# How many sprays cover the space?

Alessandro Andretta  Università degli Studi di Torino, Dipartimento di
Matematica “G. Peano”, Via Carlo Alberto 10, 10123 Torino, Italy
<EMAIL_ADDRESS>and  Ivan Izmestiev  TU Wien, Institute of
Discrete Mathematics and Geometry, Wiedner Hauptstraße 8-10/104, 1040 Vienna,
Austria<EMAIL_ADDRESS>
###### Abstract.

For all $d\geq 3$ we show that the cardinality of $\mathbb{R}$ is at most
$\aleph_{n}$ if and only if $\mathbb{R}^{d}$ can be covered with
$(n+1)(d-1)+1$ sprays whose centers are in general position in a hyperplane.
This extends previous results by Schmerl when $d=2$.

###### Key words and phrases:

Continuum hypothesis, sprays

###### 2020 Mathematics Subject Classification:

03E50, 51N20

##  1\. Introduction

A spray in the plane is a subset of $\mathbb{R}^{2}$ together with a
distinguished point (called center) such that all circles centered in that
point have finite intersection with the spray. These objects were introduced
in [Sch03], where it is observed that the continuum hypothesis implies that
the plane can be covered with three sprays. (The continuum hypothesis,
${\mathsf{CH}}$ from now on, asserts that every infinite subset of
$\mathbb{R}$ is either countable or else in bijection with $\mathbb{R}$; by
work of Gödel and Cohen ${\mathsf{CH}}$ can neither be disproved nor proved
from $\mathsf{ZFC}$, the Zermelo-Frænkel set theory with the axiom of choice.)
In [Veg09] it is shown that ${\mathsf{CH}}$ is equivalent to the plane being
covered with three sprays with collinear centers—in fact the statement “the
cardinality of $\mathbb{R}$ is at most $\aleph_{n}$” (in symbols:
$2^{\aleph_{0}}\leq\aleph_{n}$) is equivalent to $\mathbb{R}^{2}$ being
covered with $n+2$ sprays with collinear centers [Sch10]. Collinearity is
essential, since $\mathsf{ZFC}$ proves that the plane can be covered with
three sprays centered in arbitrary non-collinear points [Veg09, Sch10].

The notion of spray can be extended to $\mathbb{R}^{d}$ for any $d\geq 3$,
with $(d-1)$-dimensional spheres in place of $1$-dimensional spheres, i.e.
circles, and the natural question is the relation (if any) between the size of
the continuum and the number of sprays needed to cover $\mathbb{R}^{d}$. By
[EJM94, Theorem 2], $2^{\aleph_{0}}\leq\aleph_{n}$ implies that
$\mathbb{R}^{3}$ can be covered with $2n+3$ sprays such that their centers are
coplanar, and no three of them are collinear. In particular, ${\mathsf{CH}}$
implies that given five coplanar points such that no three of them are
collinear, then there are sprays centered around them that cover
$\mathbb{R}^{3}$. By [Sch12] $\mathbb{R}^{3}$ cannot be covered with three
sprays, but the problem whether the space can be covered with four sprays
remained open.

We prove that $2^{\aleph_{0}}\leq\aleph_{n}$ is equivalent to $\mathbb{R}^{3}$
being covered with $2n+3$ sprays whose centers are coplanar, and no three of
them are collinear. Moreover $2n+3$ is optimal—in particular $\mathbb{R}^{3}$
cannot be covered with four sprays with coplanar centers. In fact we prove a
similar result for $\mathbb{R}^{d}$ with $d>3$, namely:
$2^{\aleph_{0}}\leq\aleph_{n}$ is equivalent to $\mathbb{R}^{d}$ being covered
with $(n+1)(d-1)+1$-many sprays whose centers lie on a hyperplane $H$, and the
affine span of any $d$ of them is $H$ (Theorem 5.4). Again the number
$(n+1)(d-1)+1$ of sprays is optimal. Finally we show that, irrespective of the
size of the continuum, $\mathbb{R}^{d}$ can be covered with countably many
sprays whose centers lie on a hyperplane $H$, and the affine span of any $d$
of them is $H$ (Theorem 5.8).

We do not know if $\mathbb{R}^{d}$ can be covered with $d+1$-many sprays such
that the affine span of their centers is $\mathbb{R}^{d}$, even in the case
$d=3$. In other words: can $\mathbb{R}^{3}$ be covered with four sprays whose
centers are not coplanar, i.e. form a tetrahedron? We suspect that the answer
is affirmative and that it can be proved in $\mathsf{ZFC}$. If true, this
would be analogous to the fact that $\mathbb{R}^{2}$ can be covered with three
sprays with non-collinear centers.

The results in this paper lie on the interface between set theory and the
geometry of euclidean spaces. The archetypal results in this area are the
classical theorems of Sierpiński—see Theorem 2.6 below and the beginning of
Section 4 for some context and references. Most of these results deal with
linear objects like lines, or hyperplanes in $\mathbb{R}^{d}$, while sprays
are essentially non-linear objects. We extend a construction in [Sch10] from
$\mathbb{R}^{2}$ to $\mathbb{R}^{d}$, transforming the quadratic problem of
covering the space with sprays to the linear problem of covering the space
with sets having finite intersections with certain families of hyperplanes.
This latter problem has been studied before [Bag59, EJM94, Sim97]. Extending
these earlier results we are able to prove our results on sprays.

The paper is organized as follows. After recalling the notations and the basic
notions that will be used throughout the paper, we show in Section 3 how to
transform a covering of $\mathbb{R}^{d}$ with sprays with centers on a given
hyperplane into a covering of (an open subset of) $\mathbb{R}^{d}$ with small
intersection with certain families of hyperplanes. Section 4 is devoted to
studying the following problem: given distinct, non-zero vectors
$\mathbf{u}_{1},\dots,\mathbf{u}_{k}$ in $\mathbb{R}^{d}$, are there
$A_{1},\dots,A_{k}$ covering $\mathbb{R}^{d}$ such that every plane orthogonal
to $\mathbf{u}_{i}$ has finite intersection with $A_{i}$? It turns out that
this problem is closely related to the size of the continuum, and by the
results in Section 3 it is equivalent to the existence of sprays
$X_{1},\dots,X_{k}$ with centers on a given hyperplane $H$, and covering
$\mathbb{R}^{d}$, as the (directions of the) $\mathbf{u}_{i}$s correspond to
the position of the centers of the $X_{i}$s on $H$. Finally in Section 5 we
prove the results about the existence of sprays covering $\mathbb{R}^{d}$ and
the size of the continuum.

In order to make the present paper accessible to a wider audience, many basic
definitions in set theory and geometry are summarized in Section 2.1. The
reader who is proficient with these matters can safely skip it.

##  2\. Notation and preliminary results

###  2.1. Notation

####  2.1.1. Set theory

We work in $\mathsf{ZFC}$—this is the standard framework that most
mathematicians (implicitly, or explicitly) adopt to prove theorems. Our
notation is standard: $\mathscr{P}(X)$ is the power-set of $X$, the pointwise
image of $Z\subseteq X$ via some $f\colon X\to Y$ is
$f[Z]\coloneqq\mathopen{\\{}{f(z)}\boldsymbol{\mid}{z\in Z}\mathclose{\\}}$.

Ordinals are special sets that are well-ordered by $\in$, the membership
relation, and every well-ordered set is isomorphic to a unique ordinal. Lower-
case Greek letters range over the ordinals, $\mathord{\mathrm{Ord}}$ is the
collection of all ordinals, and for $\alpha,\beta\in\mathord{\mathrm{Ord}}$ we
have $\alpha<\beta$ if and only if $\alpha\in\beta$. Any natural number can be
construed as an ordinal, letting $0$ be $\emptyset$ the empty set, $1$ be the
set $\left\\{{0}\right\\}$, and more generally $n+1\coloneqq
n\cup\left\\{{n}\right\\}$. The least infinite ordinal $\omega$ is the set
$\mathbb{N}=\left\\{{0,1,\dots,n,\dots}\right\\}$. A cardinal is an ordinal
that is not in bijection with a smaller ordinal—every natural number is a
cardinal, and so is $\omega$. Infinite cardinals are enumerated by the
function $\aleph$ using ordinals as indexes; so $\aleph_{0}$ is $\omega$, the
smallest infinite cardinal, $\aleph_{1}$ is the smallest cardinal bigger than
$\aleph_{0}$, and more generally $\aleph_{n+1}$ is the smallest cardinal
bigger than $\aleph_{n}$. As $\omega$ is the smallest ordinal bigger than any
finite ordinal (i.e. natural number), $\aleph_{\omega}$ is the smallest
cardinal bigger than any $\aleph_{n}$. As there is no reason to stop there we
can take larger ordinals such as $\omega+1<\omega+2<\dots$ and construct even
larger cardinals $\aleph_{\omega+1}<\aleph_{\omega+2}<\dots$. And so on.

Using the axiom of choice, every set $X$ can be well-ordered, so it is in
bijection with some ordinal, and hence with a unique cardinal: this cardinal
is denoted by $\mathopen{|}X\mathclose{|}$ and it is called the cardinality of
$X$. A set $X$ is finite if it is in bijection with a natural number i.e.
$\mathopen{|}X\mathclose{|}<\aleph_{0}$; otherwise it is infinite i.e.
$\aleph_{0}\leq\mathopen{|}X\mathclose{|}$. We say $X$ is countable if either
$\mathopen{|}X\mathclose{|}<\aleph_{0}$, or else
$\mathopen{|}X\mathclose{|}=\aleph_{0}$—this can be written in a compact way
as $\mathopen{|}X\mathclose{|}\leq\aleph_{0}$ or as
$\mathopen{|}X\mathclose{|}<\aleph_{1}$. A set $X$ is uncountable if it is not
countable, that is $\aleph_{1}\leq\mathopen{|}X\mathclose{|}$. The countable
union of countable sets is countable. More generally, for any ordinal
$\delta$, if $(X_{i})_{i\in I}$ is a family of sets, each of cardinality
$\leq\aleph_{\delta}$ and $\mathopen{|}I\mathclose{|}\leq\aleph_{\delta}$,
then $\mathopen{|}\bigcup_{i\in I}X_{i}\mathclose{|}\leq\aleph_{\delta}$.

The set $\mathbb{R}$ is in bijection with
$\mathopen{\\{}{f}\boldsymbol{\mid}{f\colon\omega\to\left\\{{0,1}\right\\}}\mathclose{\\}}$,
and for this reason the cardinality of $\mathbb{R}$ is often denoted with
$2^{\aleph_{0}}$. By Cantor’s theorem
$\aleph_{1}\leq\mathopen{|}\mathbb{R}\mathclose{|}$, and Cantor’s continuum
hypothesis ${\mathsf{CH}}$ asserts that the inequality can be replaced with an
equality. In other words ${\mathsf{CH}}$ asserts that every subset of the real
line is either countable or else it is in bijection with $\mathbb{R}$. Since
$\aleph_{1}\leq 2^{\aleph_{0}}$ one can state ${\mathsf{CH}}$ as
$2^{\aleph_{0}}\leq\aleph_{1}$. By Cohen’s results, it is consistent that the
cardinality of $\mathbb{R}$ be any $\aleph_{n+1}$, or even larger cardinals,
like $\aleph_{\omega+n+1}$. (But $2^{\aleph_{0}}$ cannot be $\aleph_{\omega}$
by König’s theorem.)

####  2.1.2. Geometry

The standard basis of $\mathbb{R}^{d}$ is denoted by
$\mathbf{e}_{1},\dots,\mathbf{e}_{d}$. The following notation for hyperplanes
will be used throughout the paper.

######  Notation 2.1.

Given a vector
$\mathbf{u}\in\mathbb{R}^{d}\setminus\left\\{{\mathbf{0}}\right\\}$ and
$\mathbf{p}\in\mathbb{R}^{d}$ let

$H_{\mathbf{u}}(\mathbf{p})=\mathbf{p}+\mathopen{\\{}{\mathbf{v}}\boldsymbol{\mid}{\mathbf{v}\boldsymbol{\cdot}\mathbf{u}=0}\mathclose{\\}}$

be the hyperplane of $\mathbb{R}^{d}$ orthogonal to $\mathbf{u}$ passing
through $\mathbf{p}$, where $\boldsymbol{\cdot}$ is the (standard) inner
product. We also let
$H_{i}(x)=\mathopen{\\{}{(p_{1},\dots,p_{d})\in\mathbb{R}^{d}}\boldsymbol{\mid}{p_{i}=x}\mathclose{\\}}$.

An affine subspace $E$ of $\mathbb{R}^{d}$ is a translate of some vector
subspace of $\mathbb{R}^{d}$, that is a set of the form $\mathbf{p}+V$ where
$V$ is a vector subspace of $\mathbb{R}^{d}$; the vector subspace $V$ is
unique, and it is the vector space associated to $E$, while
$\mathbf{p}+V=\mathbf{p}^{\prime}+V$ if and only if
$\mathbf{p}-\mathbf{p}^{\prime}\in V$. In this paper, a (finite dimensional)
affine space is an affine subspace of some $\mathbb{R}^{d}$. Elements of an
affine space are called points, and since every vector subspace $V$ of
$\mathbb{R}^{d}$ is also an affine space, we can refer to elements of it as
points or vectors, depending if we privilege the affine or vector space
structure. If $E$ is an affine space of $\mathbb{R}^{d}$ and
$\mathbf{p},\mathbf{q}\in E$ then $\mathbf{q}-\mathbf{p}$ belongs to $V$, the
underlying vector space of $E$, and if $\mathbf{v}\in V$ then
$\mathbf{p}+\mathbf{v}$ belongs to $E$. The dimension of an affine space is,
by definition, the dimension of the associated vector space. Given an affine
space $E$ with associated vector space $V$, the affine envelope or affine span
of a non-empty set $S\subseteq E$ is
$\mathbf{p}+\operatorname{span}\mathopen{\\{}{\mathbf{q}-\mathbf{p}}\boldsymbol{\mid}{\mathbf{q}\in
S}\mathclose{\\}}$ where $\mathbf{p}\in S$ and $\operatorname{span}X$ is the
smallest vector subspace of $V$ containing $X\subseteq V$. It is easy to check
that the definition does not depend on the point $\mathbf{p}$, and that the
affine span of $S$ is the intersection of all affine subspaces of $E$
containing $S$. Two affine subspaces $E$ and $F$ of $\mathbb{R}^{d}$ are
complementary if their associated vector spaces $V$ and $W$ are complementary,
that is $\mathbb{R}^{d}=V+W$ and $V\cap W=\left\\{{\mathbf{0}}\right\\}$. The
intersection of two complementary spaces $E,F$ is a single point. In fact if
$\mathbf{p},\mathbf{q}\in E\cap F$, then $\mathbf{p}-\mathbf{q}\in V\cap
W=\left\\{{\mathbf{0}}\right\\}$, so $\mathbf{p}=\mathbf{q}$, that is $E\cap
F$ has at most one point. If $E=\mathbf{p}+V$ then $\mathbf{p}$ can be written
in a unique way as $\mathbf{v}+\bar{\mathbf{w}}$ with $\mathbf{v}\in V$ and
$\bar{\mathbf{w}}\in W$, and $E=\bar{\mathbf{w}}+V$; similarly
$F=\bar{\mathbf{v}}+W$ for a unique $\bar{\mathbf{v}}\in V$. Therefore
$\bar{\mathbf{v}}+\bar{\mathbf{w}}$ is the unique element of $E\cap F$.

The inner product $\boldsymbol{\cdot}$ on $\mathbb{R}^{d}$ yields the
Euclidean norm $\|\cdot\|$, and hence we have a distance on $\mathbb{R}^{d}$
and all its affine subspaces. In particular, if $E$ is an affine space then
$\|\mathbf{p}-\mathbf{q}\|$ is the distance between $\mathbf{p},\mathbf{q}\in
E$. Two vectors $\mathbf{v},\mathbf{w}$ of $\mathbb{R}^{d}$ are orthogonal if
$\mathbf{v}\boldsymbol{\cdot}\mathbf{w}=0$. If $V,W$ are complementary
subspaces of $\mathbb{R}^{d}$, then $V$ and $W$ are orthogonal if
$\forall{\mathbf{v}\in V}\,\forall{\mathbf{w}\in
W}\,(\mathbf{v}\boldsymbol{\cdot}\mathbf{w}=0)$; in this case we call one of
the two subspaces the orthogonal complement of the other. Two complementary
affine subspaces $E$ and $F$ of $\mathbb{R}^{d}$ are orthogonal if their
underlying vector spaces are orthogonal.

Let $E$ be an affine subspace of $\mathbb{R}^{d}$. The sphere in $E$ with
center $\mathbf{c}\in E$ and radius $r\in\mathbb{R}$ is the set

$\mathbb{S}(E;\mathbf{c},r)\coloneqq\mathopen{\\{}{\mathbf{x}\in
E}\boldsymbol{\mid}{\|\mathbf{x}-\mathbf{c}\|=r}\mathclose{\\}}.$

We convene that $\mathbb{S}(E;\mathbf{c},r)$ is empty if $r<0$, and it is the
singleton $\left\\{{\mathbf{c}}\right\\}$ when $r=0$. Observe that if $r>0$
then $\mathbb{S}(E;\mathbf{c},r)$ has cardinality $2$ when $\dim(E)=1$; if
$\dim(E)\geq 2$, then $\mathbb{S}(E;\mathbf{c},r)$ has cardinality
$2^{\aleph_{0}}$. Whenever the ambient space (i.e. $\mathbb{R}^{d}$, $E$, …)
is clear we will simply write $\mathbb{S}(\mathbf{c},r)$. A
$(k-1)$-dimensional sphere is a sphere in an affine subspace $E$ of some
$\mathbb{R}^{d}$ with $\dim E=k$.

###  2.2. Families with finite mesh

######  Definition 2.2.

Let $N,d\geq 2$ be natural numbers and suppose that
$\mathcal{H}_{i}\subseteq\mathscr{P}(\mathbb{R}^{d})$ with $1\leq i\leq N$ are
pairwise disjoint. The sequence $(\mathcal{H}_{i})_{i=1}^{N}$

  * •

is locally finite if there is a positive integer $s$ such that for any $s$
distinct points in $\mathbb{R}^{d}$ there are at most finitely many sets in
$\bigcup_{i=1}^{N}\mathcal{H}_{i}$ to which these points belong;

  * •

has mesh $r$, if $r\geq 2$ is the least integer such that the intersection of
$r$ sets belonging to distinct $\mathcal{H}_{i}$s is a finite set.

All sequences $(\mathcal{H}_{i})_{i=1}^{N}$ we consider are locally finite.
The reason for Definition 2.2 is the following very general result by Erdős,
Jackson, and Mauldin [EJM94, Theorem 2]:

######  Theorem 2.3.

For all $d\geq 2$, $n\geq 0$, and $\delta=0,1$ the following are equivalent:

  1. (a)

$2^{\aleph_{0}}\leq\aleph_{\delta+n}$;

  2. (b)

for any $r\geq 2$, letting $N=(n+1)(r-1)+1$, and for any sequence of pairwise
disjoint $\mathcal{H}_{i}\subseteq\mathscr{P}(\mathbb{R}^{d})$ with $1\leq
i\leq N$ locally finite and of mesh $r$, there are $A_{1},\dots,A_{N}$
covering $\mathbb{R}^{d}$ such that $\forall{1\leq i\leq
N}\,\forall{H\in\mathcal{H}_{i}}\,(\mathopen{|}H\cap
A_{i}\mathclose{|}<\aleph_{\delta})$.

