Dimension Independent Helly Theorem for Lines and Flats

Suppose for each $i\in[d+1],\;\mathcal{L}_{i}$ is a set of $k$ linearly independent vectors in the limiting directions set $LDS(\mathcal{S}_{i})$ of $\mathcal{S}_{i}$. Then there exists a linearly independent set of $k$ vectors, say $\mathcal{L}=\{z_{1},\dots,z_{k}\}$, such that for each $i\in[k],\;z_{i}\in\mathcal{L}_{i}$. Suppose $\mathcal{K}$ is the $k$-flat generated by the linear span of $\mathcal{L}$. Now for each $i\in[k]$, since $z_{i}\in\mathcal{L}_{i}$, so $\exists$ a sequence $\{S_{i,n}\}_{n\in\mathbb{N}}$ in $\mathcal{S}_{i}$ and for each $n\in\mathbb{N}$, $\exists x_{i,n}\in S_{i,n}$ such that the sequence $\{f(x_{i,n})\}_{n\in\mathbb{N}}$ converges to $z_{i}$, as $n\rightarrow\infty$.

Now if $\exists i\in[k]$ such that $\mathcal{S}_{i}$ has a $k$-transversal then there is nothing to prove. Otherwise for each $i\in[d+1],i>k$, we take any $B_{i}\in\mathcal{S}_{i}$ and take any $n\in\mathbb{N}$. Then $B_{k+1},\dots,B_{d+1}$ together with $S_{1,n},\dots,S_{k,n}$, as a colorful tuple, is pierceable by a $k$-flat. This follows that $B_{k+1},\dots,B_{d+1}$ can be pierced by a $k$-flat arbitrarily close to the direction of $\mathcal{K}$. So by compactness of $B_{i}$’s we can say that, $B_{k+1},\dots,B_{d+1}$ can be pierced by a $k$-flat in the direction of $\mathcal{K}$, i.e, parallel to $\mathcal{K}$.

Now for each $i\in[d+1],\;i>k$, suppose $\mathcal{S}^{\prime}_{i}$ is the projected family of $\mathcal{S}_{i}$ on the $(d-k)$ dimensional space $\mathcal{K}^{\perp}$, orthogonal to $\mathcal{K}$. Then every colorful $(d-k+1)$ tuple from $\mathcal{S}^{\prime}_{k+1},\dots,\mathcal{S}^{\prime}_{d+1}$ is pierceable by a point in the space $\mathcal{K}^{\perp}$. So by Colorful Helly’s Theorem, $\exists i\in[d+1],\;i>k$ such that $\mathcal{S}^{\prime}_{i}$ is pierceable by a point in $\mathcal{K}^{\perp}$. Hence there exists a $k$-flat parallel to $\mathcal{K}$ that hits all the members of the family $\mathcal{S}_{i}$. ∎

Suppose for any set $F\subset\mathbb{R}^{d}$, $\bar{F}$ is the closure of $F$. Since for any point $p\in\mathbb{R}^{d}$, $d(p,F)=d(p,\bar{F})$, it is enough to show that $\exists j\in[r]$ such that there exists a $k$-flat $K$ satisfying $\forall C\in\mathcal{F}_{j}$,

$$d(\bar{C},K)\leq\sqrt{\frac{1}{r-k}}.$$

Now without loss of generality, we assume that $J_{k}=[k]$. Then for each $j\in J$, $y_{j}\in\mathrm{LDS}(\mathcal{F}_{j})$ implies that $\exists\text{ a sequence }\{F_{j,n}\}_{n\in\mathbb{N}}$ in $\mathcal{F}_{j}$ and for each $n\in\mathbb{N},\;\exists x_{j,n}\in\mathcal{F}_{j,n}$ such that the sequence $\{f(x_{j,n})\}_{n\in\mathbb{N}}$ converges to $y_{j}$.

