Proof. Let $x \in X$ and $U \subseteq X$ be non-empty and open. Fix $L \subseteq \beta \mathbb {N}$ a minimal left ideal. As described in §2.2.3, the map $\pi : (L, \unicode{x3bb} _1) \to (X,T)$ given by $\pi (p) = T^p x$ is a factor map of minimal systems. By [Reference Hindman and Strauss29, Theorem 19.23], since x is uniformly recurrent, there exists an idempotent $p \in L$ such that $\pi (p) = x$ . It is quick to check that

$$ \begin{align*}R_T(x, U) = R_{\unicode{x3bb}_1}(p, \pi^{-1} U).\end{align*} $$

We will show that the set $R_{\unicode{x3bb} _1}(p, \pi ^{-1} U)$ is dynamically syndetic.

By the definition of the topology on $\beta \mathbb {N}$ , since $\pi ^{-1} U$ is non-empty and open, there exists $A \subseteq \mathbb {N}$ such that $\emptyset \neq \overline {A} \cap L \subseteq \pi ^{-1} U$ . In the notation defined in §2.1.3, we see that

(3.1) $$ \begin{align} R_{\unicode{x3bb}_1}(p, \pi^{-1} U) \supseteq R_{\unicode{x3bb}_1}(p, \overline{A}) = A - p. \end{align} $$

Since $\overline {A} \cap L \neq \emptyset $ and $(L,\unicode{x3bb} _1)$ is minimal, we see that the set $A-p$ is non-empty (indeed, syndetic).

Considering $1_A$ as an element of $\{0,1\}^{\mathbb {N}_0}$ , we see in the full shift $(\{0,1\}^{\mathbb {N}_0}, \sigma )$ that $A = R_{\sigma }(1_A,[1]_0)$ . It follows by (2.1) in §2.2.3 that

$$ \begin{align*}A - p = R_{\sigma}(1_A,[1]_0) - p = R_{\sigma}(\sigma^p 1_A, [1]_0).\end{align*} $$

Since p is a minimal idempotent ultrafilter, we have by [Reference Hindman and Strauss29, Theorem 19.23] that the point $\sigma ^p 1_A$ is uniformly recurrent under the shift. Since $R_{\sigma }(\sigma ^p 1_A, [1]_0)$ is non-empty, we see that the set $R_{\sigma }(\sigma ^p 1_A, [1]_0)$ , and hence the set $A - p = R_{\unicode{x3bb} _1}(p, \overline {A})$ , is dynamically syndetic. Because these families are upward closed, by (3.1), the same can be said about the set $R_{\unicode{x3bb} _1}(p, \pi ^{-1} U) = R_T(x, U)$ , as was to be shown.

In the case where $x \in U$ , we wish to show that the set $R_T(x,U)$ is dynamically central syndetic. The argument given above can be followed with some additional information. First, since $x \in U$ , the point $p \in \pi ^{-1} U$ . The set $A \subseteq \mathbb {N}$ can then be assumed to be such that $p \in \overline {A} \cap L$ . Considering $1_A$ as an element of $\{0,1\}^{\mathbb {N}_0}$ , we see

$$ \begin{align*}R_\sigma(1_A,[1]_0) = A \in p,\end{align*} $$

whereby $\sigma ^p 1_A \in [1]_0$ . Thus, the set $A-p = R_{\sigma }(\sigma ^p 1_A, [1]_0)$ is dynamically central syndetic, as desired.

Proof. Let $A \in d\mathcal {S}$ , $B \in d\mathcal {T}$ , and $n \in \mathbb {N}$ . By Lemma 2.9, we see that $A \cap \{n+1, n+2, \ldots \} \in d\mathcal {S}$ . Since $B \in d\mathcal {T}$ , we have that $A \cap \{n+1, n+2, \ldots \} \cap B \neq \emptyset $ . Therefore, $\max (A \cap B)> n$ . Since $n \in \mathbb {N}$ was arbitrary, we see that $A \cap B$ is infinite. The same argument works to show that $A \cap B$ is infinite when $A \in dc\mathcal {S}$ and $B \in dc\mathcal {T}$ .

