1 Sumsets and
$B_h$
-sets
Let A be a nonempty subset of an additive abelian group or semigroup G. For every positive integer h, the h-fold sumset of A is the set of all sums of h not necessarily distinct elements of A:
$$\begin{align*}hA = \underbrace{A+\cdots + A}_{h\text{ summands}} = \left\{ a'_1+\cdots + a'_h: a'_i \in A \text{ for all } i \in [1,h] \right\}. \end{align*}$$
For integers
$h \geq 2$
and
$n \geq 2$
, let
$$\begin{align*}\mathcal{X}_{h,n} = \left\{ \mathbf{x} = (x_1,\ldots, x_n) \in \mathbf{N}_0^n: \sum_{i=1}^n x_i = h \right\}. \end{align*}$$
For all
$\mathbf {x} = (x_1,\ldots , x_n) \in \mathcal {X}_{h,n}$
and
$\mathbf {y} = (y_1,\ldots , y_n) \in \mathcal {X}_{h,n}$
, we have
$$ \begin{align} \| \mathbf{x} - \mathbf{y}\|_{\infty} = \max\{| x_i - y_i | : i=1,\ldots, n\} \leq h. \end{align} $$
Let
$\mathbf {a} = (a_1,\ldots , a_n) \in G^n$
be a vector with distinct coordinates. Associated with the vector
$\mathbf {a}$
is the subset
$A = \{a_1,\ldots , a_n\}$
of G of cardinality n. For all
$\mathbf {x} \in \mathcal {X}_{h,n}$
and
$\mathbf {a} \in G^n$
, we define the dot product
$$\begin{align*}\mathbf{x}\cdot \mathbf{a} = \sum_{i=1}^n x_ia_i. \end{align*}$$
Then
$\mathbf {x}\cdot \mathbf {a} \in G$
. The h-fold sumset of
$A $
can be written in the form
$$\begin{align*}hA = \left\{ \mathbf{x} \cdot \mathbf{a} : \mathbf{x} \in \mathcal{X}_{h,n} \right\}. \end{align*}$$
The subset A of G is called a
$B_h$
-set if every element of G has at most one representation (up to permutation of the summands) as the sum of h not necessarily distinct elements of A.
$B_2$
-sets are also called Sidon sets. The study of
$B_h$
-sets of integers is a classical topic in combinatorial additive number theory (cf. [Reference Bose and Chowla1–Reference Ruzsa19]).
This article studies
$B_h$
-sets of real and complex numbers. Let
$K = \mathbf {R}$
or
$\mathbf {C}$
. Associated with every vector
$\mathbf {a} = (a_1,\ldots , a_n) \in K^n$
with distinct coordinates is the subset
$A = \{a_1, \ldots , a_n\}$
of K of cardinality n. The vector
$\mathbf {a}$
is called a
$B_h$
-vector if its associated set A is a
$B_h$
-set. Let
$\mathcal {B}_h$
be the set of
$B_h$
-vectors. If
$\mathbf {a} = (a_1, a_2, \ldots , a_n) \in \mathcal {B}_h$
, then
$\sigma \mathbf {a} = \left ( a_{\sigma (1)}, a_{\sigma (2)},\ldots , a_{\sigma (n)} \right ) \in \mathcal {B}_h$
for every permutation
$\sigma $
of
$\{1,\ldots , n\}$
.
The object of this article is to prove that “almost all” n-element subsets of K are
$B_h$
-sets in the sense that the set
$\mathcal {B}_h$
is an open and dense subset of
$K^n$
.
2 Open and dense subsets
Theorem 1 Let
$K = \mathbf {R}$
or
$\mathbf {C}$
. The set
$\mathcal {B}_h$
is an open subset of
$K^n$
.
Proof Let
$\mathbf {a} = (a_1,\ldots , a_n) \in K^n$
be a
$B_h$
-vector. If
$\mathbf {x},\mathbf {y} \in \mathcal {X}_{h,n}$
with
$\mathbf {x} \neq \mathbf {y}$
, then
$\mathbf {x}\cdot \mathbf {a} \neq \mathbf {y} \cdot \mathbf {a}$
and so
$$\begin{align*}0 < \Delta = \min \left\{ \left\| (\mathbf{x}\ - \mathbf{y}) \cdot \mathbf{a} \right\|_{\infty} : (\mathbf{x},\mathbf{y}) \in \mathcal{X}_{h,n}^2 \text{ and } \mathbf{x} \neq \mathbf{y} \right\}. \end{align*}$$
We shall prove that, for all vectors
$\mathbf {b} \in K^n$
with
$$\begin{align*}0 < \| \mathbf{b} \|_{\infty} < \frac{\Delta}{h}, \end{align*}$$
the vector
$$\begin{align*}\mathbf{a} + \mathbf{b} = \mathbf{c} \end{align*}$$
is a
$B_h$
-vector and so the set of
$B_h$
-vectors contains the open ball with center at
$\mathbf {a}$
and radius
$\Delta /h$
.
