Problem 1.1. When does equality replace inclusion in relation (1.1)? The basic problem is to describe, classify and understand the strictly decreasing sequences $(A_q)_{q=1}^{\infty }$ of subsets of an additive abelian semigroup such that

$$ \begin{align*} h\, \bigg(\!\bigcap_{q=1}^{\infty} A_q \bigg) = \bigcap_{q=1}^{\infty} hA_q. \end{align*} $$

Proof. For all $x \in S$ , the set $ \mathcal {R}_{S,h}(x)$ of h-tuples $(b_1,\ldots , b_h) \in S^h$ such that

$$ \begin{align*} x = b_1 + \cdots + b_h \end{align*} $$

is finite with cardinality $| \mathcal {R}_{S,h}(x)| = r_{S,h}(x)$ .

Let $x \in \bigcap _{q=1}^{\infty } hA_q$ . We must show that $x \in hA$ . For all $q\in \mathbf N $ , we have $x \in hA_q$ and so there is an h-tuple $(a_{1,q},\ldots , a_{h,q}) \in A_q^h \cap \mathcal {R}_{S,h}(x)$ such that

$$ \begin{align*} x = a_{1,q} + \cdots + a_{h,q}. \end{align*} $$

Because $ \mathcal {R}_{S,h}(x)$ is a finite set, there is some h-tuple $(b_1,\ldots , b_h) \in \mathcal {R}_{S,h}(x)$ such that

$$ \begin{align*} (b_1,\ldots, b_h) = (a_{1,q},\ldots, a_{h,q}) \end{align*} $$

for infinitely many q and so, for infinitely many q, we have $b_i = a_{i,q} \in A_q$ for all $i=1,\ldots , h$ . Because the sequence $(A_q)_{q=1}^{\infty }$ is decreasing, we have $b_i \in A_q$ for all $i=1,\ldots , h$ and $q \in \mathbf N $ . It follows that $b_i \in \bigcap _{q=1}^{\infty } A_q = A$ and $x = b_1 + \cdots + b_h \in hA$ . This completes the proof.

Proof. For all $n \in \mathbf N _0$ , the number $r_{ \mathbf N _0,h}(n)$ of representations of n as a sum of h nonnegative integers is finite. Indeed, $r_{ \mathbf N _0,h}(n) = \binom {n+h-1}{h-1}$ . For all $ \mathbf {n} = (n_1,\ldots , n_d) \in \mathbf N _0^d$ ,

$$ \begin{align*} r_{ \mathbf N _0^d,h}( \mathbf{n}) = \prod_{i=1}^d r_{ \mathbf N _o,h}(n_i) < \infty. \end{align*} $$

The result follows from Theorem 2.1.

Proof. For all $q \in \mathbf N $ , we have $A \subseteq A_q$ and so $S = h_0A \subseteq h_0A_q \subseteq S$ . It follows that $h_0A_q = S$ and

$$ \begin{align*} h_0A = S = \bigcap_{q=1}^{\infty} h_0A_q. \end{align*} $$

Let $h \geq h_0$ . If S contains an additive identity, then $h_0A \subseteq hA$ . If A is a basis of order $h_0$ for S, then $S = h_0A \subseteq hA \subseteq S$ and so A is a basis of order h for S. This completes the proof.

Proof. For all $\mathbf {x} = (x_1,\ldots , x_d) \in \mathbf R^d$ , define $\| \mathbf {x} \| = \max ( |x_i| : i = 1,\ldots , d)$ . For all ${q \in \mathbf N} $ , let

$$ \begin{align*} A_{q,d}^{\sharp} = \{ \mathbf{x} \in G: \| \mathbf{x} \| \geq q \}. \end{align*} $$

Note that $\{ \mathbf {x} \in \mathbf Z^d: \| \mathbf {x} \| \geq q \} \subseteq A_{q,d}^{\sharp }$ .

Let $h \geq 2$ and $\mathbf {x} = (x_1,\ldots , x_d) \in G$ . If $x_1 \geq 0$ , then

$$ \begin{align*} \mathbf{y} = (x_1 + (h-1)q, x_2,\ldots, x_d) \in A_{q,d}^{\sharp} \end{align*} $$

and

$$ \begin{align*} \mathbf{z} = (-q,0,\ldots, 0) \in A_{q,d}^{\sharp} \end{align*} $$

and so

$$ \begin{align*} \mathbf{x} = (x_1 + (h-1)q, x_2,\ldots, x_d) + (h-1)(-q,0,\ldots, 0) = \mathbf{y} + (h-1) \mathbf{z} \in h A_{q,d}^{\sharp}. \end{align*} $$

Similarly, if $x_1 \leq 0$ , then

$$ \begin{align*} \mathbf{y} = (x_1 - (h-1)q, x_2,\ldots, x_d) \in A_{q,d}^{\sharp} \end{align*} $$

and

$$ \begin{align*} \mathbf{z} = (q,0,\ldots, 0) \in A_{q,d}^{\sharp} \end{align*} $$

and so

$$ \begin{align*} \mathbf{x} = (x_1 - (h-1)q, x_2, \ldots, x_d) + (h-1)(q,0,\ldots, 0) = \mathbf{y} + (h-1) \mathbf{z} \in h A_{q,d}^{\sharp}. \end{align*} $$

Thus, $hA_{q,d}^{\sharp } = G$ for all $q \in \mathbf N $ and $h \geq 2$ .