Proof. (i) Suppose $A \in d\mathcal {T}$ and $n \in \mathbb {N}$ . To see that $A - n \in d\mathcal {T}$ , it suffices to show that $A-n \in d\mathcal {S}^*$ . Let $B \in d\mathcal {S}$ . By Lemma 2.10, the set $B + n \in d\mathcal {S}$ . By Lemma 3.3, the set $A \cap (B+n)$ is infinite, so $(A-n) \cap B \supseteq (A-n) \cap (B+n - n) \neq \emptyset $ , as desired.

(ii) In the language of the family algebra developed in §2.1, this statement is equivalent to $d\mathcal {T} + dc\mathcal {T} \subseteq d\mathcal {T}$ . Recall from Lemma 2.10 that $d\mathcal {S} \subseteq d\mathcal {S} + dc\mathcal {S}$ . Taking the dual and simplifying using the algebra in Lemma 2.3, we see that $d\mathcal {T} + dc\mathcal {T} \subseteq d\mathcal {T}$ .

(iii) This statement is equivalent to $dc\mathcal {T} + dc\mathcal {T} \subseteq dc\mathcal {T}$ . Taking the dual of $dc\mathcal {S} \subseteq dc\mathcal {S} + dc\mathcal {S}$ from Lemma 2.10, we see that $dc\mathcal {T} + dc\mathcal {T} \subseteq dc\mathcal {T}$ .

That condition (i) holds with $A-n$ replaced with $A+n$ follows the same argument used in proof (i). That conditions (ii) and (iii) hold with $A-n$ replaced by $A+n$ requires different reasoning, since the set algebra explanation only treats negative translates.

Suppose that A is such that the set in (3.2) (with $A+n$ instead of $A-n$ ) is dynamically thick. We wish to show that A is dynamically thick. It suffices by the definitional robustness discussed in the previous subsection to show: for all minimal, invertible $(X,T)$ , all $x \in X$ , and all non-empty, open $U \subseteq X$ , there exists $a \in A$ such that $T^a x \in U$ . Thus, let $(X,T)$ be a minimal, invertible system, $x \in X$ , and $U \subseteq X$ open and non-empty. Since the system $(X,T^{-1})$ is minimal (see, for example, [Reference Glasscock, Koutsogiannis and Richter26, Lemma 2.7]) and the set in (3.2) is dynamically thick, there exists $n \in \mathbb {N}$ such that $T^{-n} x \in U$ and $A + n \in dc\mathcal {T}$ . By the definition of set of pointwise recurrence, there exists $m \in A+n$ (so $m> n$ ) such that $T^m T^{-n} x \in U$ . We see that $m - n \in A$ and $T^{m-n} x \in U$ , as desired. The argument that condition (iii) holds with $A-n$ replaced by $A+n$ is very similar and is left to the interested reader.

Finally, for (3.3), we only prove $d\mathcal {T} = \bigcap _{n \in \mathbb {N}} (dc\mathcal {T} + n)$ as the proof for $d\mathcal {T} = \bigcap _{n \in \mathbb {N}} (dc\mathcal {T} - n)$ is similar. Now, if $A \in d\mathcal {T}$ , then by condition (i), for all $n \in \mathbb {N}$ , the set $A - n \in d\mathcal {T} \subseteq dc\mathcal {T}$ . Thus, $d\mathcal {T} \subseteq \bigcap _{n \in \mathbb {N}} (dc\mathcal {T} + n)$ . Conversely, if $A \in \bigcap _{n \in \mathbb {N}} (dc\mathcal {T} + n )$ , then we have

$$ \begin{align*} \{ n \in \mathbb{N} \ | \ A-n \text{ is a set of pointwise recurrence} \} = \mathbb{N}, \end{align*} $$

which is a dynamically thick set. By condition (ii), we have $A \in d\mathcal {T}$ as desired.

Proof. We will make use of the following analogous result for dynamically (central) syndetic sets from [Reference Glasscock and Le27, Lemma 3.4]. Let $k \in \mathbb {N}$ . If $B \subseteq \mathbb {N}$ is dynamically central syndetic, then so are the sets kB and B/k. If B is dynamically syndetic, then so is the set kB.

