Limits and Decomposition of de Bruijn’s Additive Systems

Abstract

An additive system for the nonnegative integers is a family $(A_i)_{i\in I}$ of sets of nonnegative integers with $0 \in A_i$ for all $i \in I$ such that every nonnegative integer can be written uniquely in the form $\sum _{i\in I} a_i$ with $a_i \in A_i$ for all i and $a_i \ne 0$ for only finitely many i. In 1956, de Bruijn proved that every additive system is constructed from an infinite sequence $(g_i )_{i \in \mathbf N}$ of integers with $g_i \ge 2$ for all i or is a contraction of such a system. This paper discusses limits and the stability of additive systems and also describes the “uncontractable” or “indecomposable” additive systems.

1 Additive Systems and de Bruijn’s Theorem

Let $\mathbf N_0$ and $\mathbf N$ denote the sets of nonnegative integers and positive integers, respectively. For real numbers a and b, we define the interval of integers $[a,b) = \{x\in \mathbf Z: a \le x < b\}$ and $[a,b] = \{x\in \mathbf Z: a \le x \le b\}$.

Let I be a nonempty finite or infinite set, and let $\mathcal A= (A_i)_{i\in I}$ be a family of sets of integers with $0 \in A_i$ and $|A_i| \ge 2$ for all $i \in I$. Each set $A_i$ can be finite or infinite. The sumset $S = \sum _{i\in I} A_i$ is the set of all integers n that can be represented in the form $n = \sum _{i\in I} a_i$, where $a_i \in A_i$ for all $i \in I$ and $a_i \ne 0$ for only finitely many $i \in I$. If every element of S has a unique representation in the form $n = \sum _{i\in I} a_i$, then we call $\mathcal A$ a unique representation system for S, and we write $S = \bigoplus _{i\in I} A_i$.

In a unique representation system $\mathcal A$ for S, we have $A_i \cap A_j = \{ 0 \}$ for all $i \ne j$. The condition $|A_i| \ge 2$ for all $i\in I$ implies that $A_i = S$ for some $i\in I$ if and only if $|I|=1$. Moreover, if $I^{\flat } \subseteq I$ and $S = \sum _{i\in I^{\flat }} A_i$, then $S = \bigoplus _{i\in I^{\flat }} A_i$ and $I = I^{\flat }$.

The family $\mathcal A= (A_i)_{i\in I}$ is an additive system if $\mathcal A$ is a unique representation system for the set of nonnegative integers, that is, if $\mathbf N_0 = \bigoplus _{i\in I} A_i$. The following lemma follows immediately from the definition of an additive system.

Lemma 1

Let $\mathcal B= (B_j )_{j\in J}$ be an additive system. If $\{J_i\}_{i\in I}$ is a partition of J into pairwise disjoint nonempty sets, and if

$$ A_i= \sum _{j \in J_i} B_j $$

then $\mathcal A= (A_i )_{i\in I}$ is an additive system.

The additive system $\mathcal A$ obtained from the additive system $\mathcal B$ by the partition procedure described in Lemma 1 is called a contraction of $\mathcal B$. (In [1], de Bruijn called $\mathcal A$ a degeneration of $\mathcal B$ .) If $I=J$ and if $\sigma$ is a permutation of J such that $J_i =\{\sigma (i) \}$ for all $i \in J$, then $\mathcal A$ and $\mathcal B$ contain exactly the same sets. Thus, every additive system is a contraction of itself. An additive system $\mathcal A$ is a proper contraction of $\mathcal B$ if at least one set $A_i \in \mathcal A$ is the sum of at least two sets in $\mathcal B$.

Let X be a set of integers, and let g be an integer. The dilation of X by g is the set $g*X = \{g x : x \in X \}$.

Lemma 2

Let $\mathcal B= (B_j )_{j\in J}$ be an additive system and let $I = \{i_0\} \cup J$, where $i_0 \notin J$. If

$$ A_{i_0} = [0,g) $$

and

$$ A_j = g*B_j \qquad \text {for all } j \in J $$

then $\mathcal A= (A_i )_{i\in I}$ is an additive system.

The additive system $\mathcal A$ obtained from the additive system $\mathcal B$ by the procedure described in Lemma 2 is called the dilation of $\mathcal B$ by g.