Now if $\exists j\in[r]$ such that there exists a $k$-flat $K$ satisfying $\forall C\in\mathcal{F}_{j}$, $d(C,K)\leq\sqrt{\frac{1}{r-k}}$, then there is nothing to prove. Otherwise, suppose $K$ is the $k$-flat generated by the linear span of $\{y_{1},\dots,y_{k}\}$. Now for each $i\in[r],\;i>k,$ take any $F_{i}\in\mathcal{F}_{i}$. Then for each $n\in\mathbb{N}$, there exists a $k$-flat $K_{n}$ intersecting $B(0,1)$ and piercing $F_{1,n},F_{2,n},\dots,F_{k,n},F_{k+1},\dots,F_{r}$. This implies that $F_{k+1},\dots,F_{r}$ can be pierced by a $k$-flat arbitrarily close to the direction of $K$. Now by compactness of $\bar{F_{i}}$’s, we conclude that $\exists a\in K^{\perp}$ such that the $k$-flat $a+K$ pierces $\bar{F}_{k+1},\dots,\bar{F}_{r}$. Since $a+K$ intersects $B(O,1)$, we must have $\|a\|\leq 1$.

Now let for any set $A$ in $\mathbb{R}^{d}$, $\pi(A)$ denote the orthogonal projection of $A$ onto the $(d-k)$-dimensional space $K^{\perp}$. Then for any $(C_{k+1},\dots C_{r})\in\mathcal{F}_{k+1}\times\dots\times\mathcal{F}_{r}$, we have

$$\left(\bigcap_{i=k+1}^{r}\pi\left(\bar{C_{i}}\right)\right)\bigcap B(0,1)\neq\emptyset.$$

Then, by Theorem 2, $\exists q\in K^{\perp}$ and $\exists i\in\{k+1,\dots,r\}$ such that $\forall C\in\mathcal{F}_{i}$ we have

$$d\left(q,\pi\left(\overline{C}\right)\right)<\sqrt{\frac{1}{r-k}}\;.$$

Suppose $q^{\prime}\in\pi^{-1}(q)$ and consider the $k$-flat $K^{\prime}=q^{\prime}+K$. Then we have $\forall C\in\mathcal{F}_{i}$,

$$d\left(K^{\prime},\overline{C}\right)<\sqrt{\frac{1}{r-k}}\;.$$

∎

To establish the necessity of $k$-unboundedness we will be using a construction that is closely related to the one given by Aronov et al. [AGP02].

Consider the eight shaded convex regions in Figure 1 created by four circles and four squares centered at a point $\mathcal{O}$. Let $x_{3}=0$ be the plane that contains the Figure 1, and without loss of generality assume that $\mathcal{O}$ be the origin in $\mathbb{R}^{3}$. We will call these eight shaded convex regions $a_{1}$, $a_{2}$, $\dots$, $a_{8}$, respectively. Observe that any $3$ of these convex sets can be intersected by a straight line passing $\mathcal{O}$.

We will now create additional convex sets in the following way: we choose the eight convex sets in a fixed order, and in each step elevate the sets in increasing heights in that order along the $x_{3}$-axis such that for any $n\in\mathbb{N}$ there are infinitely many sets of this collection that lie outside $B\left(\mathcal{O},n\right)$. This gives us a countably infinite sequence $\mathcal{F}$ of sets, where $\mathcal{F}$ is $1$-unbounded, but not $2$-unbounded. Clearly, for any three sets in $\mathcal{F}$, there exists a plane that passes through $\mathcal{O}$ and intersects these sets.

Let $\ell$ denote the length of the side of the smallest square in Figure 1. We show that it is not possible to find a plane that is at most a distance 1 unit away from all the sets in $\mathcal{F}$ when $\ell$ is large enough. Let $K$ be a plane for which the maximum distance from the sets in $\mathcal{F}$ is minimized, and $l_{K}$ be the intersection of $K$ with the plane $x_{3}=0$. Clearly, $K$ must be perpendicular to the plane that contains the first $8$ sets, because otherwise for any $R>0$ we would find a set $C$ in $\mathcal{F}$ for which $d(K,C)>R$. Consider the straight line $l_{K}$ that is the intersection of $K$ and the plane given by the equation $x_{3}=0$. If $l_{K}$ is moved on the $x_{3}=0$ plane closer to $\mathcal{O}$ along the line perpendicular to $l_{K}$ from $\mathcal{O}$, the quantity $\max\left\{d\left(a_{2n},l_{K}\right),d\left(a_{2n-1},l_{K}\right)\right\}$ does not increase for $n\in\left\{1,2,3,4\right\}$. Since $d(K,a_{i})=d(l_{K},a_{i})$ $\forall i\in[8]$, we can take $K$ to be passing through $\mathcal{O}$. Let the side-lengths of the $4$ squares in Figure 1 be $\ell_{1}\,(=\ell),\ell_{3},\ell_{5},\ell_{7}$, where the side of side-length $\ell_{i}$ is shared by the set $a_{i}$. Let the diagonals of the largest square in Figure 1 lie on the $x_{1}$-axis and the $x_{2}$-axis respectively. Then, if $l_{K}$ makes an angle $\theta\in[0,\pi)$ with the $x_{1}$-axis, then we have the following: for $i\in\left\{1,3,5,7\right\}$ we have