Suppose $A \in dc\mathcal {T}$ . If $B \in dc\mathcal {S}$ , then the sets $kB$ and $B/k$ are $dc\mathcal {S}$ sets, whereby $A \cap (kB) \neq \emptyset $ and $A \cap (B/k) \neq \emptyset $ . Since $(kB)/k = B$ and $k(B/k) \subseteq B$ , it follows that $A/k \cap B \neq \emptyset $ and $kA \cap B \neq \emptyset $ . Since $B \in dc\mathcal {S}$ was arbitrary, we have that $A/k$ and $kA$ are sets of pointwise recurrence.

When $A \in d\mathcal {T}$ , the same argument shows that $A/k \in d\mathcal {T}$ .

Proof. (i) $\iff $ (ii) Because $dc\mathcal {S}$ and $dc\mathcal {T}$ are dual, the set A is a set of pointwise recurrence if and only if the set $A' := \mathbb {N} \backslash A$ is not dynamically central syndetic. By Theorem 2.13, the set $A'$ is not dynamically central syndetic if and only if for all $B' \subseteq A'$ , there exists a finite set $F \subseteq B'$ such that the set

$$ \begin{align*}B' \cap \bigcap_{f \in F} (B' - f)\end{align*} $$

is not syndetic. Taking the complement of this set in $\mathbb {N}$ , we see that this happens if and only if the set

$$ \begin{align*}( \mathbb{N} \backslash B' ) \cup \bigcup_{f \in F} ( (\mathbb{N} \backslash B') - f ) = (\mathbb{N} \backslash B') \cup ( (\mathbb{N} \backslash B') - F )\end{align*} $$

is thick. Set $B := \mathbb {N} \backslash B'$ . Note that $B' \subseteq A'$ if and only if $A \subseteq B$ .

Summarizing the previous paragraph, we have shown that the set A is a set of pointwise recurrence if and only if for all $B \supseteq A$ , there exists a finite set $F \subseteq \mathbb {N} \backslash B$ such that the set $B \cup (B - F)$ is thick, as was to be shown.

(i) $\implies $ (v) Suppose that $A \in dc\mathcal {T}$ . Let $S \subseteq \mathbb {N}$ be piecewise syndetic and $L \subseteq \beta \mathbb {N}$ be a minimal left ideal such that $\overline {S} \cap L \neq \emptyset $ . Note that $\overline {S} \cap L$ is non-empty and clopen in L. By the definitional robustness of $dc\mathcal {T}$ sets regarding non-metrizable systems (see Lemma 3.1), for all $p \in \overline {S} \cap L$ , there exists $a \in A$ such that $a + p \in \overline {S}$ . Therefore,

$$ \begin{align*}\overline{S} \cap L \subseteq \bigcup_{a \in A} \overline{S - a}.\end{align*} $$

Because $\overline {S} \cap L$ is compact, there exists a finite set $F \subseteq A$ such that

$$ \begin{align*}\overline{S} \cap L \subseteq \bigcup_{f \in F} \overline{S - f} \subseteq \overline{S-F},\end{align*} $$

as was to be shown.

(v) $\implies $ (iv) Let $S \subseteq \mathbb {N}$ be syndetic and $H \subseteq \mathbb {N}$ be thick, and put $P := S \cap H$ . It follows by [Reference Hindman and Strauss29, Theorem 4.48] that there exists a minimal left ideal $L \subseteq \overline {H}$ . Note that since S is syndetic, we have that $\emptyset \neq \overline {S} \cap L = \overline {S} \cap L \cap \overline {H} = \overline {P} \cap L$ .

Let $F \subseteq A$ be the finite set guaranteed by item (v) for the piecewise syndetic set P. Let $S' \subseteq S$ be syndetic and put $P' := S' \cap H$ . Since $S'$ is syndetic and $S' \subseteq S$ , we have that $\emptyset \neq \overline {S'} \cap L = \overline {P'} \cap L$ . Since $\emptyset \neq \overline {P'} \cap L \subseteq \overline {P} \cap L \subseteq \overline {P-F}$ , we see that $\overline {P'} \cap \overline {P-F} \neq \emptyset $ , whereby $P' \cap (P-F) \neq \emptyset $ , as desired.