There are certain additive systems that de Bruijn called British number systems . A British number system is an additive system constructed from an infinite sequence of integers according to the algorithm in Theorem 1 below. de Bruijn [1] proved that British number systems are essentially the only additive systems.

Theorem 1

Let $(g_i )_{i \in \mathbf N}$ be an infinite sequence of integers such that $g_i \ge 2$ for all $i \ge 1$. Let $G_0 = 1$ and, for $i \in \mathbf N$, let $G_i = \prod _{j=1}^i g_j$ and

$$ A_i = \{ 0, G_{i-1}, 2 G_{i-1},\ldots , (g_i-1)G_{i-1} \} = G_{i-1} *[0,g_i). $$

Then $\mathcal A= (A_i )_{i \in \mathbf N}$ is an additive system.

Theorem 2

Every additive system is a British number system or a proper contraction of a British number system.

The proof of Theorem 2 depends on the following fundamental lemma.

Lemma 3

Let $\mathcal A= (A_i)_{i\in I}$ be an additive system with $|I| \ge 2$, and let $i_1$ be the unique element of I such that $1 \in I_{i_1}$. There exist an integer $g \ge 2$ and a family of sets $\mathcal B= (B_i)_{i\in I}$ such that

$$ A_{i_1} = [0, g) \oplus g*B_{i_1} $$

and, for all $i \in I \setminus \{i_1\}$,

$$ A_i = g*B_i. $$

If $B_{i_1} = \{ 0\}$, then $\mathcal B= (B_i)_{i\in I\setminus \{i_1\}}$ is an additive system, and $\mathcal A$ is the dilation of the additive system $\mathcal B$ by the integer g. If $B_{i_1} \ne \{ 0\}$, then $\mathcal B= (B_i)_{i\in I}$ is an additive system and $\mathcal A$ is a contraction of the additive system $\mathcal B$ dilated by g.

For proofs of Lemmas 1, 2, and 3 and Theorems 1 and 2, see Nathanson [4].

This paper gives a refinement of de Bruijn’s theorem. Every additive system is a contraction of a British number system, but even a British number system can be a proper contraction of another British number system. An additive system that is not a proper contraction of another number system will be called indecomposable. In Sect. 3, we describe all indecomposable British number systems. Unsurprisingly, there is a one-to-one correspondence between indecomposable British number systems and infinite sequences of prime numbers.

In Sect. 4, we define the limit of a sequence of additive systems and discuss the stability of British number systems.

Maltenfort [2] and Munagi [3] have also studied de Bruijn’s additive systems.

2 Decomposable and Indecomposable Sets

The set A of integers is a proper sumset if there exist sets B and C of integers such that $|B| \ge 2$, $|C| \ge 2$, and $A = B + C$. For example, if u and v are integers and $v-u \ge 3$, then the interval [u, v) is a proper sumset:

$$ [u, v) = [0,i) + [u, v + 1 -i) $$

for every $i \in [2,v-u)$.

The set A of integers is decomposable if there exist sets B and C such that (B, C) is a unique representation system for A, that is, if $|B| \ge 2$, $|C| \ge 2$, and $A=B \oplus C$. A decomposition $A = B \oplus C$ is also called a tiling of A by B. For example,

$$ [0,12) = \{0,3\} \oplus \{0,1,2,6,7,8\}. $$

If $A = B \oplus C$ is a decomposition, then $|A| = |B| \ |C|$ and so the integer |A| is composite.

Let $n \ge 2$ and consider the interval of integers $A = [0,n)$. A proper divisor of n is a divisor d of n such that $1< d < n$. Associated to every proper divisor d of n is the decomposition

$$\begin{aligned}{}[0,n) = [0, d) \oplus d*[0, n/d). \end{aligned}$$

(1)

This is simply the division algorithm for integers. The number of decompositions of type (1) is the number of proper divisors d of n. There is exactly one such decomposition if and only if the integer n has a unique proper divisor if and only if n is the square of a prime number.

Lemma 4

Let $n \ge 2$. The interval [0, n) is indecomposable if and only if n is prime.

Proof

If n is prime then [0, n) is indecomposable, and if n is composite, then [0, n) is decomposable.

If $A=B \oplus C$ and g is a nonzero integer, then $g *A= g *B \oplus g *C$, and so every dilation of a decomposable set is decomposable.