$\displaystyle d\left(a_{i},l_{K}\right)=d\left(a_{i+1},l_{k}\right),$

(3)

and

	$\displaystyle d(a_{1},l_{K})$	$\displaystyle=\begin{cases}\frac{\ell_{1}}{\sqrt{2}}\sin\left({\pi}/{4}-\theta\right)&\text{if }0\leq\theta\leq{\pi}/{4}\\ \frac{\ell_{1}}{\sqrt{2}}\sin\left(\pi-\theta\right)&\text{if }{3\pi}/{4}\leq\theta\leq\pi\\ 0&\text{otherwise}\end{cases}$		(4)
	$\displaystyle d(a_{3},l_{K})$	$\displaystyle=\begin{cases}\frac{\ell_{3}}{\sqrt{2}}\sin\theta&\text{if }0\leq\theta\leq{\pi}/{4}\\ \frac{\ell_{3}}{\sqrt{2}}\sin\left({\pi}/{2}-\theta\right)&\text{if }{\pi}/{4}\leq\theta\leq{\pi}/{2}\\ 0&\text{otherwise}\end{cases}$		(5)
	$\displaystyle d(a_{5},l_{K})$	$\displaystyle=\begin{cases}\frac{\ell_{5}}{\sqrt{2}}\sin\left(\theta-{\pi}/{4}\right)&\text{if }{\pi}/{4}\leq\theta\leq{\pi}/{2}\\ \frac{\ell_{5}}{\sqrt{2}}\sin\left({3\pi}/{4}-\theta\right)&\text{if }{\pi}/{2}\leq\theta\leq{3\pi}/{4}\\ 0&\text{otherwise}\end{cases}$		(6)
	$\displaystyle d(a_{7},l_{K})$	$\displaystyle=\begin{cases}\frac{\ell_{7}}{\sqrt{2}}\sin\left(\theta-{\pi}/{2}\right)&\text{if }{\pi}/{2}\leq\theta\leq{3\pi}/{4}\\ \frac{\ell_{7}}{\sqrt{2}}\sin\left(\pi-\theta\right)&\text{if }{3\pi}/{4}\leq\theta\leq\pi\\ 0&\text{otherwise}\end{cases}$		(7)

Now, observe that

	$\displaystyle\sum_{i=1}^{8}d\left(a_{i},l_{K}\right)$	$\displaystyle\geq\frac{\ell}{\sqrt{2}}\times\min_{\theta\in\left[0,{\pi}/{4}\right]}\left(\sin\theta+\sin\left({\pi}/{4}-\theta\right)\right)$
		$\displaystyle\geq\ell\sqrt{2}\sin\left({\pi}/{8}\right)$

Thus, for $\ell>\sqrt{2}/\sin\left({\pi}/{8}\right)$, there are no planes that are at most $1$ distance away from each set in $\mathcal{F}$. ∎

We provide an example where the bound given in Theorem 6 is tight.

Let $\{A,B,C,D,E,F,G,H\}$ be the $8$ vertices cube in $\mathbb{R}^{3}$ whose centroid is the origin $\mathcal{O}$ and side length is $\sqrt{2}$ and $EFGH$ is parallel to the plane $x_{3}=0$ (see Figure 2). Define $\mathcal{F}_{i}\coloneqq\{S_{i,1},S_{i,2},S_{i,3},\dots\}$ for $i=1,2,3$ in the following way: Set $S_{1,1}=\overline{AB}$, $S_{1,2}=\overline{CD}$ and for $n>2$, $S_{1,n}$ is the line segment $\overline{CD}$ raised to the height $x_{3}=n$. Similarly, set $S_{2,1}=\overline{GH}$, $S_{1,2}=\overline{EF}$ and for $n>2$, $S_{2,n}$ is the line segment $\overline{EF}$ lowered to the height $x_{3}=-n$. Now set $S_{3,1}=\overline{BC},S_{3,2}=\overline{DA},S_{3,3}=\overline{FG},S_{3,4}=\overline{HE}$, and let $S_{3,4n+j}$ be the set $S_{3,j}$, for all $j\in\{0,1,2,3\}$ (see Figure 2). Clearly, any colorful $3$-tuple $(C_{1},C_{2},C_{3})$, $C_{i}\in\mathcal{F}_{i}$, $i\in[3]$, can be hit by a straight line that is at most at a distance $1$ away from the centroid. Let $\mathcal{L}$ denote the set of all straight line transversals $l$ of colorful $3$-tuples such that $l$ is as close to $\mathcal{O}$ as possible. Let $l_{n}$ denote the straight line transversal in $\mathcal{L}$ that passes through $S_{1,n},S_{2,1},S_{3,1}$. Clearly, $d(l_{n},\mathcal{O})\to 1$ as $n\to\infty$. Then