(iv) $\implies $ (iii) Let $S \subseteq \mathbb {N}$ be syndetic. Get $F \subseteq A$ finite from item (iv) for $H = \mathbb {N}$ . Then, for all $S' \subseteq S$ syndetic,

$$ \begin{align*}S' \cap (S-F) \neq \emptyset,\end{align*} $$

whereby $F \cap (S-S') \neq \emptyset $ , as desired.

(iii) $\implies $ (i) We will prove the contrapositive: if A is not a set of pointwise recurrence, then there exists a syndetic set $S \subseteq \mathbb {N}$ such that for all finite $F \subseteq A$ , there exists a syndetic set $S' \subseteq S$ such that $F \cap (S - S') = \emptyset $ . Since $F \cap (S - S') = \emptyset $ if and only if $S' \cap (S - F) = \emptyset $ , it is equivalent to show: there exists a syndetic set $S \subseteq \mathbb {N}$ such that for all finite $F \subseteq A$ , the set $S \cap (\mathbb {N} \backslash (S-F))$ is syndetic. Taking complements and considering B as $\mathbb {N} \backslash A$ and C as $\mathbb {N} \backslash S$ , we must show: if a set $B \subseteq \mathbb {N}$ is dynamically central syndetic, then there exists a set $C \subseteq \mathbb {N}$ that is not thick such that for all finite $F \subseteq \mathbb {N} \backslash B$ , the set $(\mathbb {N} \backslash C) \cap \bigcap _{f \in F} (C - f)$ is syndetic.

Suppose $B \subseteq \mathbb {N}$ is dynamically central syndetic. If $B = \mathbb {N}$ , put $C = \emptyset $ and note that the conclusion holds since the empty intersection is $\mathbb {N}$ . Suppose $B \neq \mathbb {N}$ . By Theorem 2.13, there exists a minimal system $(X,T)$ , a point $x \in X$ , and a clopen set $U \subseteq X$ with $x \in U$ such that $R(x,U) \subseteq B$ . Define $V := X \backslash U$ and $C := \mathbb {N} \backslash R(x,U) = R(x, V)$ . Since $U \neq X$ and the set $R(x,U)$ is syndetic, the set C is non-empty but not thick. Let $F \subseteq \mathbb {N} \backslash B$ be finite. Since $F \subseteq C$ , we have that for all $f \in F$ , the set $T^{-f}V$ is an open neighborhood of x. We see that

$$ \begin{align*}(\mathbb{N} \backslash C) \cap \bigcap_{f \in F} (C - f) = R\bigg( x, U \cap \bigcap_{f \in F}T^{-f}V \bigg),\end{align*} $$

which is syndetic since the set $U \cap \bigcap _{f \in F}T^{-f}V$ is an open neighborhood of x, as desired.

Proof. (i) $\iff $ (ii) Because $d\mathcal {S}$ and $d\mathcal {T}$ are dual, the set A is dynamically thick if and only if the set $A' := \mathbb {N} \backslash A$ is not dynamically syndetic. By Theorem 2.12, the set $A'$ is not dynamically syndetic if and only if for all non-empty subsets $B' \subseteq A'$ , there exists a finite set $F \subseteq B'$ such that the set $\bigcap _{f \in F} (B' - f)$ is not syndetic. Taking the complement of this set in $\mathbb {N}$ , we see that this happens if and only if the set

$$ \begin{align*} \bigcup_{f \in F} ((\mathbb{N} \backslash B') - f) = (\mathbb{N} \backslash B') - F\end{align*} $$

is thick. Set $B := \mathbb {N} \backslash B'$ . Note that $B' \subseteq A'$ if and only if $A \subseteq B$ , and that $B' \neq \emptyset $ if and only if $B \neq \mathbb {N}$ .

Summarizing the previous paragraph, we have shown that the set A is dynamically thick if and only if for all $A \subseteq B \subsetneq \mathbb {N}$ , there exists a finite set $F \subseteq \mathbb {N} \backslash B$ such that the set $B - F$ is thick, as was to be shown.

