$.

math.LO↗

arXiv2026

Quadruples and cubes

We prove, in $\mathsf{ZFC}$, that the $\lambda$-terraced cube relation fails whenever $\lambda$ is an uncountable cardinal. The corresponding terraced relation for quadruples fails for every $\lambda$. If $\lambda$ is $\aleph_0$ then the pretinent terraced relation has consistency strength of at least one Woodin cardinal. We prove positive polarized relations at a successor and a double successor from wondrous ideals. We show, however, that there are no such ideals over two consecutive cardinals simultaneously.

math.LO↗

arXiv2026

Possibilistic Logic over a Logic of Formal Inconsistency

In this article, we have introduced a new possibilistic logic on a logic of formal inconsistency with the aim of developing a possibility theoretic framework to deal with uncertainty and inconsistency meaningfully without leading to a system collapse. We have discussed the syntax and semantics for this logic and have proved the soundness and completeness theorems. A set of new measures of consistency, contradictoriness, and triviality of a set of formulas have been defined. These have then been put to use in an example to show that this framework can provide better means of machine reasoning.

math.LO↗