$$\sup_{l\in\mathcal{L}}d(\mathcal{O},l)=1.$$

Now note that for the straight line $L$ that is perpendicular to the plane on which $ABCD$ lies and passes through $\mathcal{O}$, we have

$$\inf_{i\in[3]}\sup_{C\in\mathcal{F}_{i}}d(l,C)\geq\inf_{i\in[3]}\sup_{C\in\mathcal{F}_{i}}d(L,C),$$

for any straight line $l$ in $\mathbb{R}^{3}$. This we can show in the following way: let $l_{1}$ be a straight line such that

$$\sup_{C\in\mathcal{F}_{1}}d(l_{1},C)=\inf_{l\in\mathcal{L}}\sup_{C\in\mathcal{F}_{1}}d(C,l).$$

Then $l_{1}$ must be perpendicular to the plane on which $ABCD$ lies, otherwise, the supremum of its distances from sets in $\mathcal{F}_{1}$ would be infinity. Then, $l_{1}$ must be equidistant from both $S_{1,1}$ and $S_{1,2}$, and therefore, we can take $l_{1}$ to be $L$. Similar arguments show that

$$\sup_{C\in\mathcal{F}_{2}}d(L,C)=\inf_{l\in\mathcal{L}}\sup_{C\in\mathcal{F}_{1}}d(C,l).$$

We have

$$\inf_{i\in[3]}\sup_{C\in\mathcal{F}_{i}}d(L,C)=\frac{1}{\sqrt{2}}.$$

To see that

$$\inf_{l\in\mathcal{L}}\sup_{C\in\mathcal{F}_{3}}d(C,l)=\frac{1}{\sqrt{2}},$$

project $\overline{BC}$, $\overline{AD}$, $\overline{FG}$, $\overline{EH}$ onto the plane $P$ that contains $ABEF$. If there is a straight line $l_{3}$ such that

$$\sup_{C\in\mathcal{F}_{3}}d(l_{3},C)=\inf_{l\in\mathcal{L}}\sup_{C\in\mathcal{F}_{3}}d(C,l),$$

then let the projection of $l_{3}$ onto $P$ be $l^{\prime}_{3}$. If $L$ is not the straight line that minimizes

$$\inf_{l\in\mathcal{L}}\sup_{C\in\mathcal{F}_{6}}d(C,l),$$

then the perpendicular distance from $l^{\prime}_{3}$ to $A,B,E$ and $F$ must be smaller than $\frac{1}{\sqrt{2}}$. Let, without loss of generality, $A$ be the point from which $l^{\prime}_{3}$ is the farthest. Then, we must have another point among $B,E,$ and $F$ from which $l^{\prime}_{3}$ has the same distance as $A$. This point then must be $E$, because otherwise, we could have taken $L$ to be $l_{3}$. This means that $l^{\prime}_{3}$ passes through the centroid of the square $ABEF$ and two points from $A,B,E,$ and $F$ lie on each side of $l^{\prime}_{3}$. But this implies that $l^{\prime}_{3}$ must be parallel to $AB$ since $l^{\prime}_{3}$ has the minimum distance from both $A$ and $B$ and is at least as close to $E$ and $F$, which is a contradiction. We have

$$\inf_{i\in[3]}\sup_{C\in\mathcal{F}_{i}}d(L,C)=\frac{1}{\sqrt{2}},$$

which is what we get by plugging in the values of $r$ and $k$ in the inequality given in Theorem 6. ∎