Let
$\mathbf {x},\mathbf {y} \in \mathcal {X}_{h,n}$
with
$\mathbf {x} \neq \mathbf {y}$
. If
$\mathbf {x}\cdot \mathbf {c} = \mathbf {y} \cdot \mathbf {c}$
, then
$$\begin{align*}\mathbf{x}\cdot (\mathbf{a} + \mathbf{b}) = \mathbf{y}\cdot (\mathbf{a} + \mathbf{b}) \end{align*}$$
and
$$\begin{align*}(\mathbf{x}\ - \mathbf{y}) \cdot \mathbf{a} = (\mathbf{y} - \mathbf{x}) \cdot \mathbf{b}. \end{align*}$$
Applying inequality (1.1), we obtain
$$ \begin{align*} \Delta & \leq \|(\mathbf{x}\ - \mathbf{y}) \cdot \mathbf{a} \|_{\infty} = \| (\mathbf{y} - \mathbf{x}) \cdot \mathbf{b} \|_{\infty} \\ & \leq \| \mathbf{y} - \mathbf{x} \|_{\infty} \| \mathbf{b} \|_{\infty} \\ & < h \left(\frac{\Delta}{h} \right) = \Delta \end{align*} $$
which is absurd. Therefore,
$\mathbf {x} \neq \mathbf {y}$
implies
$\mathbf {x}\cdot \mathbf {c} \neq \mathbf {y} \cdot \mathbf {c}$
and so
$\mathbf {c}$
is a
$B_h$
-vector. This completes the proof.
Lemma 1 For all
$\delta> 0$
there is a
$B_h$
-vector
$\mathbf {b}$
such that
$\|\mathbf {b}\|_{\infty } < \delta $
.
Proof If
$\mathbf {w} = (w_1,\ldots , w_n)$
is any
$B_h$
-vector in
$\mathbf {R}^n$
or
$\mathbf {C}^n$
with associated
$B_h$
-set
$W = \{w_1,\ldots , w_n\}$
, then, for every
$\lambda \neq 0$
, the “contraction”
$\lambda \ast W = \{\lambda w_i: i=1,\ldots , n\}$
is a
$B_h$
-set and the corresponding vector
$\lambda \mathbf {w} = (\lambda w_1,\ldots , \lambda w_n)$
is also a
$B_h$
-vector. Choosing
$0 < \lambda < \delta /\|\mathbf {w}\|_{\infty }$
and
$\mathbf {b} = \lambda \mathbf {w}$
gives
$$\begin{align*}\| \mathbf{b}\|_{\infty} = \| \lambda \mathbf{w}\|_{\infty} = |\lambda | \|\mathbf{w} \|_{\infty} < \left( \frac{\delta}{\|\mathbf{w}\|_{\infty} } \right) \|\mathbf{w}\|_{\infty} = \delta. \end{align*}$$
This completes the proof.
Theorem 2 Let
$K = \mathbf {R}$
or
$\mathbf {C}$
. The set
$\mathcal {B}_h$
is a dense subset of
$K^n$
.
Proof Let
$\mathbf {a} \in K^n$
be a vector that is not a
$B_h$
-vector. We shall prove that, for every
$\varepsilon> 0$
, there is a
$B_h$
-vector
$\mathbf {c} \in K^n$
such that
$\|\mathbf {c} - \mathbf {a}\|_{\infty } < \varepsilon $
.
We partition the set of pairs of distinct vectors in
$\mathcal {X}_{h,n}$
into two disjoint finite sets as follows:
$$\begin{align*}\mathcal{U} = \{ (\mathbf{x},\mathbf{y}) \in \mathcal{X}_{h,n}^2: \mathbf{x} \neq \mathbf{y} \text{ and } \mathbf{x}\cdot \mathbf{a} \neq \mathbf{y} \cdot \mathbf{a} \} \end{align*}$$
and
$$\begin{align*}\mathcal{V}= \{ (\mathbf{x},\mathbf{y}) \in \mathcal{X}_{h,n}^2: \mathbf{x} \neq \mathbf{y} \text{ and } \mathbf{x}\cdot \mathbf{a} = \mathbf{y} \cdot \mathbf{a}\}. \end{align*}$$
The set
$\mathcal {V}$
is nonempty because
$\mathbf {a}$
is not a
$B_h$
-vector. The set
$\mathcal {U}$
is nonempty because
$n \geq 2$
and
$(\mathbf {x},\mathbf {y}) \in \mathcal {U}$
with
$\mathbf {x} = (h,0,0,\ldots , 0)$
and
$\mathbf {y} = (0,h,0,\ldots , 0)$
. Then
$$\begin{align*}0 < \Delta = \min \left\{ \left\| (\mathbf{x} - \mathbf{y}) \cdot \mathbf{a} \right\|_{\infty} : (\mathbf{x},\mathbf{y}) \in \mathcal{U} \right\}. \end{align*}$$
By Lemma 1, for all
$\varepsilon> 0$
there is a
$B_h$
-vector
$\mathbf {b}$
in
$K^n$
such that
$$\begin{align*}\|\mathbf{b}\|_{\infty} < \min \left( \varepsilon,\frac{\Delta}{h} \right). \end{align*}$$
Applying inequality (1.1), for all pairs
$(\mathbf {x},\mathbf {y}) \in \mathcal {X}_{h,n}^2$
, we have
$$ \begin{align*} \| ( \mathbf{x} - \mathbf{y} )\cdot \mathbf{b} \|_{\infty} & \leq \| \mathbf{x} - \mathbf{y} \|_{\infty} \|\mathbf{b} \|_{\infty} < h \left( \frac{\Delta}{h} \right) = \Delta. \end{align*} $$
The vector
$$\begin{align*}\mathbf{a} + \mathbf{b} = \mathbf{c} \end{align*}$$
satisfies
$$\begin{align*}\|\mathbf{c} - \mathbf{a}\|_{\infty} = \|\mathbf{b}\|_{\infty} < \varepsilon. \end{align*}$$
We shall prove that
$\mathbf {c}$
is a
$B_h$
-vector. Equivalently, we shall prove that
$\mathbf {x}\cdot \mathbf {c} \neq \mathbf {y}\cdot \mathbf {c}$
for all pairs
$(\mathbf {x}, \mathbf {y}) \in \mathcal {X}_{h,n}^2$
with
$\mathbf {x} \neq \mathbf {y}$
.