(i) $\implies $ (v) Suppose that $A \in d\mathcal {T}$ . Let $S \subseteq \mathbb {N}$ be piecewise syndetic and $L \subseteq \beta \mathbb {N}$ be a minimal left ideal such that $\overline {S} \cap L \neq \emptyset $ . Note that $\overline {S} \cap L$ is non-empty and open in L. By the definitional robustness of $d\mathcal {T}$ sets regarding non-metrizable systems (see Lemma 3.1), for all $p \in L$ , there exists $a \in A$ such that $a + p \in \overline {S}$ . Therefore,

$$ \begin{align*}L \subseteq \bigcup_{a \in A} \overline{S - a}.\end{align*} $$

Because L is compact, there exists a finite set $F \subseteq A$ such that

$$ \begin{align*}L \subseteq \bigcup_{f \in F} \overline{S - f} \subseteq \overline{S-F},\end{align*} $$

as was to be shown.

(v) $\implies $ (iv) Let $S \subseteq \mathbb {N}$ be a piecewise syndetic set. It follows by [Reference Hindman and Strauss29, Theorem 4.40] that there exists a minimal left ideal $L \subseteq \beta \mathbb {N}$ such that $\overline {S} \cap L \neq \emptyset $ . Let $F \subseteq A$ be the finite set guaranteed by item (v). Since $\overline {S-F}$ contains a minimal left ideal, by [Reference Hindman and Strauss29, Theorem 4.48], the set $S-F$ is thick, as desired.

(iv) $\implies $ (iii) Let $S \subseteq \mathbb {N}$ be syndetic. Let $F \subseteq A$ be the finite set guaranteed by item (iv). Since $S-F$ is thick, for all $S' \subseteq \mathbb {N}$ syndetic, the set $S' \cap (S-F)$ is non-empty, whereby $F \cap (S-S') \neq \emptyset $ , as desired.

(iii) $\implies $ (i) We will prove the contrapositive: if A is not dynamically thick, then there exists a syndetic set $S \subseteq \mathbb {N}$ such that for all finite $F \subseteq A$ , the set $S - F$ is not thick. Taking complements and considering B as $\mathbb {N} \backslash A$ and C as $\mathbb {N} \backslash S$ , we must show: if a set $B \subseteq \mathbb {N}$ is dynamically syndetic, then there exists a set $C \subseteq \mathbb {N}$ that is not thick such that for all finite $F \subseteq \mathbb {N} \backslash B$ , the set $\bigcap _{f \in F} (C - f)$ is syndetic.

Suppose $B \subseteq \mathbb {N}$ is dynamically syndetic. If $B = \mathbb {N}$ , put $C = \emptyset $ and note that the conclusion holds since the empty intersection is $\mathbb {N}$ . Suppose $B \neq \mathbb {N}$ . By Theorem 2.12, there exists a minimal system $(X,T)$ , a point $x \in X$ , and a non-empty, clopen set $U \subseteq X$ such that $R(x,U) \subseteq B$ . Define $V := X \backslash U$ and $C := \mathbb {N} \backslash R(x,U) = R(x, V)$ . Since $U \neq X$ and the set $R(x,U)$ is syndetic, the set C is non-empty but not thick. Let $F \subseteq \mathbb {N} \backslash B$ be finite. Since $F \subseteq C$ , we have that for all $f \in F$ , the set $T^{-f}V$ is an open neighborhood of x. We see that

$$ \begin{align*}\bigcap_{f \in F} (C - f) = R\bigg( x, \bigcap_{f \in F}T^{-f}V \bigg),\end{align*} $$

which is syndetic since the set $\bigcap _{f \in F}T^{-f}V$ is an open neighborhood of x, as desired.

Proof. We will apply Theorem B(ii). Let $S \subseteq \mathbb {N}$ be piecewise syndetic. Since piecewise syndeticity is partition regular and $S = \bigcup _{i=0}^{k-1} (S \cap (k\mathbb {N} - i))$ , there exists $i \in \{0, \ldots , k-1\}$ such that $S \cap (k \mathbb {N} - i)$ is piecewise syndetic. Since demonstrating item (ii) for a subset of $S+i$ suffices to demonstrate it for S, by replacing S with $(S+i) \cap k\mathbb {N}$ , we can proceed under the assumption that $S \subseteq k \mathbb {N}$ .