The theorem remains true if $\delta$ is taken to be an ordinal, not just $0$
or $1$. The main use of Theorem 2.3 in this paper is the direction
(a)$\Rightarrow$(b). Given $(\mathcal{H}_{i})_{i=1}^{N}$ locally finite of
mesh $r$ in $\mathbb{R}^{d}$, then
$\mathopen{|}\mathbb{R}\mathclose{|}\leq\aleph_{\delta+n}$ implies that there
are $A_{1},\dots,A_{N}\subseteq\mathbb{R}^{d}$ covering $\mathbb{R}^{d}$ such
that for all $H\in\mathcal{H}_{i}$, the set $H\cap A_{i}$ is _finite_ if
$\delta=0$ or _countable_ if $\delta=1$. Actually, the “forward implication”
(i.e. from a bound on the size of the continuum, to the existence of specific
subsets of the space) in many papers [Bag59, Dav62, Dav63, Dav63a, Bag68,
Kom01] follows from Theorem 2.3.

###  2.3. Points and vectors in general position

######  Definition 2.4.

A set of vectors of $\mathbb{R}^{d}$ is in general position if any subset of
size $\leq d$ is linearly independent—in other words: the vectors are as
linearly independent as possible.

######  Example 2.5.

If $\mathbf{u}_{1},\dots,\mathbf{u}_{N}$ are vectors in general position in
$\mathbb{R}^{d}$, and
$\mathcal{H}_{i}\coloneqq\mathopen{\\{}{H_{\mathbf{u}_{i}}(\mathbf{p})}\boldsymbol{\mid}{\mathbf{p}\in\mathbb{R}^{d}}\mathclose{\\}}$
is the family of all hyperplanes orthogonal to $\mathbf{u}_{i}$, then
$(\mathcal{H}_{i})_{i=1}^{N}$ is locally finite, and of mesh $d$.

By Example 2.5 and Theorem 2.3, if
$\mathbf{u}_{1},\dots,\mathbf{u}_{(n+1)(d-1)+1}$ are vectors in general
position in $\mathbb{R}^{d}$, then

  * •

$2^{\aleph_{0}}\leq\aleph_{n}$ implies that there are
$A_{1},\dots,A_{(n+1)(d-1)+1}$ covering $\mathbb{R}^{d}$ such that
$H_{\mathbf{u}_{i}}(\mathbf{p})\cap A_{i}$ is finite for every
$\mathbf{p}\in\mathbb{R}^{d}$ and $1\leq i\leq(n+1)(d-1)+1$;

  * •

$2^{\aleph_{0}}\leq\aleph_{n+1}$ implies that there are
$A_{1},\dots,A_{(n+1)(d-1)+1}$ covering $\mathbb{R}^{d}$ such that
$H_{\mathbf{u}_{i}}(\mathbf{p})\cap A_{i}$ is countable, for every
$\mathbf{p}\in\mathbb{R}^{d}$ and $1\leq i\leq(n+1)(d-1)+1$.

Therefore ${\mathsf{CH}}$ implies that if
$\mathbf{u}_{1},\dots,\mathbf{u}_{d}$ are a basis of $\mathbb{R}^{d}$, then
there are $A_{1},\dots,A_{d}$ covering $\mathbb{R}^{d}$ such that every
hyperplane orthogonal to $\mathbf{u}_{i}$ has countable intersection with
$A_{i}$.

In fact, the implications above are actually equivalences [KT06, pp. 71,
327–328].

######  Theorem 2.6 (Sierpiński).

${\mathsf{CH}}$ is equivalent to either one of the following:

  1. (a)

there are $A_{1},A_{2}$ covering the plane such that every vertical line has
countable intersection with $A_{1}$ and every horizontal line has countable
intersection with $A_{2}$,

  2. (b)

there are $A_{1},A_{2},A_{3}$ covering $\mathbb{R}^{3}$ such that every plane
orthogonal to $\mathbf{e}_{i}$ has countable intersection with $A_{i}$.

Definition 2.4 can be extended to points in affine spaces.

######  Definition 2.7.

Let $S$ be a set of points of an affine space $E$ of dimension $d$. We say
that $S$ is

  * •

in general position in $E$ if the affine span of any of its subset of size
$k+1$ has dimension $k$, for all $k\leq d$,

  * •

well-placed in $E$ if $S\subseteq H$ for some hyperplane $H$, and $S$ is in
general position in $H$.

######  Remarks 2.8.

Let $E$ be an affine space of dimension $d\geq 2$ and let $S\subseteq E$.

  1. (a)

If $S$ is in general position (well-placed), and $S^{\prime}\subseteq S$ then
$S^{\prime}$ is in general position (well-placed). In other words, the notions
of being in general position/well-placed are downward persistent with respect
to inclusion.

  2. (b)

Suppose $S$ is well-placed:

     * •

if $\mathopen{|}S\mathclose{|}\geq d$ then the hyperplane in the definition is
unique, being the affine span of any subset of size $d$;

     * •

if $\mathopen{|}S\mathclose{|}\leq d$ then $S$ is in general position in $E$.

  3. (c)

Suppose $H$ is a hyperplane of $E$, and $S\subseteq H$. Then $S$ is well-
placed in $E$ if and only if $S$ is in general position in $H$.

  4. (d)

$S$ is in general position in $E$ if and only if the set of vectors
$\mathopen{\\{}{\mathbf{q}-\mathbf{p}}\boldsymbol{\mid}{\mathbf{q}\in
S\setminus\left\\{{\mathbf{p}}\right\\}}\mathclose{\\}}$ is in general
position in $\mathbb{R}^{d}$, for all $\mathbf{p}\in S$.

Every set of points in $\mathbb{R}$ is in general position, a set of points in
$\mathbb{R}^{2}$ is in general position if no three of them are aligned, a set
of points in $\mathbb{R}^{3}$ is in general position if no four of them are
coplanar, and so on. A set of points in $\mathbb{R}^{2}$ is well-placed if
they are collinear, a set of points in $\mathbb{R}^{3}$ is well-placed if they
are coplanar, and no three of them are collinear, a set of points in
$\mathbb{R}^{4}$ is well-placed if they belong to the same $3$-dimensional
affine subspace, and no four of them are coplanar, and so on.

Finally observe that for any affine space $E$ of dimension $d\geq 2$ and any
$S\subseteq E$

(1) $\begin{multlined}\bigl{(}S\text{ is general position in $E$, or $S$ is
well-placed in }E\bigr{)}\Rightarrow{}\qquad\\\
\qquad\forall{k<d}\,\bigl{(}\text{the affine span of any subset of $S$ of size
}k+1\\\ \qquad\text{has dimension }k\bigr{)}.\end{multlined}\bigl{(}S\text{ is
general position in $E$, or $S$ is well-placed in
}E\bigr{)}\Rightarrow{}\qquad\\\ \qquad\forall{k<d}\,\bigl{(}\text{the affine
span of any subset of $S$ of size }k+1\\\ \qquad\text{has dimension
}k\bigr{)}.$

######  Lemma 2.9.

Suppose $\mathbf{c}_{1},\dots,\mathbf{c}_{k+1}$ are distinct points in general
position in $\mathbb{R}^{d}$, with $d\geq k$. Let $H$ be a hyperplane
orthogonal to $\mathbf{c}_{k}-\mathbf{c}_{k+1}$, let
$\pi\colon\mathbb{R}^{d}\to H$ be the orthogonal projection. Let
$\bar{\mathbf{c}}_{k}=\pi(\mathbf{c}_{k})=\pi(\mathbf{c}_{k+1})$ and let
$\bar{\mathbf{c}}_{i}=\pi(\mathbf{c}_{i})$ for $i<k$. Then
$\bar{\mathbf{c}}_{1},\dots,\bar{\mathbf{c}}_{k}$ are distinct and in general
position in $H$.

###### Proof.

If $\bar{\mathbf{c}}_{i}=\bar{\mathbf{c}}_{k}$ for some $i<k$, then
$\mathbf{c}_{i},\mathbf{c}_{k},\mathbf{c}_{k+1}$ must be collinear, and if
$\bar{\mathbf{c}}_{i}=\bar{\mathbf{c}}_{j}$ for some $i<j<k$, then
$\mathbf{c}_{i}-\mathbf{c}_{j}$ and $\mathbf{c}_{k}-\mathbf{c}_{k+1}$ must be
parallel, so the four points
$\mathbf{c}_{i},\mathbf{c}_{j},\mathbf{c}_{k},\mathbf{c}_{k+1}$ are coplanar.
In either case this contradicts the assumption that
$\mathbf{c}_{1},\dots,\mathbf{c}_{k+1}$ are in general position. Therefore the
points $\bar{\mathbf{c}}_{1},\dots,\bar{\mathbf{c}}_{k}$ are distinct.

In order to conclude the proof it is enough to show that the vectors
$\bar{\mathbf{c}}_{1}-\bar{\mathbf{c}}_{k}$,
$\bar{\mathbf{c}}_{2}-\bar{\mathbf{c}}_{k}$, …,
$\bar{\mathbf{c}}_{k-1}-\bar{\mathbf{c}}_{k}$ of $\mathbb{R}^{d}$ are linearly
independent. Towards a contradiction, suppose there are
$(r_{1},\dots,r_{k-1})\neq(0,\dots,0)$ such that

$\mathbf{0}=r_{1}(\bar{\mathbf{c}}_{1}-\bar{\mathbf{c}}_{k})+r_{2}(\bar{\mathbf{c}}_{2}-\bar{\mathbf{c}}_{k})+\dots+r_{k-1}(\bar{\mathbf{c}}_{k-1}-\bar{\mathbf{c}}_{k}).$

As $\bar{\mathbf{c}}_{i}$ is the projection of $\mathbf{c}_{i}$ along the
vector $\mathbf{c}_{k+1}-\mathbf{c}_{k}$, there are
$s_{1},\dots,s_{k}\in\mathbb{R}$ such that
$\bar{\mathbf{c}}_{i}=\mathbf{c}_{i}+s_{i}(\mathbf{c}_{k+1}-\mathbf{c}_{k})$.
Substituting in the previous formula we obtain

$\mathbf{0}=r_{1}(\mathbf{c}_{1}-\mathbf{c}_{k})+\dots+r_{k-1}(\mathbf{c}_{k-1}-\mathbf{c}_{k})\\\
{}+[r_{1}(s_{1}-s_{k})+\dots+r_{k-1}(s_{k-1}-s_{k})](\mathbf{c}_{k+1}-\mathbf{c}_{k}),$

and since $\mathbf{c}_{1},\dots,\mathbf{c}_{k+1}$ are in general position in
$\mathbb{R}^{d}$, then $r_{1}=\dots=r_{k-1}=0$ and
$[r_{1}(s_{1}-s_{k})+\dots+r_{k-1}(s_{k-1}-s_{k})]=0$, against our choice of
the $r_{i}$s. ∎

######  Example 2.10.

Suppose $N\geq d$ and $\mathbf{c}_{1},\dots,\mathbf{c}_{N}$ are distinct
points of $\mathbb{R}^{d}$ such that any $d$ of them span a subspace of
dimension $d-1$. If $\mathcal{H}_{i}$ is the collection of all spheres of
$\mathbb{R}^{d}$ with center $\mathbf{c}_{i}$, then
$(\mathcal{H}_{i})_{i=1}^{N}$ is locally finite, and of mesh $d$. Here is why.

It is enough to consider the case $N=d$. If $d=1$ this is trivial since a
sphere in $\mathbb{R}$ is just a pair of points that are symmetric with
respect to the center, and if $d=2$ it follows immediately from the fact that
the intersection of two circles with distinct centers has size at most $2$. If
$d=3$ this follows from the fact that the intersection of three spheres in
$\mathbb{R}^{3}$ with non-collinear centers has size at most $2$. The case for
$d>3$ follows from Theorem 2.14 below.

######  Lemma 2.11.

Suppose $H$ is an hyperplane of an affine space $E$ of dimension $d+1$. Let
$\pi\colon E\to H$ be the orthogonal projection, and let $h$ be the distance
between a point $\mathbf{c}\in E$ and $H$. Then

$\forall{R\in\mathbb{R}}\,\exists{r\in\mathbb{R}}\,\left(\mathbb{S}(E;\mathbf{c},R)\cap
H=\mathbb{S}(H;\pi(\mathbf{c}),r)\right)$

where

  1. (i)

$R>h\Leftrightarrow R>r>0$,

  2. (ii)

$R=h\Leftrightarrow r=0$, and in this case $\mathbb{S}(H;\pi(\mathbf{c}),r)$
is the singleton $\left\\{{\pi(\mathbf{c})}\right\\}$,

  3. (iii)

$R<h\Leftrightarrow r<0$, that is $\mathbb{S}(H;\pi(\mathbf{c}),r)=\emptyset$.

Conversely, if $\bar{\mathbf{c}}\in H$ and $\ell$ is the line orthogonal to
$H$ passing through $\bar{\mathbf{c}}$, then

$\forall{r>0}\,\forall{\mathbf{c}\in\ell}\,\exists{R>0}\,\left(\mathbb{S}(E;\mathbf{c},R)\cap
H=\mathbb{S}(H,\bar{\mathbf{c}},r)\right).$

###### Proof.

Without loss of generality we may assume that $E=\mathbb{R}^{d+1}$, that
$H=\mathbb{R}^{d}\times\mathopen{\\{}{0}\mathclose{\\}}$,
$\mathbf{c}=(0,\dots,0,h)$, and $\pi(\mathbf{c})=\mathbf{0}$. It is clear that
$R<h$ if and only if $\mathbb{S}(\mathbf{c},R)\cap H=\emptyset$, and that
$R=h$ if and only if $\mathbb{S}(\mathbf{c},R)\cap
H=\mathopen{\\{}{\pi(\mathbf{c})}\mathclose{\\}}$, so (ii) and (iii) are true.
Suppose $R>h$. Let $r=\sqrt{R^{2}-h^{2}}$ so that $r^{2}+h^{2}=R^{2}$. Then

$\begin{split}\mathbf{x}\in\mathbb{S}(E;\mathbf{c},R)\cap
H&\Leftrightarrow\mathbf{x}=(x_{1},\dots,x_{d},0)\wedge(\textstyle\sum_{i=1}^{d}x_{i}^{2})+h^{2}=R^{2}\\\
&\Leftrightarrow x_{1}^{2}+\dots+x_{d}^{2}=r^{2}\\\
&\Leftrightarrow\|\mathbf{x}\|=r\\\
&\Leftrightarrow\mathbf{x}\in\mathbb{S}(H;\pi(\mathbf{c}),r).\end{split}$

For the second part, fix $\mathbf{c}\in\ell$ and let $R=\sqrt{r^{2}+h^{2}}$
where $h=\|\mathbf{c}-\bar{\mathbf{c}}\|$ is the distance from $\mathbf{c}$ to
$H$, and proceed as before. ∎

######  Lemma 2.12.

Suppose $\mathbf{c}_{1},\mathbf{c}_{2}$ are distinct points of an affine space
$E$ of dimension $d\geq 2$, and let $r_{1},r_{2}>0$ be such that
$\mathopen{|}r_{1}-r_{2}\mathclose{|}\leq\|\mathbf{c}_{1}-\mathbf{c}_{2}\|\leq
r_{1}+r_{2}$. Then there exist and are unique $r,\mathbf{c},H$ such that

  * •

$r$ is a real number $\geq 0$,

  * •

$\mathbf{c}$ is a point on the open segment $(\mathbf{c}_{1};\mathbf{c}_{2})$,

  * •

$H$ is the hyperplane passing through $\mathbf{c}$ and orthogonal to
$\mathbf{c}_{1}-\mathbf{c}_{2}$,

and

$\mathbb{S}(E;\mathbf{c}_{1},r_{1})\cap\mathbb{S}(E;\mathbf{c}_{2},r_{2})=\mathbb{S}(H;\mathbf{c},r).$

###### Proof.

We may assume that $E=\mathbb{R}^{d}$. The hypothesis
$\mathopen{|}r_{1}-r_{2}\mathclose{|}\leq\|\mathbf{c}_{1}-\mathbf{c}_{2}\|$
guarantees that neither sphere properly contains inside the other sphere,
while the hypothesis $\|\mathbf{c}_{1}-\mathbf{c}_{2}\|\leq r_{1}+r_{2}$
ensures that if neither sphere contains the other, then they are close enough
to intersect.

Let $\mathbf{x}$ be a point of
$\mathbb{S}(\mathbf{c}_{1},r_{1})\cap\mathbb{S}(\mathbf{c}_{2},r_{2})$, and
let $P$ be the plane passing through
$\mathbf{c}_{0},\mathbf{c}_{1},\mathbf{x}$. (If $d=2$ then $P$ coincides with
$\mathbb{R}^{2}$.) Consider the triangle with vertices
$\mathbf{c}_{0},\mathbf{c}_{1},\mathbf{x}$.

If $r_{1}+r_{2}=\|\mathbf{c}_{1}-\mathbf{c}_{2}\|$ then the two spheres are
tangent in the point $\mathbf{c}\coloneqq\mathbf{x}$ which belongs to the open
segment $(\mathbf{c}_{1};\mathbf{c}_{2})$, so the triangle is degenerate and
letting $r=0$ we have that
$\mathbb{S}(H;\mathbf{c},r)=\mathopen{\\{}{\mathbf{c}}\mathclose{\\}}$.

So we may assume that $r_{1}+r_{2}>\|\mathbf{c}_{1}-\mathbf{c}_{2}\|$ and that
the triangle

$\mathbf{c}_{1}$$\mathbf{c}_{2}$$\mathbf{x}$$\mathbf{c}$$r_{1}$$r_{2}$$r$

is non-degenerate. Let $\mathbf{c}$ is the point on the open segment
$(\mathbf{c}_{1};\mathbf{c}_{2})$ whose distance from $\mathbf{c}_{i}$ is
$\sqrt{r_{i}^{2}-r^{2}}$, and let $H$ be the hyperplane orthogonal to
$\mathbf{c}_{1}-\mathbf{c}_{2}$ passing through $\mathbf{c}$. Observe that
$r$, the point $\mathbf{c}$, and the hyperplane $H$, depend only on the points
$\mathbf{c}_{1}$ and $\mathbf{c}_{2}$, and not on the point $\mathbf{x}$ or
the plane $P$. Moreover $r=\|\mathbf{x}-\mathbf{c}\|$. Therefore for all
$\mathbf{x}\in\mathbb{R}^{d}$,

$\begin{split}\mathbf{x}\in\mathbb{S}(\mathbf{c}_{1},r_{1})\cap\mathbb{S}(\mathbf{c}_{2},r_{2})&\Leftrightarrow\|\mathbf{x}-\mathbf{c}_{1}\|=r_{1}\wedge\|\mathbf{x}-\mathbf{c}_{2}\|=r_{2}\\\
&\Leftrightarrow\mathbf{x}\in H\wedge\|\mathbf{c}-\mathbf{x}\|=r.\end{split}$

The argument above shows that $r,\mathbf{c},H$ are unique. ∎

######  Proposition 2.13.

Let $E$ be an affine space of dimension $d\geq 2$, let
$\mathbf{c}_{1},\dots,\mathbf{c}_{k}$ be in general position in $E$ with
$k\leq d$, and let $K$ be the affine span of the $\mathbf{c}_{i}$s. Let
$S_{i}$ be a sphere centered in $\mathbf{c}_{i}$ for $i\leq k$.