The translate of the set A by an integer t is the set

$$ A+t = \{a+t:a\in A \}. $$

Let $t_1,t_2 \in \mathbf Z$ with $t = t_1+t_2$. If $A = B+C$, then

$$ A+t = (B+t_1) + (C+t_2). $$

In particular, $A+t = (B+t) + C$. If $A=B \oplus C$, then $A + t= (B + t) \oplus C$, and so every translate of a decomposable set is decomposable. Similarly, if $A=B \oplus C$, then $A = (B - t) \oplus (C+t)$ for every integer t.

Let A be a set of nonnegative integers with $0 \in A$, and let B and C be sets of integers with $A = B \oplus C$. Let $t = \min (B)$. Defining $B' = B-t$ and $C' = C+t$, we obtain $A = B' \oplus C'$. Because $\min ( B') = 0$, we obtain

$$ 0 = \min (A) = \min (B') + \min (C') = 0 + \min (C') = \min (C') $$

and so $B'$ and $C'$ are sets of nonnegative integers with $0 \in B' \cap C'$.

Not every set with a composite number of elements is decomposable. For example, the n-element set $\{0,1,2,2^2,\ldots , 2^{n-2}\}$ is indecomposable for every $n \ge 2$. This is a special case of the following result.

Lemma 5

Let $m \ge 2$. Let A be a set of integers that contains integers $a_0$ and $a_1$ such that $a_0 \not \equiv a_1 \pmod {m}$, and $a \equiv a_0 \pmod {m}$ for all $a \in A\setminus \{a_1\}$. The set A is indecomposable.

Proof

The distinct congruence classes $a_0 \pmod {m}$ and $a_1\pmod {m}$ contain elements of A. Let B and C be sets of integers such that $A = B +C$ with $|B|, |C| \ge 2$. If B is contained in the congruence class $r \pmod {m}$ and C is contained in the congruence class $s \pmod {m}$, then $B+C$ is contained in the congruence class $r+s \pmod {m}$, and so $A \ne B+C$ (because A intersects two congruence classes). Therefore, at least one of the sets B and C must contain elements from distinct congruence classes modulo m. Let $b_1,b_2 \in B$ with $b_1 \not \equiv b_2 \pmod {m}$, and let $c_1,c_2 \in C$ with $c_1\ne c_2$. We have $b_i+c_1 \in B+C$ for $i = 1,2$ and $b_1+c_1 \not \equiv b_2 + c_1 \pmod {m}$. Because A intersects only two congruence classes modulo m, and because the intersection with the congruence class $a_1 \pmod {m}$ contains only the integer $a_1$, we must have $b_i+c_1 = a_1$ for some $i \in \{1,2\}$.

Similarly, $b_j+c_2 \in B+C$ for $j = 1,2$ with $b_1+c_2 \not \equiv b_2 + c_2 \pmod {m}$, and so $b_j+c_2 = a_1$ for some $j \in \{1,2\}$. The equation $b_i+c_1 = b_j+c_2$ implies that $A \ne B \oplus C$. This completes the proof.

The following examples show that, in Lemma 5, the condition that the set A contains exactly one element of the congruence class $a_1 \pmod {m}$ is necessary.

Let $m \ge 2$, and let $R \subseteq [0,m)$ with $|R| \ge 2$. For every set J of integers with $|J| \ge 2$, we have

$$ A = \{jm + r : j\in J \text { and } r \in R \} = B \oplus C $$

where

$$ B = \{jm:j\in J\} {{\mathrm{\qquad \text {and}\qquad }}}C = R. $$

Let k, $\ell$, and m be integers with $k \ge 2$, $\ell \ge 2$, and $m \ge 2$, and let u and v be integers such that $u \not \equiv v \pmod {m}$. Consider the set

$$ A = \{im+u: i \in [0,\ell ) \} \cup \{ j m + v: j \in [0, k \ell ) \}. $$

The sets

$$ B = \{ u \} \cup \{ q \ell m + v: q \in [0, k ) \} $$

and

$$ C = \{im : i \in [0, \ell ) \} $$

satisfy $|B| = 1 + k\ell \ge 2$, $|C| = \ell \ge 2$ and

$$ A = B \oplus C. $$

3 Decomposition of Additive Systems

Contraction and dilation are two methods to construct new additive systems from old ones. Decomposition is a third method to produce new additive systems.