For all $q \in \mathbf N $ , let

$$ \begin{align*} A_{q,d} = A \cup A_{q,d}^{\sharp}. \end{align*} $$

Then $hA_{q,d} = hA_{q,d}^{\sharp } = G$ for all $q \in \mathbf N $ and $h \geq 2$ . The sequence $(A_{q,d})_{q=1}^{\infty }$ is a decreasing sequence of sets such that $ A = \bigcap _{q=1}^{\infty } A_{q,d}$ . If A is a nonbasis of order $h_0$ for G, then

$$ \begin{align*} h_0A \neq G = \bigcap_{q=1}^{\infty} h_0A_{q,d}. \end{align*} $$

If A is a nonbasis for G, then

$$ \begin{align*} hA \neq G = \bigcap_{q=1}^{\infty} hA_{q,d} \end{align*} $$

for all $h \geq 2$ . This completes the proof.

Proof. Note that a set of integers that is bounded below cannot be a basis for $\mathbf Z$ . By Theorem 2.2 with $d=1$ , we can assume that $a^* = \min (A) < 0$ .

For all $q \in \mathbf N $ , let

$$ \begin{align*} A^{\flat}_q = \{r\in \mathbf N : r \geq q \} \quad \text{ and } \quad A_q = A \cup A^{\flat}_q = A \cup \{r\in \mathbf N : r \geq q\}. \end{align*} $$

The sequence $(A_q)_{q=1}^{\infty }$ is decreasing and $A = \bigcap _{q=1}^{\infty } A_q$ . If $q \in \mathbf N $ and $q \notin A$ , then ${q \in A_q}$ but $q \notin A_{q+1}$ and so $A_q \neq A_{q+1}$ . Because A does not contain all sufficiently large integers, the sequence $(A_q)_{q=1}^{\infty }$ is asymptotically strictly decreasing.

For all $h \in \mathbf N $ and $q \in \mathbf N $ ,

$$ \begin{align*} hA \subseteq hA_q \subseteq hA \cup \{r\in \mathbf N : r \geq q + (h-1)a^*\} \end{align*} $$

and

$$ \begin{align*} \bigcap_{q=1}^{\infty} \{r\in \mathbf N : r \geq q + (h-1)a^*\} = \emptyset. \end{align*} $$

Therefore,

$$ \begin{align*} hA \subseteq \bigcap_{q=1}^{\infty} hA_q \subseteq hA \cup \bigcap_{q=1}^{\infty} \{r\in \mathbf N : r \geq q + (h-1)a^*\} = hA. \end{align*} $$

This completes the proof.

Proof. Let A be a maximal nonbasis of order h for S and let $(A_q)_{q=1}^{\infty }$ be a strictly decreasing sequence such that $A = \bigcap _{q=1}^{\infty } A_q$ . For all q, the set A is a proper subset of $A_q$ and so $hA_q = S$ . It follows that $hA \neq S = \bigcap _{q=1}^{\infty } hA_q$ . This completes the proof.

Problem 2.7. Let A be a subset of the infinite abelian semigroup S such that A is a nonbasis of order h but not a maximal nonbasis of order h for S. Does there exist an asymptotically strictly decreasing sequence $(A_q)_{q=1}^{\infty }$ of subsets of S such that ${A = \bigcap _{q=1}^{\infty } A_q}$ and $ hA = \bigcap _{q=1}^{\infty } h A_q$ ? In particular, consider this problem for $S = \mathbf Z$ .

Proof. For all $h \in \mathbf N $ and $q \in \mathbf N $ , the sumset $hA_q$ and the sumset intersection $\bigcap _{q=1}^{\infty } hA_q$ are nonempty closed compact sets. Let $x \in \bigcap _{q=1}^{\infty } hA_q$ . We must show that $x \in hA$ .