  1. (a)

There are $r\in\mathbb{R}$ and an affine subspace $H$ of $E$ of dimension
$d-(k-1)$, such that $H\cap K$ is a singleton $\left\\{{\mathbf{c}}\right\\}$
and $H$ and $K$ are orthogonal, so that $S_{1}\cap\dots\cap
S_{k}=\mathbb{S}(H;\mathbf{c},r)$.

  2. (b)

If $\mathbf{c}_{k+1}\in
E\setminus\mathopen{\\{}{\mathbf{c}_{1},\dots,\mathbf{c}_{k}}\mathclose{\\}}$
and
$\mathopen{\\{}{\mathbf{c}_{1},\dots,\mathbf{c}_{k},\mathbf{c}_{k+1}}\mathclose{\\}}$
is not in general position in $E$, then there is $R\in\mathbb{R}$ such that

$S_{1}\cap\dots\cap S_{k}\subseteq\mathbb{S}(\mathbf{c}_{k+1},R).$

###### Proof.

We may assume that $E=\mathbb{R}^{d}$.

If $d=2$ then the result is clear. Part (a) amounts to say that: given two
circles $C_{1},C_{2}$ in the plane with distinct centers $\mathbf{c}_{1}$,
$\mathbf{c}_{2}$, either $I\coloneqq C_{1}\cap C_{2}$ is empty, or else there
is a point $\mathbf{c}$ on the open segment $(\mathbf{c}_{1};\mathbf{c}_{2})$
such that $I=\mathopen{\\{}{\mathbf{c}}\mathclose{\\}}$, or else $I$ is the
set of two points on line orthogonal to $(\mathbf{c}_{1};\mathbf{c}_{2})$
passing through $\mathbf{c}$, and symmetric with respect to $\mathbf{c}$. Part
(b) says that if three points are collinear, then given any two circles
centered in the first two points there is a (possibly degenerate) circle
centered in the third point that passes through the intersection of the first
two circles.

Therefore we may assume that $d\geq 3$.

(a) We proceed by induction on $k$. If $k=1$ the result is trivial, so we may
assume the result holds for some $k$ towards proving it for $k+1<d$. Suppose
$S_{1},\dots,S_{k+1}$ are spheres centered in distinct points
$\mathbf{c}_{1},\dots,\mathbf{c}_{k+1}$. By an isometry we may assume that
$K$, the affine span of the $\mathbf{c}_{i}$s, is contained in
$\mathbb{R}^{d-1}\times\mathopen{\\{}{0}\mathclose{\\}}$. Let
$\bar{S}_{k}\coloneqq S_{k}\cap S_{k+1}$. We have three cases:

  1. (1)

$\bar{S}_{k}=\emptyset$,

  2. (2)

$\bar{S}_{k}$ is a singleton,

  3. (3)

$\bar{S}_{k}$ is a non-degenerate sphere in some hyperplane $H$ of
$\mathbb{R}^{d}$ (Lemma 2.12).

If either (1) or (2) holds, then $S_{1}\cap\cdots\cap S_{k+1}$ is empty or a
singleton, and the result follows trivially. So we may assume case (3). The
hyperplane $H$ is orthogonal to the open segment
$(\mathbf{c}_{k};\mathbf{c}_{k+1})$, and $\bar{\mathbf{c}}_{k}$, the center of
$\bar{S}_{k}$, belongs to this segment. Let $\pi\colon\mathbb{R}^{d}\to H$ be
the orthogonal projection. Then
$\pi(\mathbf{c}_{k})=\pi(\mathbf{c}_{k+1})=\bar{\mathbf{c}}_{k}$ and for $i<k$
let $\bar{\mathbf{c}}_{i}\coloneqq\pi(\mathbf{c}_{i})$. By Lemma 2.11 for
every $i<k$ the set $\bar{S}_{i}\coloneqq S_{i}\cap H$ is a sphere of $H$ with
center $\bar{\mathbf{c}}_{i}$. As $S_{k}\cap S_{k+1}\subseteq H$

$S_{1}\cap\dots\cap S_{k}\cap S_{k+1}=\bar{S}_{1}\cap\dots\cap\bar{S}_{k}.$

By Lemma 2.9 $\bar{\mathbf{c}}_{1},\dots,\bar{\mathbf{c}}_{k}$ are in general
position in $H\cong\mathbb{R}^{d-1}$, and since $d-1\geq 2$, by inductive
assumption we can conclude that
$\bar{S}_{1}\cap\dots\cap\bar{S}_{k}=\mathbb{S}(H^{\prime};\mathbf{c},r)$
where $H^{\prime}$ a subspace of $H$ (and hence of $\mathbb{R}^{d}$) of
dimension $(d-1)-(k-1)=d-k$, such that
$\left\\{{\mathbf{c}}\right\\}=H^{\prime}\cap K$, and $H^{\prime}$ is
orthogonal to $K$.

(b) By assumption $\mathbf{c}_{k+1}$ belongs to $K$, and $k-1=\dim(K)$. By
part (a) there is a subspace $H$ orthogonal to $K$, of dimension $d-(k-1)$,
and $r\in\mathbb{R}$ such that $S_{1}\cap\dots\cap
S_{k}=\mathbb{S}(H;\mathbf{c},r)$ where $\mathbf{c}$ is the unique element of
$H\cap K$. It is enough to show that

$\exists{R\in\mathbb{R}}\,\bigl{(}\mathbb{S}(H;\mathbf{c},r)\subseteq
S_{k+1}\coloneqq\mathbb{S}(E;\mathbf{c}_{k+1},R)\bigr{)}.$

This is immediate if $r\leq 0$, so we may assume that $r>0$, that is to say
$\mathbb{S}(H;\mathbf{c},r)$ is non-degenerate. If
$\mathbf{c}=\mathbf{c}_{k+1}$ then taking $R=r$ the result follows at once, so
we may assume that $\mathbf{c}\neq\mathbf{c}_{k+1}$. Let
$R=\sqrt{\|\mathbf{c}_{k+1}-\mathbf{c}\|^{2}+r^{2}}$. Given
$\mathbf{x}\in\mathbb{S}(H;\mathbf{c},r)$, we must show that

$\|\mathbf{x}-\mathbf{c}_{k+1}\|^{2}=\|\mathbf{x}-\mathbf{c}\|^{2}+\|\mathbf{c}_{k+1}-\mathbf{c}\|^{2}$

so that $\mathbf{x}\in\mathbb{S}(E;\mathbf{c}_{k+1},R)$. But this is
immediate, since $H,K$ are orthogonal and hence $\mathbf{c}_{k+1}-\mathbf{c}$
and $\mathbf{x}-\mathbf{c}$ are orthogonal. ∎

######  Theorem 2.14.

Let $E$ be an affine space of dimension $d\geq 2$ and let
$\mathbf{c}_{1},\dots,\mathbf{c}_{d}$ be distinct points in general position
in $E$.

  1. (a)

For all spheres $S_{i}$ in $E$ centered in $\mathbf{c}_{i}$ with $1\leq i\leq
d$, the set $S_{1}\cap\cdots\cap S_{d}$ is finite.

  2. (b)

For all $k<d$ and $1\leq i\leq k$ there are spheres $S_{i}$ in $E$ centered in
$\mathbf{c}_{i}$ such that $S_{1}\cap\cdots\cap S_{k}$ is a non-degenerate
sphere in a subspace of dimension $d-(k-1)\geq 2$.

###### Proof.

Recall that a non-degenerate sphere in space of dimension $n$ has cardinality
$2^{\aleph_{0}}$ or $2$ depending whether $n>1$ or $n=1$. Part (a) then
follows at once form Proposition 2.13.

(b) We prove by induction on $k$ a stronger statement: for any $r_{k}>0$ and
for any affine space $E$ of dimension $d>k$, there exist
$r_{1},\dots,r_{k-1}>0$ such that

$\mathbb{S}(E;\mathbf{c}_{1},r_{1})\cap\dots\cap\mathbb{S}(E;\mathbf{c}_{k},r_{k})=\mathbb{S}(H;\mathbf{c},r),$

for some $r>0$, $\mathbf{c}\in H$ and $H$ a subspace of $E$ of dimension
$d-(k-1)$.

The base case $k=1$ is immediate, so we may assume that the result holds for
some $k$ towards proving it for $k+1<d$. Fix $r_{k+1}>0$ and pick $r_{k}>0$
such that
$\mathopen{|}r_{k}-r_{k+1}\mathclose{|}<\|\mathbf{c}_{k}-\mathbf{c}_{k+1}\|<r_{k}+r_{k+1}$.
By Lemma 2.12
$\mathbb{S}(E;\mathbf{c}_{k+1},r_{k+1})\cap\mathbb{S}(E;\mathbf{c}_{k},r_{k})$
is a non-degenerate sphere
$\mathbb{S}(\bar{E};\bar{\mathbf{c}}_{k},\bar{r}_{k})$ in some hyperplane
$\bar{E}$ of $E$. Let $\pi\colon E\to\bar{E}$ be the orthogonal projection. By
Lemma 2.9 the points

$\displaystyle\bar{\mathbf{c}}_{1}$ $\displaystyle=\pi(\mathbf{c}_{1}),$
$\displaystyle\bar{\mathbf{c}}_{2}$ $\displaystyle=\pi(\mathbf{c}_{2}),$
$\displaystyle\dots$ $\displaystyle\bar{\mathbf{c}}_{k-1}$
$\displaystyle=\pi(\mathbf{c}_{k-1}),$ $\displaystyle\bar{\mathbf{c}}_{k}$
$\displaystyle=\pi(\mathbf{c}_{k})=\pi(\mathbf{c}_{k+1})$

are distinct and in general position in the affine space $\bar{E}$ of
dimension $d-1>k$. Since $\bar{r}_{k}>0$, by inductive assumption there exist
$\bar{r}_{1},\dots,\bar{r}_{k-1}>0$ such that

$S\coloneqq\mathbb{S}(\bar{E},\bar{\mathbf{c}}_{1},\bar{r}_{1})\cap\dots\cap\mathbb{S}(\bar{E},\bar{\mathbf{c}}_{k},\bar{r}_{k})$

is a non-degenerate sphere in a subspace $H$ of dimension $(d-1)-(k-1)=d-k$.
Letting $r_{i}=\sqrt{\bar{r}_{i}^{2}+h_{i}^{2}}$ where $h_{i}$ is the distance
between $\mathbf{c}_{i}$ and the hyperplane $\bar{E}$, we have that

$S=\mathbb{S}(E,\mathbf{c}_{1},r_{1})\cap\dots\cap\mathbb{S}(E,\mathbf{c}_{k},r_{k})\cap\mathbb{S}(E,\mathbf{c}_{k+1},r_{k+1})$

as required. ∎

######  Corollary 2.15.

Suppose $E$ is an affine space of dimension $d\geq 2$ and
$\mathbf{c}_{1},\dots,\mathbf{c}_{d+1}$ are distinct points that are not in
general position in $E$. Then there are spheres $S_{i}$ centered in
$\mathbf{c}_{i}$ such that $S_{1}\cap\dots\cap S_{d+1}$ is infinite.

###### Proof.

Let $k$ be the dimension of the affine span $K$ of
$\left\\{{\mathbf{c}_{1},\dots,\mathbf{c}_{d+1}}\right\\}$. By assumption
$k<d$, and without loss of generality we may assume that
$\mathbf{c}_{1},\dots,\mathbf{c}_{k+1}$ are in general position in $K$. By
part (b) of Theorem 2.14 there are spheres $S_{1},\dots,S_{k+1}$ centered in
$\mathbf{c}_{1},\dots,\mathbf{c}_{k+1}$ whose intersection is a non-degenerate
sphere in a space of dimension $\geq 2$. By repeated applications of part (b)
of Proposition 2.13, there are spheres $S_{k+2},\dots,S_{d+1}$ centered in
$\mathbf{c}_{k+2},\dots,\mathbf{c}_{d+1}$ such that

$S_{1}\cap\dots\cap S_{k+1}\subseteq S_{i}\quad\text{for }k+2\leq i\leq d+1$

and hence $S_{1}\cap\dots\cap S_{k+1}=S_{1}\cap\dots\cap S_{d+1}$ is infinite.
∎

Observe that by (1), both “$S$ is in general position” and “$S$ is well-
placed” imply the assumption of the next corollary.

######  Corollary 2.16.

Suppose $S$ is a set of at least $d$ points of an affine space $E$ of
dimension $d\geq 2$. Assume that any $d$-many points of $S$ span a subspace of
dimension $d-1$. Then the intersection of $d$-many spheres centered in these
points is a finite set.

###  2.4. Sprays

J. Schmerl in [Sch03] defined a spray to be a set $X\subseteq\mathbb{R}^{2}$
such that $C\cap X$ is finite, for all circles $C$ centered in some given
point. Thus a spray $X$ is a very sparse subset of the plane, and one may
investigate what happens if one relaxes the finiteness condition on $C\cap X$
with, say, being countable.

######  Definition 2.17.

Let $E$ be an affine subspace of $\mathbb{R}^{d}$, with $d\geq 2$ be a natural
number. An $\aleph_{\delta}$-spray in $E$ with center $\mathbf{c}$ is a set
$X\subseteq E$ such that $\mathopen{|}\mathbb{S}(E;\mathbf{c},r)\cap
X\mathclose{|}<\aleph_{\delta}$ for every $r\in\mathbb{R}$. When $\delta=0$ it
is called a spray, when $\delta=1$ we speak of $\sigma$-spray.

Note that a center need not to belong to the $\aleph_{\delta}$-spray, and that
a $\aleph_{\delta}$-spray may have more than one center.

By Example 2.10, if $\mathbf{c}_{1},\dots,\mathbf{c}_{(n+1)(d-1)+1}$ are
either in general position, or well-placed in $\mathbb{R}^{d}$, then

  * •

$2^{\aleph_{0}}\leq\aleph_{n}$ implies that there are
$X_{1},\dots,X_{(n+1)(d-1)+1}$ covering $\mathbb{R}^{d}$ such that each
$X_{i}$ is a spray centered in $\mathbf{c}_{i}$;

  * •

$2^{\aleph_{0}}\leq\aleph_{n+1}$ implies that there are
$X_{1},\dots,X_{(n+1)(d-1)+1}$ covering $\mathbb{R}^{d}$ such that each
$X_{i}$ is a $\sigma$-spray.

In particular if the points are well-placed, that is if we assume that the
$\mathbf{c}_{i}$s lie on a hyperplane, then:

######  Theorem 2.18.

Let $d\geq 2$, $n\geq 0$, and $\delta=0,1$. Assume that
$\mathbf{c}_{1},\dots,\mathbf{c}_{(n+1)(d-1)+1}$ are well-placed points in
$\mathbb{R}^{d}$. If $2^{\aleph_{0}}\leq\aleph_{\delta+n}$ then there are
$X_{1},\dots,X_{(n+1)(d-1)+1}$ covering $\mathbb{R}^{d}$ such that each
$X_{i}$ is an $\aleph_{\delta}$-spray centered in $\mathbf{c}_{i}$.

By (1) and Corollary 2.16, the thesis of Theorem 2.18 holds if
$\mathbf{c}_{1},\dots,\mathbf{c}_{(n+1)(d-1)+1}$ are in general position in
$\mathbb{R}^{d}$. The reason for focusing on well-placed points is that under
this assumption the implication in Theorem 2.18 can be reversed, and this is
the main goal of this paper. In Section 3 we argue that any covering of
$\mathbb{R}^{d}$ with sprays ($\sigma$-sprays) with well-placed centers
$\mathbf{c}_{i}$s can be transformed into a covering (of an open subset) of
$\mathbb{R}^{d}$ with sets whose intersection with the hyperplanes orthogonal
to $\mathbf{u}_{i}$ is finite (countable) for all $i$, where the vector
$\mathbf{u}_{i}$ is obtained from the point $\mathbf{c}_{i}$. Appealing to the
results from Section 4 the equivalence will be established.

The case of points in general position, but not well-placed, i.e. not
belonging to an hyperplane, is more problematic. If $d=2$ and three points are
not collinear (i.e. are in general position, but not well-placed in
$\mathbb{R}^{2}$) then by [Sch10, Theorem 1], the plane is the union of three
sprays centered in these points, irrespective of the size of the continuum.
The analogous statement for $d>2$ is open, and if true it would show that the
assumption that the points are well-placed cannot be dropped.

######  Conjecture 2.19.

For all $d\geq 3$, for all $\mathbf{c}_{1},\dots,\mathbf{c}_{d+1}$ in general
position in $\mathbb{R}^{d}$, there are $X_{1},\dots,X_{d+1}$ covering
$\mathbb{R}^{d}$, with $X_{i}$ a spray centered in $\mathbf{c}_{i}$ for $1\leq
i\leq d+1$.

###  2.5. A few elementary results about sprays

Observe the following very basic, but useful facts. Let $\kappa$ be an
infinite cardinal.

Glueing:

If $X$ and $Y$ are $\aleph_{\delta}$-sprays in an affine space $E$ with the
same center $\mathbf{c}$, then $X\cup Y$ is a $\aleph_{\delta}$-spray in $E$
with center $\mathbf{c}$.

Projection:

Suppose that $X_{1},\dots,X_{n}$ are $\aleph_{\delta}$-sprays in an affine
space $E$ of dimension $d+1$, and $X_{1}\cup\dots\cup X_{n}=E$ with centers
$\mathbf{c}_{1},\dots,\mathbf{c}_{n}$. Suppose $H$ is an hyperplane of $E$,
and let $\pi\colon E\to H$ be the orthogonal projection. Then $X_{1}\cap
H,\dots,X_{n}\cap H$ are $\aleph_{\delta}$-sprays in $H$ with centers
$\pi(\mathbf{c}_{1}),\dots,\pi(\mathbf{c}_{n})\in H$, that cover $H$ .

Projection follows from Lemma 2.11. Observe that the points
$\pi(\mathbf{c}_{1}),\dots,\pi(\mathbf{c}_{n})$ need not be distinct, even if
the $\mathbf{c}_{1},\dots,\mathbf{c}_{n}$ are distinct. This, together with
glueing, allows to transform $n$-many $\aleph_{\delta}$-sprays covering
$\mathbb{R}^{d+1}$ into $n^{\prime}$-many $\aleph_{\delta}$-sprays covering
$\mathbb{R}^{d}$, with $n^{\prime}<n$.