An additive system $\mathcal A= (A_i)_{i\in I}$ is called decomposable if the set $A_{i_0}$ is decomposable for some $i_0 \in I$ and indecomposable if $A_i$ is indecomposable for all $i \in I$. Equivalently, an indecomposable additive system is an additive system that is not a proper contraction of another additive system.

Theorem 3

Let $\mathcal A= (A_i)_{i\in I}$ be a decomposable additive system, and let $A_{i_0}$ be a decomposable set in $\mathcal A$. Choose sets B and C of nonnegative integers such that $0 \in B\cap C$, $|B| \ge 2$, $|C| \ge 2$, and $A_{i_0} = B \oplus C$. Let

$$ I' = \{ j_1,j_2\} \cup I\setminus \{ i_0\}. $$

The family of sets $\mathcal A' = (A'_i)_{i\in I'}$ defined by

$$ A'_i = {\left\{ \begin{array}{ll} A_i &{} \text { if } i \in I\setminus \{ i_0\} \\ B &{} \text { if } i=j_1 \\ C &{} \text { if } i= j_2 \end{array}\right. } $$

is an additive system.

Proof

This follows immediately from the definitions of additive system and indecomposable set.

We call $\mathcal A'$ a decomposition of the additive system $\mathcal A$.

Lemma 6

Let a and b be positive integers, and let X be a set of integers. Then

$$\begin{aligned}{}[0,ab) = [0,a) \oplus X \end{aligned}$$

(2)

if and only if

$$ X = a*[0,b). $$

Proof

The division algorithm implies that $[0,ab) = [0,a) \oplus a*[0,b)$, and so $X = a*[0,b)$ is a solution of the additive set equation (2).

Conversely, let X be any solution of (2). Let $I = \{1,2,3\}$ and let $A_1 = [0,a)$, $A_2 = X$, and $A_3 = ab*N_0$. By the division algorithm, $\mathcal A= (A_i)_{i\in I}$ is an additive system. Applying Lemma 3 to $\mathcal A$, we obtain an integer $g \ge 2$ and sets $B_1$, $B_2$, and $B_3$ such that

$$\begin{aligned}{}[0,a)&= [0,g) \oplus g *B_1 \\ X&= g*B_2 \\ ab*\mathbf N_0&= g *B_3. \end{aligned}$$

It follows that $g=a$, $B_1 = \{0\}$, $B_3 = b*\mathbf N_0$, and

$$ \mathbf N_0 = B_2 \oplus B_3 = B_2 \oplus b *\mathbf N_0. $$

This implies that $B_2 = [0,b)$ and $X = a*[0,b)$.

There is also a nice polynomial proof of Lemma 6. Let

$$\begin{aligned} f(t)&= \sum _{i\in [0,ab)} t^i \\ g(t)&= \sum _{j \in [0,a)} t^j \\ h(t)&= \sum _{k\in [0,b)} t^{ak} \\ h_X(t)&= \sum _{x\in X} t^x. \end{aligned}$$

The set equation $[0,ab) = [0,a) \oplus a*[0,b)$ implies that

$$ f(t) = g(t) h(t). $$

If $[0,ab) = [0,a) \oplus X$, then

$$ f(t) = g(t) h_X(t) $$

and so

$$ g(t)( h(t) - h_X(t)) = 0. $$

Because $g(t) \ne 0$, it follows that $h(t) = h_X(t)$ or, equivalently, $a*[0,b) = X$.

By Theorem 2, every additive system is a British number system or a proper contraction of a British number system. However, a British number system can also be a proper contraction of another British number system. Consider, for example, the British number systems $\mathcal A_2$ and $\mathcal A_4$ generated by the sequences $(2)_{i \in \mathbf N}$ and $(4)_{i \in \mathbf N}$, respectively:

$$\begin{aligned} \mathcal A_2&= (\{ 0, 2^{i-1} \} )_{i \in \mathbf N} = ( 2^{i-1}*[0,2) )_{i \in \mathbf N} \\&= \left( \{ 0, 1 \}, \{ 0, 2 \}, \{ 0, 4 \}, \{ 0, 8 \}, \ldots \right) \end{aligned}$$

and

$$\begin{aligned} \mathcal A_4&= ( \{ 0, 4^{i-1}, 2\cdot 4^{i-1}, 3\cdot 4^{i-1} \} )_{i \in \mathbf N} = ( 4^{i-1}*[0,4) )_{i \in \mathbf N} \\&= \left( \{ 0, 1, 2, 3 \}, \{ 0, 4,8,12 \}, \{ 0, 16, 32, 48 \}, \{ 0, 64,128, 192, 256 \}, \ldots \right) . \end{aligned}$$