Write $S = B \cap C$ , where B is syndetic with maximum gap length less than $\ell \in \mathbb {N}$ and C is thick. For $i \in \{0, \ldots , k-1\}$ , choose an interval $I_i$ of $H_i$ such that $|I_i|> \ell $ and $\max I_i < \min I_{i+1}$ . The set

$$ \begin{align*} F := \bigcup_{i=0}^{k-1} ((k\mathbb{N} + i) \cap I_i) \end{align*} $$

is a finite subset of A. We will show that for all $M \in \mathbb {N}$ , there is a finite subset $S' \subseteq S$ for which the set $S' - F$ contains an interval of length greater than M. It will follow that the set $S - F$ is thick.

Let $M \in \mathbb {N}$ . Choose $S' \subseteq S$ to be the set S intersected with a long interval (whose length will be specified later) on which the distance between consecutive elements of S is less than $\ell $ . Since $\ell < |I_i|$ and $S' \subseteq k\mathbb {N}$ , the set $S' - ((k\mathbb {N} + i) \cap I_i)$ contains the set $(k \mathbb {N} - i) \cap J_i$ , where

$$ \begin{align*} J_i := \{\min S' - \max I_i, \ \ldots, \max S' - \min I_i\}. \end{align*} $$

It follows that

$$ \begin{align*} S' - F \supseteq \bigcup_{i=0}^{k-1} ((k \mathbb{N} - i) \cap J_i) \supseteq \bigcap_{i=0}^{k-1} J_i = \{\min S' - \max I_0, \ \ldots, \max S' - \min I_{k-1} \}, \end{align*} $$

provided $\min S' - \max I_0 \leqslant \max S' - \min I_{k-1}$ . This interval has length $\max S' - \min S' - (\min I_{k-1} - \max I_0)$ , which is greater than M if $\max S' - \min S'> M + \min I_{k-1} - \max I_0$ .

Proof. To apply Theorem B(iii), let $\mathbb {N} \supsetneq B \supseteq A$ . We want to show there exists a finite set $F \subseteq \mathbb {N} \backslash B$ such that $B - F$ is thick. Consider two cases.

Case 1. There exists $i \in \mathbb {N}$ for which the set $\mathbb {N} \backslash B$ contains a complete residue system modulo $p_i$ . Let $F = \{f_0, \ldots , f_{p_i - 1}\} \subseteq \mathbb {N} \backslash B$ be a complete system of modulo $p_i$ residues. We see that

$$ \begin{align*}B - F \supseteq A - F \supseteq ((p_i \mathbb{N} + c_i) \cap H_i ) - F \supseteq \bigcap_{\ell = 0}^{p_i - 1} (H_i - f_{\ell}),\end{align*} $$

which is thick, as desired.

Case 2. For all $i \in \mathbb {N}$ , the set $\mathbb {N} \backslash B$ avoids some modulo $p_i$ congruence class. In this case, for all $i \in \mathbb {N}$ , there exists $a_i \in \mathbb {N}$ such that $B \supseteq p_i \mathbb {N} + a_i$ . We will show that B is thick. Indeed, let $k \in \mathbb {N}$ . It suffices to show there exists $n \in \mathbb {N}$ such that $n + i \in p_i \mathbb {N} + a_i$ for all $1 \leqslant i \leqslant k$ . This is equivalent to showing that there exists $n \in \mathbb {N}$ such that $n \equiv a_i - i \bmod p_i$ for $1 \leqslant i \leqslant k$ . This is an immediate consequence of the Chinese remainder theorem since the $p_i$ terms are distinct primes.

To see that the set A is not an IP set under the stipulated conditions, note that for all $n \in \mathbb {N}$ , the set $A \cap (A-n)$ is contained in a finite union of sets of the form $(p \mathbb {N} + c ) \cap H$ , where $c \not \equiv 0 \bmod p$ and H is thick. If A was an IP set, there would exist $n \in \mathbb {N}$ for which $A \cap (A-n)$ is an IP set. Since the family $\mathcal {IP}$ is partition regular, it would follow that a set of the form $(p \mathbb {N} + c ) \cap H$ is an IP set, which is false as every IP set contains infinitely many multiples of p.