######  Proposition 2.20.

Suppose $\mathbb{R}^{2}=X_{1}\cup X_{2}$ where $X_{1}$ is an
$\aleph_{\delta}$-spray and $X_{2}$ is an $\aleph_{\delta+1}$-spray. Then
$2^{\aleph_{0}}\leq\aleph_{\delta}$.

###### Proof.

For $i=1,2$, let $\mathbf{c}_{i}$ be the center of $X_{i}$, and for $r>0$ let

$C_{i}(r)\coloneqq\mathopen{\\{}{\mathbf{p}\in\mathbb{R}^{2}}\boldsymbol{\mid}{\|\mathbf{p}-\mathbf{c}_{i}\|=r}\mathclose{\\}}$

be the circle centered in $\mathbf{c}_{i}$ of radius $r$. If
$\mathbf{c}_{1}=\mathbf{c}_{2}$ then by glueing $X_{1}\cup X_{2}$ would be an
$\aleph_{\delta+1}$-spray in $\mathbb{R}^{2}$, and every circle centered in
$\mathbf{c}_{1}$ is contained in $X_{1}\cup X_{2}=\mathbb{R}^{2}$ is of
cardinality $\leq\aleph_{\delta}$. As any circle is in bijection with
$\mathbb{R}$, the result follows. Therefore we may assume that
$\mathbf{c}_{1}\neq\mathbf{c}_{2}$. Applying an isometry if needed, we may
assume that $\mathbf{c}_{1}=(0,0)$ and $\mathbf{c}_{2}=(a,0)$ for some $a>0$.

Towards a contradiction, suppose $2^{\aleph_{0}}>\aleph_{\delta}$. Fix
distinct reals $r_{\alpha}$ in the interval $(a/2;a)$, for
$\alpha<\aleph_{\delta}$. By assumption $X_{2}(\alpha)\coloneqq X_{2}\cap
C_{2}(r_{\alpha})$ has size $\leq\aleph_{\delta}$, so the set

$\textstyle\mathopen{\\{}{r\in(a/2;a)}\boldsymbol{\mid}{C_{1}(r)\cap(\bigcup_{\alpha\in\aleph_{\delta}}X_{2}(\alpha))\neq\emptyset}\mathclose{\\}}$

has size $\leq\aleph_{\delta}$. As
$\mathopen{|}\mathbb{R}\mathclose{|}>\aleph_{\delta}$ we may pick
$r\in(a/2;a)$ outside of this set. For each $\alpha\in\aleph_{\delta}$ the set
$C_{1}(r)\cap C_{2}(r_{\alpha})$ has size $2$, and its points belong to
$X_{1}$. As the $C_{2}(r_{\alpha})$s are disjoint, it follows that
$C_{1}(r)\cap X_{1}$ has size $\geq\aleph_{\delta}$, contradicting that
$X_{1}$ is an $\aleph_{\delta}$-spray. ∎

When $\delta=0$ we obtain at once:

######  Corollary 2.21.

$\mathbb{R}^{2}$ is not the union of a spray and a $\sigma$-spray. In
particular, $\mathbb{R}^{2}$ is not the union of two sprays.

######  Corollary 2.22.

The following are equivalent:

  1. (a)

${\mathsf{CH}}$

  2. (b)

$\mathbb{R}^{2}$ is the union of two $\sigma$-sprays with prescribed, distinct
centers.

######  Proposition 2.23.

Let $n\geq d\geq 3$, and suppose that the points
$\mathbf{c}_{1},\dots,\mathbf{c}_{n}$ belong to a hyperplane $H$ of
$\mathbb{R}^{d}$. Suppose there is $L\subseteq H$ an affine subspace of
dimension $d-2$ such that
$\mathopen{\\{}{\pi(\mathbf{c}_{1}),\dots,\pi(\mathbf{c}_{n})}\mathclose{\\}}$
has size $\leq d-1$, where $\pi\colon H\to L$ is the orthogonal projection.
Then there are no sprays $X_{1},\dots,X_{n}$ that cover $\mathbb{R}^{d}$ with
$X_{i}$ centered in $\mathbf{c}_{i}$, for $i\leq n$.

###### Proof.

Towards a contradiction, let $d\geq 3$ be least such that the statement fails,
and suppose $X_{i}$ is a spray centered in $\mathbf{c}_{i}$ such that
$\mathbb{R}^{d}=X_{1}\cup\dots\cup X_{n}$. Then $H\cong\mathbb{R}^{d-1}$ is
covered by the sprays $\pi[X_{1}],\dots,\pi[X_{n}]$ centered in
$\mathopen{\\{}{\pi(\mathbf{c}_{1}),\dots,\pi(\mathbf{c}_{n})}\mathclose{\\}}$.
This means that $\mathbb{R}^{d-1}$ can be covered by $d-1$-many sprays whose
centers lie on the hyperplane $L$. This contradicts the minimality of $d$, if
$d-1\geq 3$. If $d=3$, then we would have that $\mathbb{R}^{2}$ can be covered
by two sprays, against Corollary 2.21. ∎

In [Sch12, p. 1169] it is observed that the next result follows from results
of Sikorski.

######  Theorem 2.24.

For $d\geq 2$ the space $\mathbb{R}^{d}$ is not the union of $d$-many sprays.

###### Proof.

The case $d=2$ is Corollary 2.21. Suppose $d\geq 3$ and that
$\mathbf{c}_{1},\dots,\mathbf{c}_{d}$ are centers of sprays
$X_{1},\dots,X_{d}$ that cover $\mathbb{R}^{d}$. Let $H$ be a hyperplane
containing these points, and apply Proposition 2.23 with $n=d-1$: if $L$ is an
affine subspace of $H$ of dimension $d-2>0$ that is orthogonal to the vector
$\mathbf{c}_{d}-\mathbf{c}_{d-1}$, then
$\mathopen{\\{}{\pi(\mathbf{c}_{1}),\dots,\pi(\mathbf{c}_{d})}\mathclose{\\}}$
has size $\leq d-1$ since $\pi(\mathbf{c}_{d})=\pi(\mathbf{c}_{d-1})$. But
Proposition 2.23 implies that the $X_{i}$s cannot cover $\mathbb{R}^{d}$. ∎

A similar argument shows:

######  Theorem 2.25.

For $d\geq 2$ the space $\mathbb{R}^{d}$ is not the union of $(d-1)$-many
$\sigma$-sprays.

######  Theorem 2.26.

For $d\geq 2$ the following are equivalent:

  1. (a)

${\mathsf{CH}}$;

  2. (b)

for all well-placed $\mathbf{c}_{1},\dots,\mathbf{c}_{d}\in\mathbb{R}^{d}$
there are $X_{1},\dots,X_{d}$ covering $\mathbb{R}^{d}$ such that each $X_{i}$
is a $\sigma$-spray with center $\mathbf{c}_{i}$;

  3. (c)

there are well-placed $\mathbf{c}_{1},\dots,\mathbf{c}_{d}\in\mathbb{R}^{d}$
and $X_{1},\dots,X_{d}$ covering $\mathbb{R}^{d}$ such that each $X_{i}$ is a
$\sigma$-spray with center $\mathbf{c}_{i}$.

###### Proof.

(a)$\Rightarrow$(b) follows from Theorem 2.18; (b)$\Rightarrow$(c) is trivial,
while (c)$\Rightarrow$(a) is established by induction on $d\geq 2$.

When $d=2$ the result follows at once from Proposition 2.20. Suppose the
result holds for some $d$ towards proving it for $d+1$. Fix $\sigma$-sprays
$X_{1},\dots,X_{d+1}$ covering $\mathbb{R}^{d+1}$ and centered in well-placed
$\mathbf{c}_{1},\dots,\mathbf{c}_{d+1}$. Let $H\subseteq\mathbb{R}^{d}$ be the
hyperplane determined by the $\mathbf{c}_{i}$s. Let $H^{\prime}$ be a
hyperplane orthogonal to $\mathbf{c}_{d+1}-\mathbf{c}_{d}$, and let
$\pi\colon\mathbb{R}^{d+1}\to H^{\prime}$ be the orthogonal projection. By
projecting and glueing $X_{1}\cap H$, $X_{1}\cap H$, …, $(X_{d}\cup
X_{d+1})\cap H$ are $\sigma$-sprays centered in the points
$\pi(\mathbf{c}_{1})$, $\pi(\mathbf{c}_{2})$, …,
$\pi(\mathbf{c}_{d})=\pi(\mathbf{c}_{d+1})$ which are well-placed, and belong
to $H^{\prime}\cong\mathbb{R}^{d}$. By inductive assumption, ${\mathsf{CH}}$
holds. ∎

##  3\. Transforming sprays into linear objects

In this section we construct, for every $d\geq 2$, a continuous map $\Phi$
that transforms any spray of $\mathbb{R}^{d}$ with center $\mathbf{c}$ into a
set $A\subseteq\mathbb{R}^{d}$ such that $A\cap H$ is finite, for every
hyperplane $H$ orthogonal to some vector $\mathbf{u}$, and conversely. (The
vector $\mathbf{u}$ depends only on the point $\mathbf{c}$.) This is an
extension and an elaboration of the construction used by Schmerl when $d=2$ to
prove that if $\mathbb{R}^{2}$ is the union of $(n+2)$-many sprays with
collinear centers, then $2^{\aleph_{0}}\leq\aleph_{n}$ [Sch10, Theorem 7].

Let $d\geq 2$, and fix distinct points $\mathbf{p}_{1},\dots,\mathbf{p}_{d}$
in $\mathbb{R}^{d-1}$. For each $\mathbf{q}\in\mathbb{R}^{d-1}$ the vectors
$\mathbf{p}_{1}-\mathbf{q},\dots,\mathbf{p}_{d}-\mathbf{q}$ are linearly
dependent, and hence

(2)
$\mathcal{U}(\mathbf{q})\coloneqq\mathopen{\\{}{(u_{1},\dots,u_{d})\in\mathbb{R}^{d}}\boldsymbol{\mid}{u_{1}(\mathbf{p}_{1}-\mathbf{q})+\dots+u_{d}(\mathbf{p}_{d}-\mathbf{q})=\mathbf{0}}\mathclose{\\}}$

is a vector subspace of $\mathbb{R}^{d}$ of dimension $\geq 1$. Clearly this
space depends on the $\mathbf{p}_{i}$s, so we should write
$\mathcal{U}_{\mathbf{p}_{1},\dots,\mathbf{p}_{d}}(\mathbf{q})$, but the
$\mathbf{p}_{i}$s will be clear from the context. Observe that if
$\mathbf{e}_{i}\in\mathcal{U}(\mathbf{q})$ then $\mathbf{q}=\mathbf{p}_{i}$,
and hence if $\mathbf{q}$ is distinct from
$\mathbf{p}_{1},\dots,\mathbf{p}_{d}$ then no
$\mathbf{u}\in\mathcal{U}(\mathbf{q})\setminus\mathopen{\\{}{\mathbf{0}}\mathclose{\\}}$
is parallel to a basis vector of $\mathbb{R}^{d}$.

######  Theorem 3.1.

Suppose $\mathbf{p}_{1},\dots,\mathbf{p}_{d},\mathbf{q}\in\mathbb{R}^{d-1}$.
For every
$(u_{1},\dots,u_{d})\in\mathcal{U}(\mathbf{q})\setminus\mathopen{\\{}{\mathbf{0}}\mathclose{\\}}$
letting $b\coloneqq-\sum_{i=1}^{d}u_{i}$ and
$c\coloneqq-\sum_{i=1}^{d}u_{i}(\|\mathbf{p}_{i}\|^{2}-\|\mathbf{q}\|^{2})$ we
have that

$\forall{\mathbf{x}\in\mathbb{R}^{d-1}}\,\bigl{(}u_{1}\|\mathbf{x}-\mathbf{p}_{1}\|^{2}+\dots+u_{d}\|\mathbf{x}-\mathbf{p}_{d}\|^{2}+b\|\mathbf{x}-\mathbf{q}\|^{2}+c=0\bigr{)}.$

###### Proof.

For notational ease let $F_{i}(\mathbf{x})=\|\mathbf{x}-\mathbf{p}_{i}\|^{2}$
and $F(\mathbf{x})=\|\mathbf{x}-\mathbf{q}\|^{2}$. For
$\mathbf{x}\in\mathbb{R}^{d-1}$ and $1\leq i\leq d$

$\begin{split}F_{i}(\mathbf{x})-F(\mathbf{x})&=\|\mathbf{x}-\mathbf{p}_{i}\|^{2}-\|\mathbf{x}-\mathbf{q}\|^{2}\\\
&=\|\mathbf{x}\|^{2}+\|\mathbf{p}_{i}\|^{2}-2\mathbf{x}\boldsymbol{\cdot}\mathbf{p}_{i}-\|\mathbf{x}\|^{2}-\|\mathbf{q}\|^{2}+2\mathbf{x}\boldsymbol{\cdot}\mathbf{q}\\\
&=\|\mathbf{p}_{i}\|^{2}-\|\mathbf{q}\|^{2}-2\mathbf{x}\boldsymbol{\cdot}(\mathbf{p}_{i}-\mathbf{q}).\end{split}$

This implies

$\sum_{i=1}^{d}u_{i}(F_{i}(\mathbf{x})-F(\mathbf{x}))=2\mathbf{x}\boldsymbol{\cdot}\sum_{i=1}^{d}u_{i}(\mathbf{p}_{i}-\mathbf{q})+\sum_{i=1}^{d}u_{i}(\|\mathbf{p}_{i}\|^{2}-\|\mathbf{q}\|^{2})=-c.$

By definition of $b$ one obtains
$(\sum_{i=1}^{d}u_{i}F_{i}(\mathbf{x}))+bF(\mathbf{x})+c=0$ for all
$\mathbf{x}\in\mathbb{R}^{d-1}$, as required. ∎

The definition of $\aleph_{\delta}$-spray was given for $\mathbb{R}^{d}$, but
it can be adapted to the space

$\mathbb{H}^{d}\coloneqq\mathbb{R}^{d-1}\times(0;+\infty)\subseteq\mathbb{R}^{d}$

as follows: an $\aleph_{\delta}$-spray in $\mathbb{H}^{d}$ is a set
$X\subseteq\mathbb{H}^{d}$ together with a point $\mathbf{c}$, the center of
$X$, belonging to the hyperplane $\mathbb{R}^{d-1}\times\left\\{{0}\right\\}$
of $\mathbb{R}^{d}$, such that $\mathopen{|}S\cap
X\mathclose{|}<\aleph_{\delta}$ for any sphere $S$ of $\mathbb{R}^{d}$
centered in $\mathbf{c}$. (Observe that $\mathbf{c}$ does not belong to
$\mathbb{H}$.)

If $X_{1},X_{2},\dots$ are $\aleph_{\delta}$-sprays in $\mathbb{R}^{d}$ with
centers
$\mathbf{c}_{1},\mathbf{c}_{2},\ldots\in\mathbb{R}^{d-1}\times\mathopen{\\{}{0}\mathclose{\\}}$,
then $X_{1}\cap\mathbb{H}^{d},X_{2}\cap\mathbb{H}^{d},\dots$ are
$\aleph_{\delta}$-sprays in $\mathbb{H}^{d}$ with the same centers; moreover,
if $X_{1},X_{2},\dots$ cover $\mathbb{R}^{d}$, then
$X_{1}\cap\mathbb{H}^{d},X_{2}\cap\mathbb{H}^{d},\dots$ cover
$\mathbb{H}^{d}$. Therefore we may focus on covering $\mathbb{H}^{d}$ with
$\aleph_{\delta}$-sprays in $\mathbb{H}^{d}$ with centers on
$\mathbb{R}^{d-1}\times\mathopen{\\{}{0}\mathclose{\\}}$.

Let $\mathbf{c}_{1},\dots,\mathbf{c}_{d}$ be points in general position in
$\mathbb{R}^{d}$. These points belong to a hyperplane, and without loss of
generality we may assume that
$\mathbf{c}_{1},\dots,\mathbf{c}_{d}\in\mathbb{R}^{d-1}\times\mathopen{\\{}{0}\mathclose{\\}}$.
The set

$E^{d}\coloneqq\mathopen{\\{}{(r_{1},\dots,r_{d})}\boldsymbol{\mid}{\exists{\mathbf{x}\in\mathbb{H}^{d}}\,\forall{i\leq
d}\,(r_{i}=\|\mathbf{x}-\mathbf{c}_{i}\|^{2})}\mathclose{\\}}$

is an open subset of
$\mathbb{R}^{d}_{+}\coloneqq\mathopen{\\{}{(r_{1},\dots,r_{d})\in\mathbb{R}^{d}}\boldsymbol{\mid}{\forall{i\leq
d}\,(r_{i}>0)}\mathclose{\\}}$, and

(3) $\Phi\colon\mathbb{H}^{d}\to
E^{d},\qquad\mathbf{x}\mapsto(\|\mathbf{x}-\mathbf{c}_{1}\|^{2},\dots,\|\mathbf{x}-\mathbf{c}_{d}\|^{2})$

is a homeomorphism. Every element of $E=E^{d}$ determines a unique point in
$\mathbb{H}^{d}$, and this is the reason for retreating form $\mathbb{R}^{d}$
to $\mathbb{H}^{d}$.

The map $\Phi$ transforms spheres centered around $\mathbf{c}_{i}$ into
hyperplanes of $\mathbb{R}^{d}$ orthogonal to $\mathbf{e}_{i}$, and
conversely. To be specific—and recalling Notation 2.1—if $S$ is the sphere of
$\mathbb{R}^{d}$ centered in $\mathbf{c}_{i}$ of radius $r$, then
$\Phi[S\cap\mathbb{H}^{d}]$ is the intersection between $E$ and the hyperplane
$H_{i}(r^{2})$ of $\mathbb{R}^{d}$; conversely, $\Phi^{-1}[H_{i}(r^{2})\cap
E]=S\cap\mathbb{H}^{d}$. Therefore if $X\subseteq\mathbb{H}^{d}$ is an
$\aleph_{\delta}$-spray centered in $\mathbf{c}_{i}$, then $\Phi[X]\subseteq
E$ is such that $\mathopen{|}\Phi[X]\cap
H_{i}(r)\mathclose{|}<\aleph_{\delta}$ for every $r>0$; conversely if
$Y\subseteq E$ intersects all hyperplanes orthogonal to $\mathbf{e}_{i}$ in a
set of size $<\aleph_{\delta}$ then $\Phi^{-1}[Y]\subseteq\mathbb{H}^{d}$ is
an $\aleph_{\delta}$-spray centered in $\mathbf{c}_{i}$.