If
$(\mathbf {x},\mathbf {y}) \in \mathcal {U}$
, then
$\mathbf {x}\cdot \mathbf {a} \neq \mathbf {y} \cdot \mathbf {a}$
and
$$ \begin{align*} \| \mathbf{x} \cdot \mathbf{c} - \mathbf{y} \cdot \mathbf{c}\|_{\infty} & = \| (\mathbf{x} \cdot \mathbf{a} - \mathbf{y} \cdot \mathbf{a} ) + ( \mathbf{x} \cdot \mathbf{b} - \mathbf{y} \cdot \mathbf{b} ) \|_{\infty} \\ & \geq \| (\mathbf{x} - \mathbf{y} ) \cdot \mathbf{a} \|_{\infty} - \| ( \mathbf{x} - \mathbf{y}) \cdot \mathbf{b} \|_{\infty} \\ &> \Delta - \Delta = 0. \end{align*} $$
If
$(\mathbf {x},\mathbf {y}) \in \mathcal {V}$
, then
$\mathbf {x} \cdot \mathbf {a} = \mathbf {y} \cdot \mathbf {a}$
. Because
$\mathbf {x} \neq \mathbf {y}$
and
$\mathbf {b}$
is a
$B_h$
-vector, we have
$\mathbf {x} \cdot \mathbf {b} \neq \mathbf {y} \cdot \mathbf {b}$
. It follows that
$$ \begin{align*} \| \mathbf{x} \cdot \mathbf{c} - \mathbf{y} \cdot \mathbf{c}\|_{\infty} & = \| (\mathbf{x} \cdot \mathbf{a} - \mathbf{y} \cdot \mathbf{a} ) + ( \mathbf{x} \cdot \mathbf{b} - \mathbf{y} \cdot \mathbf{b} ) \|_{\infty} \\ & = \| \mathbf{x} \cdot \mathbf{b} - \mathbf{y} \cdot \mathbf{b} \|_{\infty} \\ &> 0. \end{align*} $$
This completes the proof.
Combining Theorems 1 and 2 gives the following result.
Theorem 3 Let
$K = \mathbf {R}$
or
$\mathbf {C}$
. The set
$\mathcal {B}_h$
is a dense open subset of
$K^n$
.
Theorem 4 Let
$\mathcal {B}_h$
be the set of
$B_h$
-vectors in
$K^n$
. The set
$$\begin{align*}\mathcal{B}_{\infty} = \bigcap_{h=1}^{\infty} \mathcal{B}_h \end{align*}$$
is a dense subset of
$K^n$
.
Proof This is simply an application of Baire’s theorem in functional analysis.
3
$B_h[g]$
-sets
Let g and h be positive integers. The subset A of an additive abelian semigroup G is called a
$B_h[g]$
-set if every element of G has at most g representations as the sum of h not necessarily distinct elements of A. The
$B_2[1]$
-sets are the
$B_2$
-sets. If A is a
$B_h[g]$
-set, then A is also a
$B_h[g']$
-set for all
$g' \geq g$
. In particular, every
$B_h$
-set is a
$B_h[g]$
-set.
Let
$K = \mathbf {R}$
or
$\mathbf {C}$
. The vector
$\mathbf {a} = (a_1,\ldots , a_n) \in K^n$
is a
$B_h[g]$
-vector if the set
$A = \{a_1,\ldots , a_n\}$
is an n-element
$B_h[g]$
-set in K. Every
$B_h$
-vector is a
$B_h[g]$
-vector. Because the set of
$B_h$
-vectors is dense in
$K^n$
, it follows that the set of
$B_h[g]$
-vectors is also dense in
$K^n$
.
Problem Is the set of
$B_h[g]$
-vectors also open in
$K^n$
?