Because

$$ 4^{i-1}*[0,4) = \{ 0, 2^{2i-2} \} + \{ 0, 2^{2i-1} \} = 2^{2i-2} *[0,2) + 2^{2i-1} *[0,2) $$

we see that $\mathcal A_4$ is a contraction of $\mathcal A_2$.

de Bruijn [1] asserted the following necessary and sufficient condition for one British number system to be a contraction of another British number system.

Theorem 4

Let $\mathcal B= (B_j)_{j \in \mathbf N}$ be the British number system constructed from the integer sequence $(h_j)_{j \in \mathbf N}$, and let $\mathcal A= (A_i)_{i \in \mathbf N}$ be the contraction of $\mathcal B$ constructed from a partition $(J_i)_{i \in \mathbf N}$ of $\mathbf N$ into nonempty finite sets. Then, $\mathcal A$ is a British number system if and only if $J_i$ is a finite interval of integers for all $i \in \mathbf N$.

Proof

Let $(J_i)_{i \in \mathbf N}$ be a partition of $\mathbf N$ into nonempty finite intervals of integers. After re-indexing, there is a strictly increasing sequence $(u_i)_{i \in \mathbf N_0}$ of integers with $u_0 = 0$ such that $J_i = [ u_{i-1}+1 , u_i]$ for all $i \in \mathbf N$.

If $\mathcal B= (B_j)_{j \in \mathbf N}$ is the British number system constructed from the integer sequence $(h_j)_{j \in \mathbf N}$, then $B_j = H_{j-1} *[0,h_j)$, where $H_0=1$ and $H_j = \prod _{k=1}^j h_k$. Let $G_0 = 1$. For $i \in \mathbf N$ we define

$$ g_i = \frac{H_{u_i}}{H_{u_{i-1}}} $$

and

$$ G_i = \prod _{j=1}^i g_j = \prod _{j=1}^i \frac{H_{u_j}}{H_{u_{j-1}}} = H_{u_i}. $$

We have

$$\begin{aligned} A_i&= \bigoplus _{j \in J_i} B_j = \bigoplus _{j = u_{i-1}+1}^{u_i} H_{j-1}*[0,h_j) \\&= H_{u_{i-1}} *\bigoplus _{j = u_{i-1}+1}^{u_i} \frac{H_{j-1} }{ H_{u_{i-1}}} *[0,h_j) \\&= H_{u_{i-1}} *\left( [0, h_{u_{i-1} + 1} ) + h_{u_{i-1} + 1} *[0, h_{u_{i-1} + 2} ) \right. \\&\qquad \left. +\, h_{u_{i-1} + 1} h_{u_{i-1} + 2}*[0, h_{u_{i-1} + 3} ) + \cdots \right. \\&\qquad \left. +\, h_{u_{i-1} + 1} \cdots h_{u_i -1}*[0, h_{u_i} ) \right) \\&= H_{u_{i-1}} *\left[ 0, \frac{H_{u_i} }{ H_{u_{i-1}}} \right) \\&= G_{i-1} *\left[ 0, g_i\right) \end{aligned}$$

and so $\mathcal A= (A_i)_{i \in \mathbf N}$ is the British number system constructed from the integer sequence $(g_i)_{i \in \mathbf N}$.