Suppose
$\mathbf{c}_{d+1}\in\mathbb{R}^{d-1}\times\mathopen{\\{}{0}\mathclose{\\}}$ is
distinct from $\mathbf{c}_{1},\dots,\mathbf{c}_{d}$. Let

$\check{E}=\mathopen{\\{}{(r_{1},\dots,r_{d+1})\in\mathbb{R}^{d+1}_{+}}\boldsymbol{\mid}{\exists{\mathbf{x}\in\mathbb{H}^{d}}\,\forall{i\leq
d+1}\,(r_{i}=\|\mathbf{x}-\mathbf{c}_{i}\|^{2})}\mathclose{\\}}.$

Then $\check{\Phi}\colon\mathbb{H}^{d}\to\check{E}$,
$\mathbf{x}\mapsto(\|\mathbf{x}-\mathbf{c}_{1}\|^{2},\dots,\|\mathbf{x}-\mathbf{c}_{d+1}\|^{2})$,
is a homeomorphism. Applying Theorem 3.1with $\mathbf{p}_{i}=\mathbf{c}_{i}$
and $\mathbf{q}=\mathbf{c}_{d+1}$, there are
$\mathbf{u}=(u_{1},\dots,u_{d})\in\mathcal{U}(\mathbf{c}_{d+1})\setminus\mathopen{\\{}{\mathbf{0}}\mathclose{\\}}$
and $b,c\in\mathbb{R}$, such that

$P=\mathopen{\\{}{(w_{1},\dots,w_{d+1})}\boldsymbol{\mid}{u_{1}w_{1}+\dots+u_{d}w_{d}+bw_{d+1}+c=0}\mathclose{\\}}$

is a hyperplane of $\mathbb{R}^{d+1}$. As $\mathbf{c}_{d+1}\neq\mathbf{c}_{i}$
for $i\leq d$, the vector $\mathbf{u}$ is not orthogonal to any
$\mathbf{e}_{i}$ for $1\leq i\leq d$, and the same is true of $P$. Therefore
the projection $\pi\colon\mathbb{R}^{d}\times\mathbb{R}\to\mathbb{R}^{d}$ is a
bijection between $P$ and $\mathbb{R}^{d}$. Let

$\check{E}=\mathopen{\\{}{(r_{1},\dots,r_{d+1})\in\mathbb{R}^{d+1}_{+}}\boldsymbol{\mid}{(r_{1},\dots,r_{d})\in
E\mathbin{\,\wedge\,}(r_{1},\dots,r_{d+1})\in P}\mathclose{\\}},$

that is: $\check{E}$ is the subset of $P$ that projects onto $E$, and
$\pi\mathop{\upharpoonright}\check{E}$ is a homeomorphism between $\check{E}$
and $E$. For any $k\in\mathbb{R}$ the set

$\begin{split}L(\mathbf{u},k)&\coloneqq\mathopen{\\{}{(r_{1},\dots,r_{d})}\boldsymbol{\mid}{u_{1}r_{1}+\dots+u_{d}r_{d}+bk+c=0}\mathclose{\\}}\\\
&=\mathopen{\\{}{(r_{1},\dots,r_{d})}\boldsymbol{\mid}{(r_{1},\dots,r_{d},k)\in
P}\mathclose{\\}}\end{split}$

is a hyperplane of $\mathbb{R}^{d}$ orthogonal to $\mathbf{u}$, and all
hyperplanes of $\mathbb{R}^{d}$ orthogonal to $\mathbf{u}$ are of this form.
Arguing as before, if $X\subseteq\mathbb{H}^{d}$ is an $\aleph_{\delta}$-spray
centered in $\mathbf{c}_{d+1}$ then $\Phi[X]$ is a subset of $E$ that
intersects every $L(\mathbf{u},k)$ in a set of size $<\aleph_{\delta}$;
conversely if $Y\subseteq E$ is such that every $L(\mathbf{u},k)$ intersects
$Y$ in $<\aleph_{\delta}$-many points, then $\Phi^{-1}[Y]$ is an
$\aleph_{\delta}$-spray centered in $\mathbf{c}_{d+1}$.

Let us summarize what we proved so far.

######  Theorem 3.2.

Suppose $\mathbf{c}_{1},\dots,\mathbf{c}_{d}$ are in general position in
$\mathbb{R}^{d}$, and without loss of generality we may assume that they
belong to $\mathbb{R}^{d-1}\times\mathopen{\\{}{0}\mathclose{\\}}$. There is
an open set $\emptyset\neq E^{d}\subseteq\mathbb{R}^{d}$ and a homeomorphism
$\Phi\colon\mathbb{H}^{d}\to E^{d}$ that transforms any
$\aleph_{\delta}$-spray of $\mathbb{H}^{d}$ centered in $\mathbf{c}_{i}$ into
a subset of $E^{d}$ that intersects any hyperplane orthogonal to
$\mathbf{e}_{i}$ in a set of size $<\aleph_{\delta}$, and conversely.

Let $\mathbf{c}\in\mathbb{R}^{d-1}\times\mathopen{\\{}{0}\mathclose{\\}}$ be
distinct from the $\mathbf{c}_{i}$s, and let
$(u_{1},\dots,u_{d})\in\mathcal{U}(\mathbf{c})\setminus\mathopen{\\{}{\mathbf{0}}\mathclose{\\}}$.
Then $\Phi$ maps any $\aleph_{\delta}$-spray centered in $\mathbf{c}$ into a
subset of $E$ that intersects any hyperplane orthogonal to
$\mathbf{u}=(u_{1},\dots,u_{d})$ in a set of size $<\aleph_{\delta}$, and
conversely.

The following picture when $d=2$ may help to visualize the previous
construction.

$\mathbf{c}_{0}$$\mathbf{c}_{1}$$\mathbf{c}$$\mathbb{H}$$\overset{\Phi}{\leadsto}$$E$

We can now sketch Schmerl’s argument that if the plane is covered with three
sprays with collinear centers, then ${\mathsf{CH}}$ holds. First of all we may
assume that the centers lie on the $x$-axis so that the three sprays cover
$\mathbb{H}^{2}$. Then $E^{2}$ can be covered with three sets
$A_{1},A_{2},A_{3}$ such that the intersection of $A_{i}$ with any line
orthogonal to some vector $\mathbf{u}_{i}$ is finite ($i=1,2,3$), and by a
theorem of Bagemihl [Bag61] this implies ${\mathsf{CH}}$. Our strategy is to
replace lines in $\mathbb{R}^{2}$ with hyperplanes in $\mathbb{R}^{d}$, and
this is the topic of the next section.

##  4\. Hyperplane sections

Over the last century several results were obtained, establishing connections
between the size of the continuum and elementary properties of the euclidean
spaces. The first such result is Sierpiński’s theorem form 1919, asserting
that ${\mathsf{CH}}$ is equivalent to a covering of the plane with two sets
such that the intersection of the first set with any vertical line is
countable and the intersection of every horizontal line with the second set is
countable (see Theorem 2.6(a)). In [Sie51] Sierpiński sharpened his previous
result by replacing “countable” with “finite”, but at the cost of increasing
the dimension: ${\mathsf{CH}}$ is equivalent to a decomposition
$A_{1},A_{2},A_{3}$ of $\mathbb{R}^{3}$ such that the intersection of any line
with direction $\mathbf{e}_{i}$ with $A_{i}$ is finite (Theorem 2.6(b)), and
this was quickly generalized by Kuratowski to higher dimensions [Kur51]:
$2^{\aleph_{0}}\leq\aleph_{n}$ if and only if there is a decomposition
$A_{1},\dots,A_{n+2}$ of $\mathbb{R}^{n+2}$ such that every line parallel to
$\mathbf{e}_{i}$ has finite intersection with $A_{i}$. In the sixties of the
last century, Bagemihl and Davies showed that Kuratowski’s result could be
proved for $\mathbb{R}^{2}$ by taking intersections with lines of prescribed
directions [Bag60, Bag61, Dav62, Dav63a, Bag68]. A line is a hyperplane in
$\mathbb{R}^{2}$, so Sierpiński’s result form 1919 could be stated as: if
$d=2$ then ${\mathsf{CH}}$ is equivalent to a decomposition of
$\mathbb{R}^{d}$ into $A_{1},\dots,A_{d}$ such that every hyperplane
orthogonal to $\mathbf{e}_{i}$ has countable intersection with $A_{i}$. This
result holds for all $d\geq 2$; Sierpiński states and proves it for $d=3$ in
[Sie51], see Corollary 4.7 below, and observes that the result for
$\mathbb{R}^{3}$ is false if “countable” is replaced by “finite”. Erdős,
Jackson, and Mauldin prove that ${\mathsf{CH}}$ is equivalent to
$\mathbb{R}^{3}$ being decomposable in five pieces $A_{1},\dots,A_{5}$ so that
every plane orthogonal to $\mathbf{u}_{i}$ has finite intersection with
$A_{i}$, where $\mathbf{u}_{1},\dots,\mathbf{u}_{5}$ are in general position
in $\mathbb{R}^{3}$, but no decomposition exists if we just allow four vectors
and four sets [EJM94, Corollary 6, Lemma 1]. In that paper it is stated (but
not proved) that an analogous result holds if ${\mathsf{CH}}$ is weakened to
$2^{\aleph_{0}}\leq\aleph_{n}$. Since we need a detailed analysis of the
positions of the various (hyper)planes we will state and prove these results
in full generality below. This section is devoted to the following

######  Problem 4.1.

Given $\mathbf{u}_{1},\dots,\mathbf{u}_{n}$ distinct, non-zero vectors of
$\mathbb{R}^{d}$, what conditions on the cardinality of $\mathbb{R}$ are
equivalent to the existence of $A_{1},\dots,A_{n}$ covering $\mathbb{R}^{d}$
such that each $A_{i}\cap H_{\mathbf{u}_{i}}(\mathbf{p})$ is finite (or
countable)?

First of all we need a result allowing us to replace the standard basis of
$\mathbb{R}^{d}$ with any basis.

######  Lemma 4.2.

Suppose $\mathbf{u}_{1},\dots,\mathbf{u}_{d}\in\mathbb{R}^{d}$ are linearly
independent, and $d\geq 2$. There is a linear isomorphism
$T\colon\mathbb{R}^{d}\to\mathbb{R}^{d}$ that maps every hyperplane orthogonal
to $\mathbf{u}_{i}$ to a hyperplane orthogonal to $\mathbf{e}_{i}$, and
conversely, for any $i\leq d$.

###### Proof.

Any linear $T$ preserves translations, so it is enough to consider hyperplanes
passing through the origin. In other words it is enough to find an injective
$T$ such that $T[H_{\mathbf{u}_{i}}(\mathbf{0})]\subseteq
H_{\mathbf{e}_{i}}(\mathbf{0})$. For $1\leq i\leq d$ let
$\mathbf{u}_{i}=(u_{i,1},\dots,u_{i,d})$. Consider the $d\times d$-matrix
$M=(u_{i,j})^{\mathrm{t}}$ that is the transpose of the matrix of the linear
transformation such that $\mathbf{e}_{i}\mapsto\mathbf{u}_{i}$ for
$i=1,\dots,d$. The matrix $M$ is invertible, as the $\mathbf{u}_{i}$s are
linearly independent, so the transformation $T$ induced by $M$ is an
isomorphism. If $\mathbf{x}=(x_{1},\dots,x_{d})\in
H_{\mathbf{u}_{i}}(\mathbf{0})$, then
$\mathbf{u}_{i}\boldsymbol{\cdot}\mathbf{x}^{\mathrm{t}}=\sum_{j=1}^{d}u_{i,j\,}x_{j}=0$,
and this is the $i$-th row of $M\mathbf{x}^{\mathrm{t}}$. We obtain that
$M\mathbf{x}^{\mathrm{t}}\in H_{\mathbf{e}_{i}}(\mathbf{0})$. Therefore
$T[H_{\mathbf{u}_{i}}(\mathbf{0})]\subseteq H_{\mathbf{e}_{i}}(\mathbf{0})$
for all $i<d$, and this is what we had to prove. ∎

######  Lemma 4.3.

Let $\varepsilon>0$. For all $X\subseteq\mathbb{R}$ with
$\aleph_{0}\leq\mathopen{|}X\mathclose{|}<\kappa\leq 2^{\aleph_{0}}$, there is
$S\subseteq(-\varepsilon;\varepsilon)$ such that
$\mathopen{|}S\mathclose{|}=\kappa$ and
$(X-X)\cap(S-S)=\mathopen{\\{}{0}\mathclose{\\}}$.

###### Proof.

Let $\tilde{X}$ be the subgroup of $(\mathbb{R},+)$ generated by $X$. Let
$T\subseteq\mathbb{R}$ be a transversal for the quotient, i.e. a set picking
exactly one element from each coset of $\mathbb{R}/\tilde{X}$. Then
$2^{\aleph_{0}}=\mathopen{|}\mathbb{R}/\tilde{X}\mathclose{|}=\mathopen{|}T\mathclose{|}=\mathopen{|}T\cap(-\varepsilon;\varepsilon)\mathclose{|}$.
Let $S\subseteq T\cap(-\varepsilon;\varepsilon)$ be of size $\kappa$.

If $s_{1}-s_{2}=x_{1}-x_{2}$ for some $s_{1},s_{2}\in S$ and $x_{1},x_{2}\in
X$, since $x_{1}-x_{2}\in\tilde{X}$ this means that the cosets
$s_{1}+\tilde{X}$ and $s_{2}+\tilde{X}$ are the same, so $s_{1}=s_{2}$ by
definition of $S$. Therefore
$(S-S)\cap(X-X)=\mathopen{\\{}{0}\mathclose{\\}}$. ∎

The next result asserts that if the sets $A_{i}\subseteq\mathbb{R}^{d}$ have
small intersection with the hyperplanes orthogonal to $\mathbf{u}_{i}$, and if
$2^{\aleph_{0}}$ is large enough, then there is a set
$Z\subseteq\mathbb{R}^{d}$ such that no translate of it can be covered by the
$A_{i}$s—in particular the $A_{i}$s do not cover $\mathbb{R}^{d}$.

######  Theorem 4.4.

Let $\delta\in\mathord{\mathrm{Ord}}$, $d\geq 3$ and let $N=2(d-1)$. Suppose
$\mathbf{u}_{1},\dots,\mathbf{u}_{N}$ are distinct, non-zero vectors of
$\mathbb{R}^{d}$. Suppose $A_{1},\dots,A_{N}$ are subsets of $\mathbb{R}^{d}$
such that for all $\mathbf{p}\in\mathbb{R}^{d}$, and for all $1\leq i\leq N$

$\mathopen{|}H_{\mathbf{u}_{i}}(\mathbf{p})\cap
A_{i}\mathclose{|}<\aleph_{\delta}.$

If $2^{\aleph_{0}}>\aleph_{\delta}$ then for every $\varepsilon>0$ there is
$Z\subseteq(-\varepsilon;\varepsilon)^{d}$ of size $\aleph_{\delta+1}$ such
that

$\forall{\mathbf{p}\in\mathbb{R}^{d}}\,(\mathbf{p}+Z\nsubseteq\textstyle\bigcup_{i=1}^{N}A_{i}).$

Before proving this let us draw a few corollaries. With the same notation as
before:

######  Theorem 4.5.

Let $\delta\in\mathord{\mathrm{Ord}}$, $d\geq 3$ and $N=2(d-1)$. Suppose
$\mathbf{u}_{1},\dots,\mathbf{u}_{N}$ are non-zero vectors of
$\mathbb{R}^{d}$. Suppose $D\subseteq\mathbb{R}^{d}$ is such that
$\operatorname{Int}(D)\neq\emptyset$. If $A_{1},\dots,A_{N}$ cover $D$ and for
all $\mathbf{p}\in\mathbb{R}^{d}$ and for all $1\leq i\leq N$

$\mathopen{|}H_{\mathbf{u}_{i}}(\mathbf{p})\cap
A_{i}\mathclose{|}<\aleph_{\delta}.$

Then $2^{\aleph_{0}}\leq\aleph_{\delta}$.

###### Proof.

Towards a contradiction, suppose $2^{\aleph_{0}}\geq\aleph_{\delta+1}$. Let
$\varepsilon>0$ be small enough so that
$\mathbf{p}+(-\varepsilon;\varepsilon)^{d}\subseteq D$ for some
$\mathbf{p}\in\mathbb{R}^{d}$, and, towards a contradiction, suppose
$A_{1},\dots,A_{N}$ are as in the statement. Let
$Z\subseteq(-\varepsilon;\varepsilon)^{d}$ be as in Theorem 4.4: then
$\mathbf{p}+Z$ is contained in $D$ but on other hand it is not contained in
$A_{1}\cup\dots\cup A_{N}$, a contradiction. ∎

The presence of the set $D$ in the statement of Theorem 4.5 may seem
unwarranted right now, but it will be crucial for the results in Section 5.
For the time being the reader can safely replace $D$ with $\mathbb{R}^{d}$
without losing much.

######  Theorem 4.6.

For all $d\geq 3$ and all $D\subseteq\mathbb{R}^{d}$ such that
$\emptyset\neq\operatorname{Int}(D)$, the following are equivalent:

  1. (a)

${\mathsf{CH}}$.

  2. (b)

For all non-zero vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{d}$ spanning
$\mathbb{R}^{d}$, there are $A_{1},\dots,A_{d}$ covering $D$ such that
$H_{\mathbf{u}_{i}}(\mathbf{p})\cap A_{i}$ is countable for all
$\mathbf{p}\in\mathbb{R}^{d}$ and $1\leq i\leq d$.

  3. (c)

There are non-zero vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{2d-2}$ and there
are $A_{1},\dots,A_{2d-2}$ covering $D$ such that
$H_{\mathbf{u}_{i}}(\mathbf{p})\cap A_{i}$ is countable for all
$\mathbf{p}\in\mathbb{R}^{d}$ and $1\leq i\leq 2d-2$.

###### Proof.

The implication (a)$\Rightarrow$(b) follows from Theorem 2.3 and Example 2.5,
while (b)$\Rightarrow$(c) is trivial—take $A_{d+1}=\dots=A_{2d-2}=\emptyset$.
The implication (c)$\Rightarrow$(a) follows from Theorem 4.5 when $\delta=1$.
∎

######  Corollary 4.7.

For all $d\geq 3$, ${\mathsf{CH}}$ is equivalent to the existence of
$A_{1},\dots,A_{d}$ covering $\mathbb{R}^{d}$ such that $H_{i}(x)\cap A_{i}$
is countable, for all $x\in\mathbb{R}$ and all $1\leq i\leq d$.

Cantor’s theorem says that $2^{\aleph_{0}}>\aleph_{\delta}$ if $\delta=0$, so
when $d=3$ Theorem 4.4 implies the following result, which is Lemma 1 of
[EJM94]:

######  Corollary 4.8.

Suppose $\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ are non-zero vectors of
$\mathbb{R}^{3}$. There are no $A_{1},\dots,A_{4}$ covering $\mathbb{R}^{3}$
such that $H_{\mathbf{u}_{i}}(\mathbf{p})\cap A_{i}$ is finite, for all
$\mathbf{p}\in\mathbb{R}^{3}$ and all $1\leq i\leq 4$.

