No Krasnoselskii number for general sets in $\mathbb{R}^2$

TL;DR

This paper proves that no Krasnoselskii number exists for the family of all sets in , meaning the property of visibility from a single point cannot be characterized by a finite number of points, even when restricting to polygonal paths.

Contribution

It answers Peterson's long-standing question negatively, showing no Krasnoselskii number exists for general sets in , including those with visibility through polygonal paths of bounded length.

Findings

No Krasnoselskii number exists for all sets in .

Counterexamples are finite unions of line segments.

The result holds even with visibility restricted to paths of length  or more.

Abstract

For a family $\mathcal{F}$ of sets in $\mathbb{R}^d$, the Krasnoselskii number of $\mathcal{F}$ is the smallest $m$ such that for any $S \in \mathcal{F}$, if every $m$ points of $S$ are visible from a common point in $S$, then any finite subset of $S$ is visible from a single point. More than 35 years ago, Peterson asked whether there exists a Krasnoselskii number for general sets in $\mathbb{R}^d$. Excluding results for special cases of sets with strong topological restrictions, the best known result is due to Breen, who showed that if such a Krasnoselskii number in $\mathbb{R}^2$ exists, then it is larger than $8$. In this paper we answer Peterson's question in the negative by showing that there is no Krasnoselskii number for the family of all sets in $\mathbb{R}^2$. The proof is non-constructive, and uses transfinite induction and the well ordering theorem. In addition, we consider Krasnoselskii numbers with respect to visibility through polygonal paths of length $ \leq n$, for which an analogue of Krasnoselskii's theorem was proved by Magazanik and Perles. We show, by an explicit construction, that for any $n \geq 2$, there is no Krasnoselskii number for the family of general sets in $\mathbb{R}^2$ with respect to visibility through paths of length $\leq n$. (Here the counterexamples are finite unions of line segments.)