Conversely, let $\mathcal A= (A_i)_{i \in \mathbf N}$ be a contraction of $\mathcal B$ constructed from a partition $(J_i)_{i \in \mathbf N}$ of $\mathbf N$ in which some set $J_{i_0}$ is a not a finite interval of integers. Let $u = \min \left( J_{i_0}\right)$ and $w = \max \left( J_{i_0} \right)$. Because $J_{i_0}$ is not an interval, there is a smallest integer v such that

$$ u< v < w $$

and $[u,v-1] \subseteq J_{i_0}$, but $v \notin I_{j_0}$. Because

$$ [u,v-1] \cup \{ w\} \subseteq J_{i_0} \subseteq [u,v-1] \cup [v+1,w] $$

and

$$ A_{i_0} = \sum _{j \in J_{i_0}} H_{j-1}*[0,h_j) $$

we have

$$\begin{aligned} H_{u-1}\,*\,&[0,h_u) \cup H_{w-1}*[0,h_w) \subseteq A_{i_0} \\&\subseteq \sum _{ j \in [u,v-1]} H_{j-1}*[0,h_j) + \sum _{ j \in [v+1,w]} H_{j-1}*[0,h_j) \\&\subseteq H_{u-1} *\left[ 0,\frac{H_{v-1}}{H_{u-1}} \right) + H_v *\left[ 0,\frac{H_w}{H_v} \right) \end{aligned}$$

Because $h_u \ge 2$ and $h_v \ge 2$, it follows that

$$ H_{u-1} \in A_{i_0} $$

and

$$ H_{w-1} = H_{u-1}\left( \frac{H_{w-1}}{H_{u-1}} \right) \in A_{i_0}. $$

The largest multiple of $H_{u-1}$ in $H_{u-1} *\left[ 0,H_{v-1}/H_{u-1} \right)$ is $H_{u-1} (H_{v-1}/H_{u-1} -1)$. The smallest positive multiple of $H_{u-1}$ in $H_v*\left[ 0,H_w/H_v \right)$ is $H_v = H_{u-1}(H_v/H_{u-1})$. The inequality

$$ 1 \le \frac{H_{v-1}}{H_{u-1}} - 1< \frac{H_{v-1}}{H_{u-1}} < \frac{H_v}{H_{u-1}} \le \frac{H_{w-1}}{H_{u-1}} $$

implies that the set $A_{i_0}$ does not contain the integer $H_{u-1}(H_{v-1}/H_{u-1})$. In a British number system, every set consists of consecutive multiples of its smallest positive element. Because the set $A_{i_0}$ lacks this property, it follows that $\mathcal A$ is not a British number system. This completes the proof.

Theorem 5

There is a one-to-one correspondence between sequences $(p_i)_{i \in \mathbf N}$ of prime numbers and indecomposable British number systems. Moreover, every additive system is either indecomposable or a contraction of an indecomposable system.

Proof

Let $\mathcal A$ be a British number system generated by the sequence $(g_i)_{i \in \mathbf N}$, so that

$$ \mathcal A= \left( G_{i-1}*[0,g_i) \right) _{i \in \mathbf N}. $$

Suppose that $g_k$ is composite for some $k \in \mathbf N$. Then $g_k = rs$, where $r \ge 2$ and $s \ge 2$ are integers. Construct the sequence $(g'_i)_{i \in \mathbf N}$ as follows:

$$ g'_i = {\left\{ \begin{array}{ll} g_i &{} \text {if}~i \le k-1 \\ r &{} \text {if}~i = k \\ s &{} \text {if}~i = k+1 \\ g_{i-1} &{} \text {if}~i \ge k+2. \end{array}\right. } $$

Then,

$$ G'_i = \prod _{j=1}^i g'_j = {\left\{ \begin{array}{ll} G_i &{} \text {if}~i \le k-1 \\ r G_{k-1}&{} \text {if}~i=k \\ G_k &{} \text {if}~i=k+1 \\ G_{i-1} &{} \text {if}~i \ge k+2 \end{array}\right. } $$

and

$$ \mathcal A' = \left( G'_{i-1}*[0,g'_i) \right) _{i \in \mathbf N} $$

is the British number system generated by the sequence $(g'_i)_{i \in \mathbf N}$. We have

$$ G_{i-1}*[0,g_i) = {\left\{ \begin{array}{ll} G'_{i-1}*[0,g'_i) &{} \text {if}~i \le k-1 \\ G'_i *[0,g'_{i+1}) &{} \text {if}~i \ge k+1. \end{array}\right. } $$

The identity

$$ [0,g_k) = [0,rs) = [0,r) \oplus r*[0,s) = [0,g'_k)+ \frac{G'_k}{G_{k-1}}*[0,g'_{k+1}) $$

implies that

$$ G_{k-1}*[0,g_k) = G'_{k-1} *[0,g'_k)+ G'_k *[0,g'_{k+1}) = \sum _{i\in \{k,k+1\} } G'_{i-1} *[0,g'_i) $$

and so the British number system $\mathcal A$ is a contraction of the British number system $\mathcal A'$.