For all $q\in \mathbf N $ , we have $x \in hA_q$ and so there is an h-tuple

$$ \begin{align*} (a_{1,q},\ldots, a_{h,q}) \in A_q^h \subseteq A_1^h \end{align*} $$

such that

$$ \begin{align*} x = a_{1,q} + \cdots + a_{h,q}. \end{align*} $$

Consider the sequences $(a_{i,q})_{q=1}^{\infty }$ for $i = 1, \ldots , h$ . Because the set $A_1$ is compact and $a_{i,q} \in A_1$ for all $i = 1, \ldots , h$ and $q\in \mathbf N $ , there is an infinite strictly increasing sequence of positive integers $(q_j)_{j=1}^{\infty }$ such that the h subsequences $(a_{i,q_j})_{j=1}^{\infty }$ converge:

$$ \begin{align*} \lim_{j\rightarrow \infty} a_{i,q_j} = x_i \end{align*} $$

for all $i = 1, \ldots , h$ . Then

$$ \begin{align*} x = \lim_{j\rightarrow \infty} ( a_{1,q_j} + \cdots + a_{h,q_j} ) = \Big( \lim_{j\rightarrow \infty} a_{1,q_j} \Big) + \cdots + \Big( \lim_{j\rightarrow \infty} a_{h,q_j} \Big) = x_1 + \cdots + x_h. \end{align*} $$

Suppose that $x_i \notin A$ for some i. Let $q_0=0$ . Construct the sequence $( x_{i,q} )_{q=1}^{\infty }$ as follows:

$$ \begin{align*} x_{i,q} = a_{i,q_j} \quad \text{for } q \in [q_{j-1}+1,q_j]. \end{align*} $$

Because $(A_q)_{q=1}^{\infty }$ is a decreasing sequence, we have $x_{i,q} \in A_q$ for all $q \in \mathbf N $ and

$$ \begin{align*} \lim_{i\rightarrow \infty} x_{i,q} = x_i. \end{align*} $$

The group G is locally compact. Because $A = \bigcap _{q=1}^{\infty } A_q$ is closed and $x_i \notin A$ , there is a compact neighbourhood K of $x_i$ that is disjoint from A such that $x_{i,q} \in K$ for all sufficiently large q. Then $x_{i,q} \in K \cap A_q$ for all sufficiently large q. It follows that, for all $Q \in \mathbf N $ , there exists $q_1 \geq Q$ with $x_{i,q_1} \in K \cap A_{q_1} \subseteq K \cap A_q $ for all $q \in [1,Q]$ , and so

$$ \begin{align*} x_{i,q_1} \in \bigcap_{q=1}^Q ( K \cap A_q ) \neq \emptyset. \end{align*} $$

Because the sets $K \cap A_q$ are compact, the finite intersection property implies

$$ \begin{align*} \emptyset \neq \bigcap_{q=1}^{\infty} ( K \cap A_q ) = K \cap \bigg(\!\bigcap_{q=1}^{\infty} A_q \bigg) = K \cap A = \emptyset \end{align*} $$

which is absurd. Therefore, $x_i \in A$ for all $i = 1, \ldots , h$ and $x \in hA$ . This completes the proof.

Proof. This follows from Theorem 3.1 with $G = \mathbf R^d$ .

Proof. For all $h \in \mathbf N $ , the sequence $(hA_q)_{q=1}^{\infty }$ is a decreasing sequence of compact subsets of G. By Theorem 3.1,

$$ \begin{align*} hA = \bigcap_{q=1}^{\infty} hA_q \end{align*} $$

and so

$$ \begin{align*} \mu(hA) = \mu \bigg(\!\bigcap_{q=1}^{\infty} hA_q \bigg) = \lim_{q\rightarrow \infty} \mu(hA_q) = \lim_{q\rightarrow \infty} \theta_{h,q} = \theta_h. \end{align*} $$

This completes the proof.

Problem 4.1. Let $(A_q)_{q=1}^{\infty }$ be a strictly decreasing sequence of sets in an additive abelian semigroup and let $A = \bigcap _{q=1}^{\infty }A_q$ . Describe the set $\{h \in \mathbf N : hA = \bigcap _{q=1}^{\infty } hA_q\}$ . What sets of integers arise in this way?

Problem 4.2. Let $(A_q)_{q=1}^{\infty }$ be a strictly decreasing sequence of sets in an additive abelian semigroup and let $A = \bigcap _{q=1}^{\infty }A_q$ .

• • When does $hA = \bigcap _{q=1}^{\infty } hA_q$ imply $(h+1)A = \bigcap _{q=1}^{\infty } (h+1)A_q$ ?

• • When does $hA = \bigcap _{q=1}^{\infty } hA_q$ imply $(h-1)A = \bigcap _{q=1}^{\infty } (h-1)A_q$ ?

Problem 4.3. For what sets A of integers does there exist a decreasing sequence $(A_q)_{q=1}^{\infty }$ of sets of integers such that $A = \bigcap _{q=1}^{\infty } A_q$ and $hA \neq \bigcap _{q=1}^{\infty } hA_q$ for all $h \in \mathbf N $ ?