######  Remark 4.9.

Corollary 4.8 is a negative result, asserting that $\mathbb{R}^{3}$ is not the
union of four sets such that each intersects any plane orthogonal to a given
vector in a finite set. But looking at the proof of Theorem 4.4, it could be
recast in a positive way as:

> Suppose $\Bbbk$ is an infinite field. If there are non-zero vectors
> $\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ of $\Bbbk^{3}$ and sets
> $A_{1},\dots,A_{4}$ covering $\Bbbk^{3}$ such that every hyperplane
> orthogonal to $\mathbf{u}_{i}$ has finite intersection with $A_{i}$, then
> $\mathopen{|}\Bbbk\mathclose{|}=\aleph_{0}$.

###### Proof of Theorem 4.4.

Let $A_{1},\dots,A_{N}$ and $\varepsilon>0$ be as in the statement. We
consider two cases, depending whether the $\mathbf{u}_{i}$s span
$\mathbb{R}^{d}$.

Case 1: the vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{N}$ do not span
$\mathbb{R}^{d}$.

Let $\mathbf{v}$ be a non-zero vector such that
$\mathbf{v}\boldsymbol{\cdot}\mathbf{u}_{k}=0$ for all $1\leq k\leq N$, and
let
$V=\mathopen{\\{}{r\mathbf{v}}\boldsymbol{\mid}{r\in\mathbb{R}}\mathclose{\\}}$.
Since $\mathopen{|}V\mathclose{|}=2^{\aleph_{0}}\geq\aleph_{\delta+1}$ we can
take $Z\subseteq V\cap(-\varepsilon;\varepsilon)^{d}$ of size
$\aleph_{\delta+1}$. Since $\mathbf{p}+V\subseteq
H_{\mathbf{u}_{i}}(\mathbf{p})$ we have that
$\mathopen{|}(\mathbf{p}+V)\cap\bigcup_{i=1}^{N}A_{i}\mathclose{|}<\aleph_{\delta}$,
and hence $\mathbf{p}+Z\nsubseteq\bigcup_{i=1}^{N}A_{i}$.

Case 2: the vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{N}$ span
$\mathbb{R}^{d}$.

Without loss of generality we may assume that
$(\mathbf{u}_{1},\dots,\mathbf{u}_{d})$ is a basis of $\mathbb{R}^{d}$. We
first prove the result under the additional assumption that

$(\mathbf{u}_{1},\dots,\mathbf{u}_{d})\text{ is the standard basis
}(\mathbf{e}_{1},\dots,\mathbf{e}_{d}).$

Let

$\displaystyle U$
$\displaystyle=\operatorname{span}\mathopen{\\{}{\mathbf{u}_{d+1},\dots,\mathbf{u}_{N}}\mathclose{\\}},$
$\displaystyle U_{j}$
$\displaystyle=\operatorname{span}\mathopen{\\{}{\mathbf{e}_{j},\mathbf{u}_{d+1},\dots,\mathbf{u}_{N}}\mathclose{\\}}\text{,
for }1\leq j\leq d.$

As $\dim U\leq d-2$ then $\dim U_{j}\leq d-1$, so there is a non-zero vector
orthogonal to $U_{j}$.

######  Claim 4.9.1.

There is $1\leq j\leq d$ such that there is $\mathbf{v}\neq\mathbf{0}$
orthogonal to $U_{j}$, and $\mathbf{v}$ not collinear to any $\mathbf{e}_{i}$.

###### Proof of the Claim.

Suppose $\dim U_{j}\leq d-2$ for some $j$, and that every
$\mathbf{v}\neq\mathbf{0}$ orthogonal to $U_{j}$ is collinear to some
$\mathbf{e}_{i}$. As the orthogonal complement of $U_{j}$ has dimension $\geq
2$, there are $\mathbf{e}_{i_{1}},\mathbf{e}_{i_{2}}$ orthogonal to $U_{j}$,
and hence $\mathbf{v}=\mathbf{e}_{i_{1}}+\mathbf{e}_{i_{2}}$ is orthogonal to
$U_{j}$ and is not collinear to any vector of the standard basis, against our
assumption.

Therefore we may assume that $\dim U_{j}=d-1$ for all $j$, and hence $\dim
U=d-2$. Towards a contradiction, suppose that for every
$j\in\mathopen{\\{}{1,\dots,d}\mathclose{\\}}$ there is a (necessarily unique)
$j^{*}\neq j$ such that $\mathbf{e}_{j^{*}}$ is orthogonal to $U_{j}$. The map
$j\mapsto j^{*}$ is not constant, so fix $i\neq j$ such that $i^{*}\neq
j^{*}$. As $d\geq 3$ pick $k\in\mathopen{\\{}{1,\dots,d}\mathclose{\\}}$
distinct from $i^{*},j^{*}$. Both vectors $\mathbf{e}_{i^{*}}$ and
$\mathbf{e}_{j^{*}}$ are orthogonal to $U$ and to $\mathbf{e}_{k}$, so both
are orthogonal to $U_{k}$, which is impossible as $\dim U_{k}=d-1$. ∎

Fix $j$ and $\mathbf{v}=(a_{1},\dots,a_{d})$ as in the Claim. By a re-
indexing, if needed, we may assume that $j=d$, that is $a_{d}=0$, and that
$a_{1},a_{2}$ are non-zero. By rescaling we may assume that

$\mathbf{v}=(1,a_{2},\dots,a_{d-1},0),\qquad a_{2}\neq 0$

and that

(4) $d\leq k\leq N\Rightarrow\mathbf{v}\boldsymbol{\cdot}\mathbf{u}_{k}=0.$

Let
$\nu\coloneqq\max\left\\{{1,\mathopen{|}a_{2}\mathclose{|},\dots,\mathopen{|}a_{d-1}\mathclose{|}}\right\\}$.
For $1\leq i\leq d$ let $X_{i}\subseteq(-\varepsilon/2;\varepsilon/2)$ such
that $\mathopen{|}X_{1}\mathclose{|}=\aleph_{\delta+1}$ and
$\mathopen{|}X_{i}\mathclose{|}=\aleph_{\delta}$ for all $2\leq i\leq d$. By
Lemma 4.3, let
$S\subseteq(-\frac{\varepsilon}{2\nu};\frac{\varepsilon}{2\nu})$ be of size
$\aleph_{\delta+1}$ such that
$(S-S)\cap(a_{2}^{-1}X_{2}-a_{2}^{-1}X_{2})=\mathopen{\\{}{0}\mathclose{\\}}$,
and let

$V=\mathopen{\\{}{s\mathbf{v}}\boldsymbol{\mid}{s\in S}\mathclose{\\}}.$

By (4), if $\mathbf{q}\in\mathbb{R}^{d}$ and $s\in S$ then

(5) $d\leq k\leq N\Rightarrow\mathbf{q}+V\subseteq
H_{\mathbf{u}_{k}}(\mathbf{q}+s\mathbf{v}).$

As $V\subseteq(-\varepsilon/2;\varepsilon/2)^{d}$ it follows that

$Z\coloneqq V+\textstyle\prod_{i=1}^{d}X_{i}$

is a subset of $(-\varepsilon;\varepsilon)^{d}$, and it is of cardinality
$\aleph_{\delta+1}$. Fix $\mathbf{p}=(p_{1},\dots,p_{d})\in\mathbb{R}^{d}$
and, towards a contradiction, suppose that
$\mathbf{p}+Z\subseteq\bigcup_{i=1}^{N}A_{i}$. Observe that $Z=\bigcup_{x\in
X_{1}}Z_{x}$, where

$Z_{x}\coloneqq
V+(\mathopen{\\{}{x}\mathclose{\\}}\times\textstyle\prod_{i=2}^{d}X_{i}).$

######  Claim 4.9.2.

The sets $Z_{x}$ are pairwise disjoint.

###### Proof.

If $Z_{x^{\prime}}\cap Z_{x^{\prime\prime}}\neq\emptyset$, then there are
$s^{\prime},s^{\prime\prime}\in S$, $x_{i}^{\prime},x_{i}^{\prime\prime}\in
X_{i}$ for $2\leq i\leq d$ such that

$s^{\prime}\mathbf{v}+(x^{\prime},x^{\prime}_{2},\dots,x^{\prime}_{d})=s^{\prime\prime}\mathbf{v}+(x^{\prime\prime},x_{2}^{\prime\prime},\dots,x^{\prime\prime}_{d}).$

The $2$-nd component yields that
$s^{\prime}a_{2}+x_{2}^{\prime}=s^{\prime\prime}a_{2}+x_{2}^{\prime\prime}$,
and hence

$s^{\prime}-s^{\prime\prime}=a_{2}^{-1}(x_{2}^{\prime\prime}-x_{2}^{\prime})\in(S-S)\cap(a_{2}^{-1}X_{2}-a_{2}^{-1}X_{2})=\left\\{{0}\right\\}.$

It follows that $s^{\prime}=s^{\prime\prime}=s$, and hence

$s\mathbf{v}+(x^{\prime},x_{2}^{\prime},\dots,x^{\prime}_{d})=s\mathbf{v}+(x^{\prime\prime},x_{2}^{\prime\prime},\dots,x^{\prime\prime}_{d}).$

Looking at the $1$-st component we obtain $s+x^{\prime}=s+x^{\prime\prime}$
and hence $x^{\prime}=x^{\prime\prime}$. ∎

Summarizing:

(6) $\mathbf{p}+Z=\bigcup_{x\in X_{1}}\mathbf{p}+Z_{x}\text{, and
}x^{\prime}\neq
x^{\prime\prime}\Rightarrow\mathbf{p}+Z_{x^{\prime}}\cap\mathbf{p}+Z_{x^{\prime\prime}}=\emptyset.$

######  Claim 4.9.3.

$\mathopen{|}A_{d}\cap(\mathbf{p}+Z)\mathclose{|}<\aleph_{\delta+1}$.

###### Proof.

As the last component of $\mathbf{v}$ is $0$, the set $V$ does not contribute
to $Z$ on the $d$-th coordinate, so for all $w\in\mathbb{R}$

$\begin{split}(\mathbf{p}+Z)\cap
H_{d}(p_{d}+w)&=(\mathbf{p}+\textstyle\prod_{i=1}^{d}X_{i})\cap
H_{d}(p_{d}+w)\\\ &=\mathbf{p}+\bigl{(}(\textstyle\prod_{i=1}^{d}X_{i})\cap
H_{d}(w)\bigr{)}\end{split}$

and this set is
$\mathbf{p}+(\prod_{i=1}^{d-1}X_{i})\times\left\\{{w}\right\\}$ if $w\in
X_{d}$, and it is empty if $w\notin X_{d}$. Therefore
$\mathbf{p}+Z\subseteq\bigcup_{w\in X_{d}}H_{d}(p_{d}+w)$ and

$A_{d}\cap(\mathbf{p}+Z)\subseteq\textstyle\bigcup_{w\in X_{d}}A_{d}\cap
H_{d}(p_{d}+w).$

By assumption $\mathopen{|}A_{d}\cap
H_{d}(p_{d}+w)\mathclose{|}<\aleph_{\delta}$ for any $w\in\mathbb{R}$, and
$\mathopen{|}X_{d}\mathclose{|}=\aleph_{\delta}$, so
$\mathopen{|}A_{d}\cap(\mathbf{p}+Z)\mathclose{|}\leq\aleph_{\delta}<\aleph_{\delta+1}$.
∎

As $\mathopen{|}X_{1}\mathclose{|}=\aleph_{\delta+1}$ and by (6), there is
$\bar{x}_{1}\in X_{1}$ such that $\mathbf{p}+Z_{\bar{x}_{1}}$ is disjoint from
$A_{d}$. This implies that

$\mathbf{p}+Z_{\bar{x}_{1}}\subseteq A_{1}\cup\dots\cup
A_{d-1}\cup\textstyle\bigcup_{k=d+1}^{N}A_{k}.$

######  Claim 4.9.4.

There is an $\bar{s}\in S$ such that
$\mathbf{p}+\bar{s}\mathbf{v}+(\mathopen{\\{}{\bar{x}_{1}}\mathclose{\\}}\times\prod_{i=2}^{d}X_{i})$
is disjoint from $\bigcup_{k=d+1}^{N}A_{k}$.

###### Proof.

Towards a contradiction, suppose that for all $s\in S$ and all $2\leq i\leq d$
there are $x_{i}(s)\in X_{i}$ such that

$\mathbf{p}+s\mathbf{v}+(\bar{x}_{1},x_{2}(s),\dots,x_{d}(s))\in\textstyle\bigcup_{k=d+1}^{N}A_{k}.$

As $\mathopen{|}X_{i}\mathclose{|}=\aleph_{\delta}$ for $2\leq i\leq d$, there
are $x_{i}^{\prime}\in X_{i}$ and $S^{\prime}\subseteq S$ of size
$\aleph_{\delta+1}$ such that $x_{i}(s)=x_{i}^{\prime}$ for all $s\in
S^{\prime}$. Therefore

$\forall{s\in
S^{\prime}}\,\bigl{(}\mathbf{p}+s\mathbf{v}+(\bar{x}_{1},x_{2}^{\prime},\dots,x_{d}^{\prime})\in\textstyle\bigcup_{k=d+1}^{N}A_{k}\bigr{)}.$

As $\mathopen{|}S^{\prime}\mathclose{|}=\aleph_{\delta+1}$ there is
$S^{*}\subseteq S^{\prime}$ of size $\aleph_{\delta+1}$, and there is $k$ with
$d<k\leq N$ such that

(7) $\forall{s\in
S^{*}}\,\bigl{(}\mathbf{p}+s\mathbf{v}+(\bar{x}_{1},x_{2}^{\prime},\dots,x_{d}^{\prime})\in
A_{k}\bigr{)}.$

Fix $s^{*}\in S^{*}$ and let

$Q_{k}\coloneqq
H_{\mathbf{u}_{k}}\bigl{(}\mathbf{p}+s^{*}\mathbf{v}+(\bar{x}_{1},x_{2}^{\prime},\dots,x_{d}^{\prime})\bigr{)}.$

By (5) with
$\mathbf{q}=\mathbf{p}+(\bar{x}_{1},x_{2}^{\prime},\dots,x_{d}^{\prime})$ we
have that

$\forall{s\in
S^{*}}\,\bigl{(}\mathbf{p}+s\mathbf{v}+(\bar{x}_{1},x_{2}^{\prime},\dots,x_{d}^{\prime})\in
Q_{k}\bigr{)}.$

Therefore by (7)
$\mathbf{p}+s\mathbf{v}+(\bar{x}_{1},x_{2}^{\prime},\dots,x_{d}^{\prime})\in
A_{k}\cap Q_{k}$ for all $s\in S^{*}$, and this is a contradiction, since
$\mathopen{|}A_{k}\cap Q_{k}\mathclose{|}<\aleph_{\delta}$, and yet it must
contain $\aleph_{\delta+1}$ points. ∎

Let $\bar{s}$ be as in Claim 4.9.4. Then
$\mathopen{|}\mathbf{p}+\bar{s}\mathbf{v}+(\left\\{{\bar{x}_{1}}\right\\}\times\prod_{i=2}^{d}X_{i})\mathclose{|}=\aleph_{\delta}$
and

(8)
$W_{1}\coloneqq\mathbf{p}+\bar{s}\mathbf{v}+(\left\\{{\bar{x}_{1}}\right\\}\times\textstyle\prod_{i=2}^{d}X_{i})\subseteq
A_{1}\cup\dots\cup A_{d-1}.$

As $W_{1}$ is included in $H_{1}(p_{1}+\bar{s}+\bar{x}_{1})$, and
$\mathopen{|}H_{1}(p_{1}+\bar{s}+\bar{x}_{1})\cap
A_{1}\mathclose{|}<\aleph_{\delta}$, then $\mathopen{|}W_{1}\cap
A_{1}\mathclose{|}<\aleph_{\delta}$. As
$\mathopen{|}X_{2}\mathclose{|}=\aleph_{\delta}$, there is some
$\bar{x}_{2}\in X_{2}$ such that the set
$W_{2}\coloneqq\mathbf{p}+\bar{s}\mathbf{v}+(\mathopen{\\{}{\bar{x}_{1}}\mathclose{\\}}\times\mathopen{\\{}{\bar{x}_{2}}\mathclose{\\}}\times\prod_{i=3}^{d}X_{i})$
is disjoint from $A_{1}$, and hence it is contained in $A_{2}\cup\dots\cup
A_{d-1}$ and in the hyperplane $H_{2}(p_{2}+\bar{s}a_{2}+\bar{x}_{2})$. As
before $\mathopen{|}W_{2}\cap A_{2}\mathclose{|}<\aleph_{\delta}$. Repeating
this argument we construct $\bar{x}_{3}\in X_{3}$, …, $\bar{x}_{d}\in X_{d}$
such that

$\mathbf{p}+\bar{s}\mathbf{v}+(\bar{x}_{1},\dots,\bar{x}_{d})\notin
A_{1}\cup\dots\cup A_{d-1}$

against (8). This concludes the proof, assuming that
$\mathbf{u}_{1},\dots,\mathbf{u}_{d}$ is the standard basis.

If $\mathbf{u}_{1},\dots,\mathbf{u}_{d}$ is an arbitrary basis of
$\mathbb{R}^{d}$, by Lemma 4.2 there is a linear injective transformation
$T\colon\mathbb{R}^{d}\to\mathbb{R}^{d}$ that maps every hyperplane orthogonal
to $\mathbf{u}_{i}$ to a hyperplane orthogonal to $\mathbf{e}_{i}$, for $1\leq
i\leq d$. The transformation $T$ maps parallel hyperplanes to parallel
hyperplanes, so for $d\leq k<N$ let $\bar{\mathbf{u}}_{k}$ be a vector such
that $T$ maps every hyperplane orthogonal to $\mathbf{u}_{k}$ to a hyperplane
orthogonal $\bar{\mathbf{u}}_{k}$. For $1\leq k\leq d$ set
$\bar{\mathbf{u}}_{k}$ to be $\mathbf{e}_{k}$. As $T^{-1}$ is continuous, pick
$\bar{\varepsilon}>0$ small enough so that
$T^{-1}[(-\bar{\varepsilon};\bar{\varepsilon})^{d}]\subseteq(-\varepsilon;\varepsilon)^{d}$.
Arguing as above there is a
$\bar{Z}\subseteq(-\bar{\varepsilon};\bar{\varepsilon})^{d}$ such that
$\mathbf{p}+\bar{Z}\nsubseteq\bigcup_{i=1}^{N}T[A_{i}]$ for all
$\mathbf{p}\in\mathbb{R}^{d}$. Then
$Z=T^{-1}[\bar{Z}]\subseteq(-\varepsilon;\varepsilon)^{d}$ is such that
$\mathbf{p}+Z\nsubseteq\bigcup_{i=1}^{N}A_{i}$ for all
$\mathbf{p}\in\mathbb{R}^{d}$. ∎

######  Theorem 4.10.

Let $\delta\in\mathord{\mathrm{Ord}}$, $d\geq 3$, $n\geq 0$, and let
$N=(n+1)(d-1)+1$. Suppose $\mathbf{u}_{1},\dots,\mathbf{u}_{N}$ are distinct,
non-zero vectors of $\mathbb{R}^{d}$, and that
$A_{1},\dots,A_{N}\subseteq\mathbb{R}^{d}$ are such that for all
$\mathbf{p}\in\mathbb{R}^{d}$

$\forall{1\leq k\leq N}\,\left(\mathopen{|}H_{\mathbf{u}_{k}}(\mathbf{p})\cap
A_{k}\mathclose{|}<\aleph_{\delta}\right).$

If $2^{\aleph_{0}}>\aleph_{\delta+n}$ then for every $\varepsilon>0$ there is
$Z_{n,\varepsilon}\subseteq(-\varepsilon;\varepsilon)^{d}$ of size
$\aleph_{\delta+n+1}$ such that

$\forall{\mathbf{p}\in\mathbb{R}^{d}}\,\left(\mathbf{p}+Z_{n,\varepsilon}\nsubseteq\textstyle\bigcup_{k=1}^{N}A_{k}\right).$

Let us draw some consequences from Theorem 4.10.

