On Visibility Problems with an Infinite Discrete, set of Obstacles

TL;DR

This paper investigates visibility in Euclidean spaces with infinite discrete obstacles, establishing conditions under which points are visible or not, and generalizing previous conjectures about dark forests.

Contribution

It proves that for dense sets in the plane, not all points are visible, and constructs examples where the entire space is visible, extending prior work on visibility problems.

Findings

Dense sets in $ extbf{R}^2$ are not fully visible.

Existence of dense subsets of $ extbf{Z}^d$ with full visibility.

Results on how the size of obstacle sets influences visibility.

Abstract

This paper studies visibility problems in Euclidean spaces $\mathbb{R}^d$ where the obstacles are the points of infinite discrete sets $Y\subseteq\mathbb{R}^d$. A point $x\in\mathbb{R}^d$ is called $\varepsilon$-visible for $Y$ (notation: $x\in\mathbf{vis}(Y, \varepsilon))$ if there exists a ray $L\subseteq\mathbb{R}^d$ emanating from $x$ such that $||y-z||\geq\varepsilon$, for all $y\in Y\setminus\{x\}$ and $z\in L$. A point $x\in\mathbb{R}^d$ is called visible for $Y$ (notation: $x\in\mathbf{vis}(Y))$ if $x\in\mathbf{vis}(Y, \varepsilon))$, for some $\varepsilon>0$.\\ Our main result is the following. For every $\varepsilon>0$ and every relatively dense set $Y\subseteq\mathbb{R}^2$, $\mathbf{vis}(Y, \varepsilon))\neq\mathbb{R}^2$. This result generalizes a theorem of Dumitrescu and Jiang, which settled Mitchell's dark forest conjecture. On the other hand, we show that there exists a relatively dense subset $Y\subseteq \mathbb{Z}^d$ such that $\mathbf{vis}(Y)=\mathbb{R}^d$. (One easily verifies that $\mathbf{vis}(\mathbb{Z}^d)=\mathbb{R}^d\setminus\mathbb{Z}^d$, for all $d\geq 2$). We derive a number of other results clarifying how the size of a sets $Y\subseteq\mathbb{R}^d$ may affect the sets $\mathbf{vis}(Y)$ and $\mathbf{vis}(Y,\varepsilon)$. We present a Ramsey type result concerning uniformly separated subsets of $\mathbb{R}^2$ whose growth is faster than linear.