Proof. Let $(X,T)$ be minimal and $\varepsilon> 0$ . Suppose for a contradiction that there are infinitely many i values for which there exists a joining $(J_i, T \times R_i)$ of $(X,T)$ and $(Z_i,R_i)$ and a point $y_i \in Z_i$ for which $\pi _X(J_i \cap (X \times \{y_i\}))$ is not $\varepsilon $ dense in X. To save on notation, by relabeling, we will ignore those $(Z_i,R_i)$ terms which do not fall into this infinite set. By passing to minimal subsystems, we may assume without loss of generality that the joinings $(J_i, T \times R_i)$ are minimal. Since $J_i$ is closed, there exists an $\varepsilon $ -ball $U_i \subseteq X$ and an open set $V_i \subseteq Z_i$ containing $y_i$ such that $J_i \cap (U_i \times V_i) = \emptyset $ .

Fix $x_0 \in X$ . Since $(J_i,T \times R_i)$ is a joining of $(X,T)$ and $(Z_i,R_i)$ , and is minimal, there exists $z_i \in Z_i$ such that $J_i = \overline {(T \times R_i)^{\mathbb {N}}(x_0,z_i)}$ . Since $J_i \cap (U_i \times V_i) = \emptyset $ , we have that

(4.1) $$ \begin{align} R_{R_i}(z_i,V_i) \subseteq R_T(x_0, X \backslash U_i). \end{align} $$

Let $c_1, c_2, \ldots , c_m \in X$ be an $\varepsilon /2$ dense subset of X. For $\ell \in \{1, \ldots , m\}$ , define

$$ \begin{align*} E_\ell = \{i \in \mathbb{N} \ | \ B_{\varepsilon/2}(c_\ell) \subseteq U_i \}. \end{align*} $$

Note that $\mathbb {N} = \bigcup _{\ell =1}^m E_\ell $ . There exists $\ell \in \{1, \ldots , m\}$ for which $E_\ell $ is infinite. Put $U = B_{\varepsilon /2}(c_\ell )$ . It follows now from (4.1) that

(4.2) $$ \begin{align} H := \bigcup_{i \in E_\ell} R_{R_i}(z_i,V_i) \subseteq \bigcup_{i \in E_\ell} R_T(x_0, X \backslash U_i) \subseteq R_T(x_0, X \backslash U). \end{align} $$

We will show that the set H is thickly syndetic and, hence, thick. This will yield a contradiction since, by the minimality of $(X,T)$ , the set $R_T(x_0, X \backslash U)$ cannot be thick.

Let $k \in \mathbb {N}$ and let $e_1, \ldots , e_k \in E_\ell $ . We see that

$$ \begin{align*}\bigcap_{i=1}^k (H-i) \supseteq \bigcap_{i=1}^k R_{R_{e_i}}(R_{e_i}^{i}z_{e_i},V_{e_i}) = R_{R_{e_1} \times \cdots \times R_{e_k}}((R_{e_1}^1z_{e_1}, \ldots, R_{e_k}^{k}z_{e_k}),V_{e_1} \times \cdots \times V_{e_k}),\end{align*} $$

which is syndetic by the minimality of the product system $(Z_{e_1} \times \cdots \times Z_{e_k},R_{e_1} \times \cdots \times R_{e_k})$ . Therefore, intervals of length k appear in H syndetically. Since $k \in \mathbb {N}$ was arbitrary, this shows that the set H is thickly syndetic, as desired.

Proof. Let $(Y,S)$ be minimal and $V \subseteq Y$ be non-empty and open. Let $\varepsilon> 0$ be such that V contains an $\varepsilon $ -ball. According to Theorem 4.3, there exists $i \in \mathbb {N}$ such that all joinings $(J,T_i \times S)$ of $(X_i,T_i)$ and $(Y,S)$ satisfy: for all $x \in X_i$ , the set $\pi _Y(J \cap (\{x\} \times Y))$ is $\varepsilon $ -dense in Y.