######  Theorem 4.11.

Let $\delta\in\mathord{\mathrm{Ord}}$, let $n\geq 1$, and let $d\geq 3$. Let
also $(n+1)(d-1)<N\leq(n+2)(d-1)$. For all $D\subseteq\mathbb{R}^{d}$ such
that $\operatorname{Int}(D)\neq\emptyset$, the following are equivalent:

  1. (a)

$2^{\aleph_{0}}\leq\aleph_{\delta+n}$.

  2. (b)

For all $\mathbf{u}_{1},\dots,\mathbf{u}_{(n+1)(d-1)+1}$ in general position
there are $A_{1},\dots,A_{(n+1)(d-1)+1}$ covering $D$ such that
$\mathopen{|}A_{k}\cap
H_{\mathbf{u}_{k}}(\mathbf{p})\mathclose{|}<\aleph_{\delta}$ for all
$\mathbf{p}\in\mathbb{R}^{d}$ and all $1\leq k\leq(n+1)(d-1)+1$.

  3. (c)

For all distinct non-zero vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{N}$ such
that $(n+1)(d-1)+1$ of them are in general position, there are
$A_{1},\dots,A_{N}$ covering $D$ such that $\mathopen{|}A_{k}\cap
H_{\mathbf{u}_{k}}(\mathbf{p})\mathclose{|}<\aleph_{\delta}$ for all
$\mathbf{p}\in\mathbb{R}^{d}$ and all $1\leq k\leq N$.

  4. (d)

There are distinct non-zero vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{N}$ and
there are $A_{1},\dots,A_{N}$ covering $D$ such that $\mathopen{|}A_{k}\cap
H_{\mathbf{u}_{k}}(\mathbf{p})\mathclose{|}<\aleph_{\delta}$ for all
$\mathbf{p}\in\mathbb{R}^{d}$ and all $1\leq k\leq N$.

###### Proof.

The implication (a)$\Rightarrow$(b) follows from the Erdős-Jackson-Mauldin
Theorem 2.3 and Example 2.5.

The implication (b)$\Rightarrow$(c) is easy. Let
$\mathbf{u}_{1},\dots,\mathbf{u}_{N}$ be as in (c). Without loss of generality
we may assume that $\mathbf{u}_{1},\dots,\mathbf{u}_{(n+1)(d-1)+1}$ are in
general position in $\mathbb{R}^{d}$, and let $A_{1},\dots,A_{(n+1)(d-1)+1}$
be as in (b). Letting $A_{(n+1)(d-1)+2}=\dots=A_{N}=\emptyset$ we have sets as
in (c).

The implication (c)$\Rightarrow$(d) is trivial, as any set of vectors in
general position spans the space, and any covering of $\mathbb{R}^{d}$ is a
covering of $D$.

Assume (d) and towards a contradiction suppose that
$2^{\aleph_{0}}>\aleph_{n+1}$. By a translation, we may assume that
$(-\varepsilon;\varepsilon)^{d}\subseteq D$. The hypotheses of Theorem 4.10
are satisfied so there is a set $Z\subseteq D$ that is not contained in
$\bigcup_{k=1}^{N}A_{k}$, so the sequence $A_{1},\dots,A_{N}$ does not cover
$D$. ∎

When $d=3$ Theorem 4.11 yields the following results—recall that vectors in
$\mathbb{R}^{3}$ are in general position if any three of them are linearly
independent. First we consider the case when the intersection of a plane with
a set is finite.

######  Corollary 4.12.

Let $D\subseteq\mathbb{R}^{3}$ be such that
$\emptyset\neq\operatorname{Int}(D)$, and let $n\geq 1$. The following are
equivalent:

  1. (a)

$2^{\aleph_{0}}\leq\aleph_{n}$;

  2. (b)

for all $\mathbf{u}_{1},\dots,\mathbf{u}_{2n+3}$ vectors in general position
in $\mathbb{R}^{3}$, then there are $A_{1},\dots,A_{2n+3}$ covering $D$ so
that every plane orthogonal to $\mathbf{u}_{i}$ intersects $A_{i}$ in a finite
set, for all $1\leq i\leq 2n+3$;

  3. (c)

there are non-zero vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{2n+4}$ in
$\mathbb{R}^{3}$, and sets $A_{1},\dots,A_{2n+4}$ covering $D$ such that every
plane orthogonal to $\mathbf{u}_{i}$ intersects $A_{i}$ in a finite set, for
all $1\leq i\leq 2n+4$.

In particular [EJM94, Sim97] ${\mathsf{CH}}$ is equivalent to

  * •

for all $\mathbf{u}_{1},\dots,\mathbf{u}_{5}$ in general position there are
$A_{1},\dots,A_{5}$ covering $\mathbb{R}^{3}$ such that any plane orthogonal
to $\mathbf{u}_{i}$ has finite intersection with $A_{i}$,

  * •

there are non-zero vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{6}$ and
$A_{1},\dots,A_{6}$ covering $\mathbb{R}^{3}$ such that any plane orthogonal
to $\mathbf{u}_{i}$ has finite intersection with $A_{i}$.

If the intersections between planes and sets are taken to be countable we have
a result analogous to Corollary 4.12, with a shift of $2$ in the number of
vectors/pieces.

######  Corollary 4.13.

Let $D\subseteq\mathbb{R}^{3}$ be such that
$\emptyset\neq\operatorname{Int}(D)$, let $n\geq 1$ and let $N$ be such that
$2n+2\leq N$. The following are equivalent:

  1. (a)

$2^{\aleph_{0}}\leq\aleph_{n}$;

  2. (b)

for all $\mathbf{u}_{1},\dots,\mathbf{u}_{2n+1}$ vectors in general position
there are $A_{1},\dots,A_{2n+1}$ covering $D$ so that every plane orthogonal
to $\mathbf{u}_{k}$ intersects $A_{k}$ in a countable set, for all $1\leq
k\leq 2n+1$;

  3. (c)

for all vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{2n+2}$ non-zero vectors such
that at least $2n+1$ of them are in general position, there are
$A_{1},\dots,A_{2n+2}$ covering $D$ such that every plane orthogonal to
$\mathbf{u}_{k}$ intersects $A_{k}$ in a countable set, for all $1\leq k\leq
2n+2$;

  4. (d)

there are vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{2n+2}$ and sets
$A_{1},\dots,A_{2n+2}$ covering $D$ such that every plane orthogonal to
$\mathbf{u}_{k}$ intersects $A_{k}$ in a countable set, for all $1\leq k\leq
2n+2$.

The tables in Figure 1 summarize the results proved so far. The table on the
left, for any dimension $d$ and any $n$ such that
$2^{\aleph_{0}}\leq\aleph_{n}$, gives $(n+1)(d-1)+1$ the minimum number of
vectors/pieces of a covering of $\mathbb{R}^{d}$ so that each piece has
_finite_ intersection with any hyperplane orthogonal to the given vector. The
maximum number for such decomposition is $(n+2)(d-1)$, the integer in the
square with coordinates $(d,n+1)$ decreased by $1$. The table on the right is
similar, but the intersections with the hyperplanes are _countable_.

$d$ $n$ | $1$ | $2$ | $3$ | $4$ | …
---|---|---|---|---|---
$2$ | $3$ | $4$ | $5$ | $6$ | …
$3$ | $5$ | $7$ | $9$ | $11$ | …
$4$ | $7$ | $10$ | $13$ | $16$ | …
$5$ | $9$ | $13$ | $16$ | $20$ | …
$\vdots$ |  |  |  |  | $\ddots$

$d$ $n$ | $1$ | $2$ | $3$ | $4$ | …
---|---|---|---|---|---
$2$ | $2$ | $3$ | $4$ | $5$ | …
$3$ | $3$ | $5$ | $7$ | $9$ | …
$4$ | $4$ | $7$ | $10$ | $13$ | …
$5$ | $5$ | $9$ | $13$ | $17$ | …
$\vdots$ |  |  |  |  | $\ddots$

Figure 1. Tables of the minimum number of pieces of a decomposition of
$\mathbb{R}^{d}$ equivalent to $2^{\aleph_{0}}\leq\aleph_{n}$: the one on the
left for _finite_ intersections, the one on the right for _countable_
intersections.

###### Proof of Theorem 4.10.

Fix $\delta\in\mathord{\mathrm{Ord}}$ and $d\geq 3$, and proceed by induction
on $n$.

The case $n=0$ is Theorem 4.4, so assume the result holds for some $\bar{n}$
towards proving it for $\bar{n}+1$.

Suppose $2^{\aleph_{0}}>\aleph_{\delta+\bar{n}+1}$. Let
$N=(\bar{n}+2)(d-1)+1$, and let $\mathbf{u}_{1},\dots,\mathbf{u}_{N}$ and
$A_{1},\dots,A_{N}$ be as in the statement. Given $\varepsilon>0$ we must
construct $Z=Z_{\bar{n}+1,\varepsilon}$ such that

$Z\subseteq(-\varepsilon;\varepsilon)^{d},\quad\mathopen{|}Z\mathclose{|}=\aleph_{\delta+\bar{n}+2},\quad\forall{\mathbf{p}\in\mathbb{R}^{d}}\,\bigl{(}\mathbf{p}+Z\nsubseteq\textstyle\bigcup_{k=1}^{N}A_{k}\bigr{)}.$

Let $\bar{N}=(\bar{n}+1)(d-1)+1$. Then
$\mathbf{u}_{1},\dots,\mathbf{u}_{\bar{N}}$ and $A_{1},\dots,A_{\bar{N}}$
satisfy the hypotheses of the statement for $\bar{n}$. By inductive assumption
there is
$\bar{Z}=Z_{\bar{n},\varepsilon/2}\subseteq(-\varepsilon/2;\varepsilon/2)^{d}$
of size $\aleph_{\delta+\bar{n}+1}$ such that

(9)
$\forall{\mathbf{p}\in\mathbb{R}^{d}}\,\bigl{(}\mathbf{p}+\bar{Z}\nsubseteq\textstyle\bigcup_{k=1}^{\bar{N}}A_{k}\bigr{)}.$

As $N-\bar{N}=d-1$, there is a non-zero vector $\mathbf{v}$ such that

(10) $\bar{N}<k\leq N\Rightarrow\mathbf{v}\boldsymbol{\cdot}\mathbf{u}_{k}=0.$

The subspace
$V=\mathopen{\\{}{r\mathbf{v}}\boldsymbol{\mid}{r\in\mathbb{R}}\mathclose{\\}}$
is of cardinality $2^{\aleph_{0}}\geq\aleph_{\delta+\bar{n}+2}$, and
$X=\mathopen{\\{}{r\in\mathbb{R}}\boldsymbol{\mid}{r\mathbf{v}\in\bar{Z}}\mathclose{\\}}$
is of cardinality $\aleph_{\delta+\bar{n}+1}$, so by Lemma 4.3 we obtain
$S\subseteq(-\varepsilon/2;\varepsilon/2)$ of size $\aleph_{\delta+\bar{n}+2}$
such that $(S-S)\cap(X-X)=\mathopen{\\{}{0}\mathclose{\\}}$. Letting
$Y=\mathopen{\\{}{s\mathbf{v}}\boldsymbol{\mid}{s\in S}\mathclose{\\}}$ then
$\mathopen{|}Y\mathclose{|}=\aleph_{\delta+\bar{n}+2}$ and
$Y\subseteq(-\varepsilon/2;\varepsilon/2)^{d}$ and

$(Y-Y)\cap(\bar{Z}-\bar{Z})=\left\\{{\mathbf{0}}\right\\}.$

Let

$Z\coloneqq Y+\bar{Z}=\textstyle\bigcup_{\mathbf{y}\in
Y}\mathbf{y}+\bar{Z}=\bigcup_{\mathbf{z}\in\bar{Z}}\mathbf{z}+Y.$

Observe that $Z\subseteq(-\varepsilon;\varepsilon)^{d}$ and is of cardinality
$\aleph_{\delta+\bar{n}+2}$. We must argue that
$\mathbf{p}+Z\nsubseteq\bigcup_{k=1}^{N}A_{k}$ for all
$\mathbf{p}\in\mathbb{R}^{d}$.

Towards a contradiction, let $\hat{\mathbf{p}}\in\mathbb{R}^{d}$ be such that
$\hat{\mathbf{p}}+Z\subseteq\bigcup_{k=1}^{N}A_{k}$.

######  Claim 4.13.1.

If $\mathbf{y},\mathbf{y}^{\prime}\in Y$ and
$(\mathbf{y}+\bar{Z})\cap(\mathbf{y}^{\prime}+\bar{Z})\neq\emptyset$ then
$\mathbf{y}=\mathbf{y}^{\prime}$.

###### Proof.

Suppose $\mathbf{y}+\mathbf{z}=\mathbf{y}^{\prime}+\mathbf{z}^{\prime}$ with
$\mathbf{y},\mathbf{y}^{\prime}\in Y$ and
$\mathbf{z},\mathbf{z}^{\prime}\in\bar{Z}$. Then
$\mathbf{y}-\mathbf{y}^{\prime}=\mathbf{z}^{\prime}-\mathbf{z}\in(Y-Y)\cap(\bar{Z}-\bar{Z})=\mathopen{\\{}{\mathbf{0}}\mathclose{\\}}$,
so $\mathbf{y}=\mathbf{y}^{\prime}$. ∎

Recall that $Y\subseteq V$ and by (10) $V$ is contained in
$H_{\mathbf{u}_{k}}(\mathbf{0})$, for $\bar{N}<k\leq N$. By assumption
$\mathopen{|}A_{k}\cap
H_{\mathbf{u}_{k}}(\mathbf{q})\mathclose{|}<\aleph_{\delta}$, so
$\mathopen{|}A_{k}\cap(\mathbf{q}+Y)\mathclose{|}<\aleph_{\delta}$, for each
$\mathbf{q}\in\mathbb{R}^{d}$. Therefore for any $\bar{N}<k\leq N$,

(11)
$\mathopen{|}A_{k}\cap(\hat{\mathbf{p}}+Z)\mathclose{|}=\mathopen{|}A_{k}\cap(\hat{\mathbf{p}}+Y+\bar{Z})\mathclose{|}=\left|\textstyle\bigcup_{\mathbf{z}\in\bar{Z}}A_{k}\cap(\hat{\mathbf{p}}+\mathbf{z}+Y)\right|\leq\aleph_{\delta+\bar{n}+1}.$

######  Claim 4.13.2.

There is $\hat{\mathbf{y}}\in Y$ such that
$(\hat{\mathbf{p}}+\hat{\mathbf{y}}+\bar{Z})\cap\bigcup_{\bar{N}<k\leq
N}A_{k}=\emptyset$.

###### Proof.

Towards a contradiction, suppose that for all $\mathbf{y}\in Y$ there are
$\mathbf{z}(\mathbf{y})\in\bar{Z}$ and $\bar{N}<k(\mathbf{y})\leq N$ such that
$\hat{\mathbf{p}}+\mathbf{y}+\mathbf{z}(\mathbf{y})\in A_{k(\mathbf{y})}$. As

$\aleph_{\delta+\bar{n}+2}=\mathopen{|}Y\mathclose{|}>\aleph_{\delta+\bar{n}+1}=\bar{Z}>\mathopen{|}\mathopen{\\{}{k}\boldsymbol{\mid}{\bar{N}<k\leq
N}\mathclose{\\}}\mathclose{|}$

there is $\hat{Y}\subseteq Y$ of size $\aleph_{\delta+\bar{n}+2}$, and
$\hat{\mathbf{z}}\in\bar{Z}$ and $\bar{N}<\hat{k}\leq N$ such that
$\mathbf{z}(\mathbf{y})=\hat{\mathbf{z}}$ and $k(\mathbf{y})=\hat{k}$ for all
$\mathbf{y}\in\hat{Y}$. Therefore
$\forall{\mathbf{y}\in\hat{Y}}\,(\hat{\mathbf{p}}+\mathbf{y}+\hat{\mathbf{z}}\in
A_{\hat{k}})$. By Claim 4.13.1 the map $\hat{Y}\to A_{\hat{k}}$,
$\mathbf{y}\mapsto\hat{\mathbf{p}}+\mathbf{y}+\hat{\mathbf{z}}$ is injective,
so
$\mathopen{|}A_{\hat{k}}\cap(\hat{\mathbf{p}}+Y+\bar{Z})\mathclose{|}\geq\mathopen{|}\hat{Y}\mathclose{|}=\aleph_{\delta+\bar{n}+2}$,
against (11). ∎

Fix $\hat{\mathbf{y}}$ as in Claim 4.13.2. Then
$(\hat{\mathbf{p}}+\hat{\mathbf{y}})+\bar{Z}\subseteq A_{1}\cup\dots\cup
A_{\bar{N}}$ against (9). Having reached a contradiction, we conclude that
$\mathbf{p}+Z\nsubseteq\bigcup_{k=1}^{N}A_{k}$ for all
$\mathbf{p}\in\mathbb{R}^{d}$. ∎

######  Remark 4.14.

Theorems 4.4 and 4.10 (together with Theorem 2.3) provide a complete solution
to Problem 4.1 when the vectors are in general position, yet some further
generalizations are possible. For the sake of readability we have opted for
less generality, and here we would like to mention some of these extensions.
(The proof of these results will appear elsewhere.)

Focusing on $d=3$ and $\delta=0$, the requirement on the size of the
intersections in Theorem 4.4 could be relaxed to

$\displaystyle\mathopen{|}A_{i}\cap
H_{\mathbf{u}_{i}}(\mathbf{p})\mathclose{|}<\aleph_{0}\quad i$
$\displaystyle=1,2,$ $\displaystyle\mathopen{|}A_{i}\cap
H_{\mathbf{u}_{i}}(\mathbf{p})\mathclose{|}\leq\aleph_{0}\quad i$
$\displaystyle=3,4,$

strengthening a theorem of Bagemihl [Bag59] that there are no
$A_{1},A_{2},A_{3}$ covering $\mathbb{R}^{3}$ so that all planes orthogonal to
$\mathbf{e}_{2},\mathbf{e}_{3}$ have countable intersections with
$A_{2},A_{3}$ and all planes orthogonal to $\mathbf{e}_{1}$ have finite
intersections with $A_{1}$. (By Theorem 2.6 and Lemma 4.2 ${\mathsf{CH}}$ is
equivalent to the fact that the sets $A_{i}\cap
H_{\mathbf{u}_{i}}(\mathbf{p})$ are countable for three distinct $i$s, if
$\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ are in general position.)