Conversely, if $\mathcal A$ is a contraction of a British number system $\mathcal A' = \left( G'_{i-1}*[0,g'_i) \right) _{i \in \mathbf N}$, then there are a positive integer k and a set $I_k$ of positive integers with $|I_k| \ge 2$ such that

$$ G_{k-1}*[0,g_k) = \sum _{i\in I_k} G'_{i-1} *[0,g'_i). $$

Therefore,

$$ g_k = \left| G_{k-1}*[0,g_k) \right| = \left| \sum _{i\in I_k} G'_{i-1} *[0,g'_i) \right| = \prod _{i\in I_k} g'_i. $$

Because $|I_k| \ge 2$ and $|g'_i| \ge 2$ for all $i\in \mathbf N$, it follows that the integer $g_k$ is composite. Thus, the British number system generated by $(g_i)_{i\in \mathbf N}$ is decomposable if and only if $g_i$ is composite for at least one $i \in \mathbf N$. Equivalently, the British number system generated by $(g_i)_{i\in \mathbf N}$ is indecomposable if and only if $(g_i)_{i\in \mathbf N}$ is a sequence of prime numbers. This completes the proof.

Theorem 5 has also been observed by Munagi [3].

4 Limits of Additive Systems

Let $\mathcal A= (A_i)_{i\in \mathbf N_0}$ be an additive system, and let $(g_i)_{i \in [1,n]}$ be a finite sequence of integers with $g_i \ge 2$ for all $i \in [1,n]$. The dilation of $\mathcal A$ by the sequence $(g_i)_{i \in [1,n]}$ is the additive system defined inductively by

$$ (g_i)_{i \in [1,n]} *\mathcal A= g_1 *\left( (g_i)_{i \in [2,n]} *\mathcal A\right) . $$

For $n = 1$, we have

$$\begin{aligned} \mathcal A^{(1)}&= (g_i)_{i \in [1,1]} *\mathcal A= g_1*\mathcal A\\&= [0,g_1) \cup (g_1*A_i)_{i\in \mathbf N_0} \\&= \left( A^{(1)}_i \right) _{i\in \mathbf N_0} \end{aligned}$$

where

$$ A^{(1)}_1 = [0, g_1) $$

and

$$ A^{(1)}_i = g_1*A_{i-1} \text { for} i \ge 2. $$

For $n = 2$, we have

$$\begin{aligned} \mathcal A^{(2)}&= (g_i)_{i \in [1,2]} *\mathcal A= g_1 *(g_2 *\mathcal A) \\&= g_1 *\left( [0, g_2) \cup (g_2 *A_i)_{i\in \mathbf N_0} \right) \\&= [0, g_1) \cup \left( g_1 *[0, g_2) \right) \cup (g_1g_2*A_i)_{i\in \mathbf N_0} \\&= \left( A^{(2)}_i \right) _{i\in \mathbf N_0} \end{aligned}$$

where

$$\begin{aligned} A^{(2)}_1&= [0, g_1) \\ A^{(2)}_2&= g_1 *[0,g_2) \end{aligned}$$

and

$$ A^{(2)}_i = g_1g_2*A_{i-2} \text { for}~i \ge 3. $$

For $n = 3$, we have

$$\begin{aligned} g_3 *\mathcal A&= [0, g_3) \cup (g_3*A_i)_{i\in \mathbf N_0} \\ g_2*(g_3 *\mathcal A)&= [0, g_2) \cup g_2 *[0, g_3) \cup (g_2g_3*A_i)_{i\in \mathbf N_0} \end{aligned}$$

and

$$\begin{aligned} \mathcal A^{(3)}&= (g_i)_{i \in [1,3]} *\mathcal A= g_1 *\left( g_2*(g_3 *\mathcal A) \right) \\&= [0, g_1) \cup \left( g_1 *[0, g_2) \right) \cup \left( g_1g_2 *[0, g_3) \right) \cup (g_1g_2g_3*A_i)_{i\in \mathbf N_0}\\&= \left( A_i^{(3)} \right) _{i \in \mathbf N_0} \end{aligned}$$

where

$$\begin{aligned} A_1^{(3)}&= [0,g_1) \\ A_2^{(3)}&= g_1 *[0,g_2) \\ A_3^{(3)}&= g_1g_2 *[0,g_3) \\ A_i^{(3)}&= g_1g_2g_3A_{i-3} \qquad \text {for}~i \ge 4. \end{aligned}$$