Let $U \subseteq X$ be non-empty and open. We claim that for all $(x,y) \in X_i \times Y$ , there exists $n \in \mathbb {N}$ such that $(T_i \times S)^n (x,y) \in U \times V$ . Indeed, let $(x,y) \in X_i \times Y$ . By a theorem of Auslander [Reference Auslander1] and Ellis [Reference Ellis16], the point $(x,y)$ is proximal to a point $(x_0,y_0) \in X_i \times Y$ that is uniformly recurrent under $T_i \times S$ . Since $J := \overline {(T_i \times S)^{\mathbb {N}_0}(x_0,y_0)}$ is a (minimal) joining of $(X_i,T_i)$ and $(Y,S)$ , it is $\varepsilon $ -dense in every fiber over $X_i$ . Since V contains an $\varepsilon $ -ball, we see that $J \cap (U \times V) \neq \emptyset $ . Because $(x,y)$ is proximal to $(x_0,y_0)$ and $R_{T_i \times S}((x_0,y_0), U \times V)$ is syndetic, we see that $R_{T_i \times S}((x,y),U \times V)$ is piecewise syndetic and, hence, non-empty.

We have shown that for all $(x,y) \in X_i \times Y$ , there exists $n \in \mathbb {N}$ such that $(T_i \times S)^n (x,y) \in U \times V$ . It follows that $\bigcup _{n=1}^\infty (T_i \times S)^{-n} (U \times V) = X_i \times Y$ . By compactness, there exists $s \in \mathbb {N}$ such that $\bigcup _{n=1}^s (T_i \times S)^{-n} (U \times V) = X_i \times Y$ . It follows that for all $(x,y) \in X_i \times Y$ , there exists $n \in \{1, \ldots , s\}$ such that $(T_i \times S)^n(x,y) \in U \times V$ , whereby for all $(x,y) \in X_i \times Y$ , the set $R_{T_i \times S}((x,y), U\times V)$ is syndetic with gap size bounded by s, as was to be shown.

Proof. To see that $A \in d\mathcal {T}$ , we must show that for all minimal $(Y,S)$ , all non-empty, open $V \subseteq Y$ , and all $y \in Y$ , the set $A \cap R_S(y,V)$ is non-empty. This follows immediately from Proposition 4.4.

Proof. Since the set P is piecewise syndetic, there exists $k \in \mathbb {N}$ such that $H := \bigcup _{i=1}^k (P-i)$ is thick. Since $(X,T)$ is minimal, the set $T^H x$ is dense. We see that

$$ \begin{align*}X = \overline{\bigcup_{i=1}^k T^{-i} T^{P}x} = \bigcup_{i=1}^k T^{-i} \overline{T^{P}x}.\end{align*} $$

By the Baire category theorem, there exits $i \in \{1, \ldots , k\}$ such that the set $T^{-i} \overline {T^{P}x}$ has non-empty interior. By semiopenness of minimal maps [Reference Kolyada, Snoha and Trofimchuk34, Theorem 2.4], applying i many times the map T, the set $\overline {T^{P}x}$ has non-empty interior, as desired.

Proof. If the set $\bigcup _{i=1}^\infty U_i$ is not dense in X, then it is disjoint from a non-empty, open set $V \subseteq X$ . The set A is clearly disjoint from the set $R(x,V)$ , whereby A is not dynamically thick.

Suppose that $\bigcup _{i=1}^\infty U_i$ is dense. Let $(Y,S)$ be a minimal system, $y \in Y$ , and $V \subseteq Y$ be non-empty and open. Since the set $R_S(y,V)$ is syndetic, by Lemma 4.6, the set $T^{R_S(y,V)} x$ is somewhere dense. Therefore, there exists $i \in \mathbb {N}$ such that $T^{R_S(y,V)} x \cap U_i \neq \emptyset $ . It follows that the set $R_{T \times S} ((x,y), U_i \times V)$ is non-empty. Since x is a distal point, by [Reference Furstenberg20, Theorem 9.11], the point $(x,y)$ is uniformly recurrent under $T \times S$ . Since $U_i \times V$ is open, we have that $R_{T \times S} ((x,y), U_i \times V)$ is (dynamically) syndetic. Therefore, the set $R_{T \times S} ((x,y), U_i \times V) \cap H_i$ is non-empty, which implies that $A \cap R_S(y,V)$ is non-empty, as desired.