Another possible generalization of Theorem 4.4 is that for any
$\mathbf{u}_{1},\dots,\mathbf{u}_{n}\in\operatorname{span}(\mathbf{e}_{1},\mathbf{e}_{2})\setminus\left\\{{\mathbf{0}}\right\\}$
and
$\mathbf{v}_{1},\dots,\mathbf{v}_{m}\in\operatorname{span}(\mathbf{e}_{1},\mathbf{e}_{3})\setminus\left\\{{\mathbf{0}}\right\\}$,
there are no $A_{1},A_{2},A_{3}$, $B_{1},\dots,B_{n}$, $C_{1},\dots,C_{m}$
covering $\mathbb{R}^{3}$ such that each $A_{i}\cap
H_{\mathbf{e}_{i}}(\mathbf{p})$, $B_{j}\cap H_{\mathbf{u}_{j}}(\mathbf{p})$,
$C_{k}\cap H_{\mathbf{v}_{k}}(\mathbf{p})$ is finite. (Clearly
$(\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3})$ can be replaced with any
other basis.) This shows that Corollary 4.12 can fail badly, if the vectors
are not in general position.

On the other hand, not being in general position does not preclude a positive
result. For example ${\mathsf{CH}}$ is equivalent to the existence of
$A_{1},\dots,A_{6}$ covering $\mathbb{R}^{3}$ such that each $A_{i}\cap
H_{\mathbf{u}_{i}}(\mathbf{p})$ is finite, where
$\mathbf{u}_{1}=\mathbf{e}_{1}$, $\mathbf{u}_{2}=\mathbf{e}_{2}$,
$\mathbf{u}_{3}=\mathbf{e}_{3}$,
$\mathbf{u}_{4}\in\operatorname{span}(\mathbf{e}_{1},\mathbf{e}_{2})\setminus\mathopen{\\{}{\mathbf{0}}\mathclose{\\}}$,
$\mathbf{u}_{5}\in\operatorname{span}(\mathbf{e}_{1},\mathbf{e}_{4})\setminus\mathopen{\\{}{\mathbf{0}}\mathclose{\\}}$,
$\mathbf{u}_{6}\in\operatorname{span}(\mathbf{e}_{2},\mathbf{e}_{3})\setminus\mathopen{\\{}{\mathbf{0}}\mathclose{\\}}$.

##  5\. The main results

Combining Theorem 3.2 together with the results from Section 4 we are ready to
prove the results about sprays in $\mathbb{R}^{d}$.

######  Theorem 5.1.

Let $\mathbf{c}_{1},\dots,\mathbf{c}_{4}$ be four coplanar points in
$\mathbb{R}^{3}$. There are no $X_{1},\dots,X_{4}$ covering $\mathbb{R}^{3}$
such that $X_{i}$ is a spray with centers $\mathbf{c}_{i}$.

###### Proof.

Towards a contradiction, suppose $\mathbb{R}^{3}=X_{1}\cup\dots\cup X_{4}$.

If $\mathbf{c}_{1},\dots,\mathbf{c}_{4}$ belong to the same line $\ell$, let
$P$ be the plane orthogonal to $\ell$, and let $\pi\colon\mathbb{R}^{3}\to P$
be the orthogonal projection. Then $Y\coloneqq\bigcup_{1\leq i\leq
4}\pi[X_{i}]$ is a spray of $P$ centered in
$\pi(\mathbf{c}_{1})=\dots=\pi(\mathbf{c}_{4})\in P$, and since the $X_{i}$
cover $\mathbb{R}^{3}$, then $Y$ should cover $P\cong\mathbb{R}^{2}$, which is
absurd.

Therefore we may assume that $\mathbf{c}_{1},\dots,\mathbf{c}_{4}$ are not
collinear, so either $\mathbf{c}_{1},\mathbf{c}_{2},\mathbf{c}_{3}$ are not
collinear or $\mathbf{c}_{1},\mathbf{c}_{2},\mathbf{c}_{4}$ are not collinear.
If $\mathbf{c}_{1},\mathbf{c}_{2},\mathbf{c}_{3}$ are not collinear, then by
Theorem 3.2 there is an open $E^{3}\subseteq\mathbb{R}^{3}$ and there are
$A_{1},\dots,A_{4}$ covering $E^{3}$ such that $A_{i}\cap H_{i}(r)$ are finite
for $i=1,2,3$, and $A_{4}\cap H_{\mathbf{u}}(\mathbf{p})$ is finite, where
$\mathbf{u}=(u_{1},u_{2},u_{3})\neq\mathbf{0}$ is such that
$u_{1}(\mathbf{c}_{1}-\mathbf{c}_{4})+u_{2}(\mathbf{c}_{2}-\mathbf{c}_{4})+u_{3}(\mathbf{c}_{3}-\mathbf{c}_{4})=\mathbf{0}$
and $x$ and $\mathbf{p}$ range in $\mathbb{R}$ and $\mathbb{R}^{3}$,
respectively. The result follows from Corollaries 4.8. Finally if
$\mathbf{c}_{0},\mathbf{c}_{1},\mathbf{c}_{3}$ are not collinear, then
swapping $\mathbf{c}_{2}$ with $\mathbf{c}_{3}$ we fall back into the previous
case. ∎

Suppose $\mathbf{c}_{1},\dots,\mathbf{c}_{d}$ are distinct, well-placed points
in $\mathbb{R}^{d}$, belonging to some hyperplane $H$, and recall the vector
space $\mathcal{U}$ of (2).

######  Proposition 5.2.

Suppose $n\geq d$ and $\mathbf{c}_{1},\dots,\mathbf{c}_{n}$ are well-placed
points in $\mathbb{R}^{d}$. For $k\leq n$ let
$\mathbf{u}_{k}\in\mathcal{U}(\mathbf{c}_{k})\setminus\mathopen{\\{}{\mathbf{0}}\mathclose{\\}}$.
Then the vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{n}$ are in general position
in $\mathbb{R}^{d}$.

###### Proof.

Let $H$ be the hyperplane passing through the points $\mathbf{c}_{i}$s. As the
definition of $\mathcal{U}(\mathbf{c}_{k})$ is not affected by a translation
of $H$, we may assume that $\mathbf{0}\notin H$, so that
$(\mathbf{c}_{1},\dots,\mathbf{c}_{d})$ forms a basis of $\mathbb{R}^{d}$. Let
$\mathbf{u}_{k}=(u_{k}^{1},\dots,u_{k}^{d})$. We claim that
$\sum_{i=1}^{d}u_{k}^{i}\neq 0$, for any $k\leq n$. Otherwise

$\mathbf{0}=\sum_{i=1}^{d}u_{k}^{i}(\mathbf{c}_{i}-\mathbf{c}_{k})=\sum_{i=1}^{d}u_{k}^{i}\mathbf{c}_{i}=u_{k}^{1}\mathbf{c}_{1}+\sum_{i=2}^{d}u_{k}^{i}\mathbf{c}_{i}\\\
=-(\sum_{i=2}^{d}u_{k}^{i})\mathbf{c}_{1}+\sum_{i=2}^{d}u_{k}^{i}\mathbf{c}_{i}=\sum_{i=2}^{d}u_{k}^{i}(\mathbf{c}_{i}-\mathbf{c}_{1})$

shows that $\mathbf{c}_{1},\dots,\mathbf{c}_{d}$ are not in general position
in $H$, against our assumption. Scaling of vectors does not affect their
general position, so we may assume that $\sum_{i=1}^{d}u_{k}^{i}=1$ for all
$k\leq n$. Then
$\sum_{i=1}^{d}u_{k}^{i}(\mathbf{c}_{i}-\mathbf{c}_{k})=\mathbf{0}$ implies
that

$\mathbf{c}_{k}=\sum_{i=1}^{d}u_{k}^{i}\mathbf{c}_{i}$

that is to say: $(u_{k}^{1},\dots,u_{k}^{d})$ are the components of the vector
$\mathbf{c}_{k}$ with respect to the basis
$(\mathbf{c}_{1},\dots,\mathbf{c}_{d})$. Therefore the general position of the
$\mathbf{u}_{k}$s follows from the general position of the $\mathbf{c}_{k}$s
in $H$. ∎

The very same proof yields:

######  Proposition 5.3.

Let $H$ be a hyperplane of $\mathbb{R}^{d}$, and suppose that for all $k\geq
1$ the points $\mathbf{c}_{k}\in H$ are in general position in $H$. For all
$k\geq 1$ let
$\mathbf{u}_{k}\in\mathcal{U}(\mathbf{c}_{k})\setminus\mathopen{\\{}{\mathbf{0}}\mathclose{\\}}$.
Then the vectors $\mathbf{u}_{1},\mathbf{u}_{2},\dots$ are in general position
in $\mathbb{R}^{d}$.

######  Theorem 5.4.

Fix $d\geq 2$ and $n\geq 1$, and let $N=(n+1)(d-1)+1$ and $M=(n+2)(d-1)$. The
following are equivalent:

  1. (a)

$2^{\aleph_{0}}\leq\aleph_{n}$.

  2. (b)

For all distinct, well-placed points
$\mathbf{c}_{1},\dots,\mathbf{c}_{N}\in\mathbb{R}^{d}$ there are sprays
$X_{1},\dots,X_{N}$ covering $\mathbb{R}^{d}$ with $X_{i}$ centered in
$\mathbf{c}_{i}$.

  3. (c)

There are sprays $X_{1},\dots,X_{M}$ covering $\mathbb{R}^{d}$ with $X_{i}$
centered in $\mathbf{c}_{i}$ such that $\mathbf{c}_{1},\dots,\mathbf{c}_{N}$
are distinct and well-placed.

###### Proof.

(a)$\Rightarrow$(b) follows from Theorem 2.18 with $\delta=0$, and
(b)$\Rightarrow$(c) is trivial—set $X_{N+1}=\dots=X_{M}=\emptyset$ and
$\mathbf{c}_{N+1},\dots,\mathbf{c}_{M}$ any points belonging to the same
hyperplane passing through $\mathbf{c}_{1},\dots,\mathbf{c}_{N}$. So it is
enough to prove (c)$\Rightarrow$(a).

If $d=2$ then $N=M=n+2$ and (c) says that $\mathbb{R}^{2}$ is the union of
$n+2$ sprays with aligned centers, so $2^{\aleph_{0}}\leq\aleph_{n}$ follows
from [Sch10, Theorem 7]. Therefore we may assume that $d\geq 3$. Towards a
contradiction, assume $2^{\aleph_{0}}\geq\aleph_{n+1}$ and suppose that
$\mathbf{c}_{1},\dots,\mathbf{c}_{M}\in\mathbb{R}^{d}$ are well-placed, and
that $X_{1},\dots,X_{M}$ are sprays as above. For ease of notation, let
$\mathbf{u}_{i}=\mathbf{e}_{i}$ for $1\leq i\leq d$. By Proposition 5.2 the
vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{N}$ are in general position in
$\mathbb{R}^{d}$, and the map $\Phi\colon\mathbb{H}^{d}\to E^{d}$ transforms
the $X_{i}$ into a covering $A_{1},\dots,A_{M}$ of $E^{d}$ such that every
hyperplane orthogonal to $\mathbf{u}_{i}$ intersects $A_{i}$ in a finite set.
But this contradicts Theorem 4.11 with $\delta=0$. ∎

In particular:

######  Theorem 5.5.

The following are equivalent.

  1. (a)

${\mathsf{CH}}$.

  2. (b)

For all well-placed $\mathbf{c}_{1},\dots,\mathbf{c}_{5}\in\mathbb{R}^{3}$
there are sprays $X_{1},\dots,X_{5}$ covering $\mathbb{R}^{3}$ with $X_{i}$
centered in $\mathbf{c}_{i}$.

  3. (c)

There are sprays $X_{1},\dots,X_{6}$ covering $\mathbb{R}^{3}$ with $X_{i}$
centered in $\mathbf{c}_{i}$ such that $\mathbf{c}_{1},\dots,\mathbf{c}_{6}$
are well-placed.

More generally:

######  Theorem 5.6.

Fix $d\geq 2$ and $n\geq 1$, and let $M=(n+2)(d-1)$, $N=(n+1)(d-1)+1$,
$\bar{M}=(n+1)(d-1)=N-1$, and $\bar{N}=n(d-1)+1$. The following are
equivalent:

  1. (a)

$2^{\aleph_{0}}\leq\aleph_{n}$.

  2. (b)

For all well-placed points
$\mathbf{c}_{1},\dots,\mathbf{c}_{N}\in\mathbb{R}^{d}$ there are sprays
$X_{1},\dots,X_{N}$ covering $\mathbb{R}^{d}$ with $X_{i}$ centered in
$\mathbf{c}_{i}$.

  3. (c)

There are sprays $X_{1},\dots,X_{M}$ covering $\mathbb{R}^{d}$ with $X_{i}$
centered in $\mathbf{c}_{i}$ such that $\mathbf{c}_{1},\dots,\mathbf{c}_{M}$
are well-placed.

  4. (d)

For all well-placed points
$\mathbf{c}_{1},\dots,\mathbf{c}_{\bar{N}}\in\mathbb{R}^{d}$ there are
$\sigma$-sprays $X_{1},\dots,X_{N}$ covering $\mathbb{R}^{d}$ with $X_{i}$
centered in $\mathbf{c}_{i}$.

  5. (e)

There are $\sigma$-sprays $X_{1},\dots,X_{\bar{M}}$ covering $\mathbb{R}^{d}$
with $X_{i}$ centered in $\mathbf{c}_{i}$ such that
$\mathbf{c}_{1},\dots,\mathbf{c}_{\bar{M}}$ are well-placed.

Proposition 5.3 is the bridge connecting Problem 4.1 with

######  Problem 5.7.

Given $\mathbf{c}_{1},\dots,\mathbf{c}_{n}$ distinct points of $H$, a
hyperplane of $\mathbb{R}^{d}$, what conditions on the cardinality of
$\mathbb{R}$ are equivalent to the existence of $X_{1},\dots,X_{n}$ covering
$\mathbb{R}^{d}$, each $X_{i}$ a spray (or $\sigma$-spray) centered in
$\mathbf{c}_{i}$?

Theorem 5.6 yields a complete solution to Problem 5.7 when the
$\mathbf{c}_{i}$s are well-placed, i.e. in general position in $H$. Using the
results mentioned in Remark 4.14 it is possible to distill a few more results
on sprays in $\mathbb{R}^{3}$. Let us mention just two of them. The first is
that given four coplanar points
$\mathbf{c}_{1},\mathbf{c}_{2},\mathbf{c}_{3},\mathbf{c}_{4}$ in
$\mathbb{R}^{3}$, there exist no $X_{1},X_{2},X_{3},X_{4}$ covering
$\mathbb{R}^{3}$ such that $X_{1},X_{2}$ are sprays centered in
$\mathbf{c}_{1},\mathbf{c}_{2}$, and $X_{3},X_{4}$ are $\sigma$-sprays
centered in $\mathbf{c}_{3},\mathbf{c}_{4}$. The second result is that the six
points in $\mathbb{R}^{2}\times\left\\{{0}\right\\}$ of Figure 2 are not in
general position in the plane, and

  * •

no five sprays centered in these points can cover $\mathbb{R}^{3}$, but

  * •

${\mathsf{CH}}$ is equivalent to the existence of six sprays, centered in
these points, covering $\mathbb{R}^{3}$.

Figure 2.

###  5.1. Covering the space with infinitely many sprays

We have seen how covering the space with sprays with well-placed centers is
equivalent to giving an upper bound for the size of the continuum, the larger
the number of sprays, the weaker the bound. Next we show that, irrespective of
the size of the continuum, the space can be covered with $\aleph_{0}$-many
sprays with well-placed centers.

A drizzle in $\mathbb{R}^{d}$ with center $\mathbf{c}$ is a
$X\subseteq\mathbb{R}^{d}$ such that any sphere centered in $\mathbf{c}$
intersects $X$ in at most one point—thus a drizzle is a very sparse spray.

######  Theorem 5.8.

If $\mathopen{\\{}{\mathbf{c}_{n}}\boldsymbol{\mid}{n\geq 1}\mathclose{\\}}$
are well-placed points in $\mathbb{R}^{d}$, with $d\geq 2$, then there are
$X_{n}$ covering $\mathbb{R}^{d}$ such that $X_{n}$ is a drizzle centered in
$\mathbf{c}_{n}$.

###### Proof.

Without loss of generality we may assume that the $\mathbf{c}_{n}$s belong to
$\mathbb{R}^{d-1}\times\left\\{{0}\right\\}$. Pick
$\mathbf{u}_{n}\in\mathcal{U}(\mathbf{c}_{n})\setminus\mathopen{\\{}{\mathbf{0}}\mathclose{\\}}$,
where
$\mathcal{U}(\mathbf{q})=\mathcal{U}_{\mathbf{c}_{1},\dots,\mathbf{c}_{d}}(\mathbf{q})$.
By Proposition 5.3 the vectors $\mathbf{u}_{n}$ ($n\geq 1$) are in general
position in $\mathbb{R}^{d}$. By [Dav74] there are sets $A_{k}$ ($k\geq 1$)
such that for all $\mathbf{p}\in\mathbb{R}^{d}$,
$H_{\mathbf{u}_{k}}(\mathbf{p})\cap A_{k}$ has at most one point, and such
that $\mathbb{R}^{d}=\bigcup_{n\geq 1}A_{2n}=\bigcup_{n\geq 0}A_{2n+1}$. The
map $\Phi\colon\mathbb{H}^{d}\to E^{d}$ of (3) can be extended to the closures
of $\mathbb{H}^{d}$ and $E^{d}$, so we can assume that
$X_{2n}\coloneqq\Phi^{-1}[A_{2n}]$ is a subset of
$\operatorname{Cl}(\mathbb{H}^{d})=\mathbb{R}^{d-1}\times[0;+\infty)$, and
that $\bigcup_{n\geq 1}X_{2n}=\operatorname{Cl}(\mathbb{H}^{d})$. Letting
$\tau\colon\mathbb{R}^{d}\to\mathbb{R}^{d}$ be the reflection with respect to
the hyperplane $\mathbb{R}^{d-1}\times\mathopen{\\{}{0}\mathclose{\\}}$, let
$X_{2n+1}=\tau[\Phi^{-1}[A_{2n+1}]]$. Then $\bigcup_{n\geq
0}X_{2n+1}=\tau[\mathbb{H}^{d}]=\mathbb{R}^{d-1}\times(-\infty;0)$. Therefore
$\mathbb{R}^{d}=\bigcup_{n}X_{n}$, and by construction each $X_{n}$ is a
drizzle centered in $\mathbf{c}_{n}$. ∎

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