Lemma 7

Let $(g_i)_{i =1}^n$ be a sequence of integers such that $g_i \ge 2$ for all i. For every additive system $\mathcal A= (A_i)_{i \in \mathbf N}$,

$$ \mathcal A^{(n)} = (g_i)_{i =1}^n *\mathcal A= \left( A^{(n)}_i \right) _{i \in \mathbf N} $$

where

$$ A^{(n)}_i = g_1g_2 \cdots g_{i-1} *[0,g_i) \qquad \text {for}~i=1,\ldots , n $$

and

$$ A^{(n)}_i = g_1g_2 \cdots g_n *A_{i-n-1} \qquad \text {for}~i \ge n+1. $$

Proof

Induction on n.

Let $(\mathcal A^{(n)})_{n\in \mathbf N}$ be a sequence of additive systems. The additive system $\mathcal A$ is the limit of the sequence $(\mathcal A^{(n)})_{n\in \mathbf N}$ if it satisfies the following condition: The set S belongs to $\mathcal A$ if and only if S belongs to $\mathcal A^{(n)}$ for all sufficiently large n. We write

$$ \lim _{n\rightarrow \infty } \mathcal A^{(n)} = \mathcal A$$

if $\mathcal A$ is the limit of the sequence $(\mathcal A^{(n)})_{n\in \mathbf N}$. The following result indicates the remarkable stability of a British number system.

Theorem 6

Let $(g_i)_{i \in \mathbf N}$ be a sequence of integers such that $g_i \ge 2$ for all $i \in \mathbf N$, and let $\mathcal G$ be the British number system generated by $(g_i)_{i \in \mathbf N}$. Let $\mathcal A$ be an additive system and let $\mathcal A^{(n)} = (g_i)_{i \in [1,n]} *\mathcal A$. Then,

$$ \lim _{n\rightarrow \infty } \mathcal A^{(n)} = \mathcal G. $$

Proof

If S is a set in $\mathcal G$, then $S = g_1g_2 \cdots g_{i-1} *[0,g_i)$ for some $i \in \mathbf N$. By Lemma 7, S is a set in $\mathcal A^{(n)}$ for all $n \ge i$, and so $S \in \lim _{n\rightarrow \infty } \mathcal A^{(n)}$.

Conversely, let S be a set that is in $\mathcal A^{(n)}$ for all sufficiently large n. If S is finite, then $\max (S) < g_1g_2\cdots g_k$ for some integer k. If $n \ge k$ and $i \ge n+1$, then

$$ \max \left( A^{(n)}_i\right) \ge g_1\dots g_n \ge g_1\dots g_k $$

and so $S \ne A^{(n)}_i$. Therefore, $S = A^{(n)}_i$ for some $i \le n$, and so $S = g_1g_2 \cdots g_{i-1} *[0,g_i)$ for some $i \le n$.

If T is an infinite set in $\mathcal A^{(n)}$, then $T = g_1g_2 \cdots g_n *A_{i-n-1}$ for some $i \ge n+1$, and so $\min (T \setminus \{0\}) \ge g_1g_2 \cdots g_n \ge 2^n$. If $T \in \mathcal A^{(n)}$ for all $n \ge N$, then $\min (T \setminus \{0\}) \ge 2^n$ for all $n \ge N$, which is absurd. It follows that the set S is in $\mathcal A^{(n)}$ for all sufficiently large n if and only if S is finite and S is a set in the British number system generated by $(g_i)_{i \in \mathbf N}$. This completes the proof.

Corollary 8

Let $(g_i)_{i \in \mathbf N}$ be a sequence of integers such that $g_i \ge 2$ for all $i \in \mathbf N$, and let $\mathcal G$ be the British number system generated by $(g_i)_{i \in \mathbf N}$. If $\mathcal G_n = (g_i)_{i \in [1,n]} *\mathbf N_0$, then

$$ \lim _{n\rightarrow \infty } \mathcal G_n = \mathcal G. $$