1 Introduction

Let be a complex and separable Banach space and () be the set of all bounded linear operators on . Let ℕ⁰ be the set of non-negative integers and let ℕ be the set of natural numbers. We say that is topologically transitive if for any non-empty open sets ⊆ and ⊆ , there exists ∈ ℕ⁰ such that −() ∩ ≠ ∅. One can read more information about these operators in [1- 3]. In the statement of [3] and [4], an operator ∈ () is called mixing, if for any two non-empty open sets ⊆ and ⊆ , there exists ∈ ℕ such that () ∩ non-empty for every ≥ .

Costakis and Sambarino showed in [4] that = is mixing, where is a scalar with || > 1 and is the backward shift on 2 . It is interesting that one can construct mixing operators on every infinite-dimensional separable Banach space [5].

Theorem 1.1. If is any infinite-dimensional separable Banach space, then supports a mixing operator [5].

Let : → be an operator. Then for any sets ⊆ and ⊆ , the return set from to is defined as:

\[N_T(A,B) = \{ n \in \mathbb{N}_0; T^n(A) \cap B \neq \emptyset \}.\]

So, if an operator T is topologically transitive, then \(N_T(U,V)\) is non-empty for any open sets \(U \subseteq X\) and \(V \subseteq X\). As it mentioned in [1], if T is topologically transitive, then \(N_T(U,V)\) is infinite. Note that if T is mixing, then \(N_T(U,V)\) is cofinite. Remember that we say a set S is cofinite if \(\mathbb{N} \setminus S\) is finite.

We call a strictly increasing sequence \((n_k)_k\) of positive integers syndetic if

\[sup_{k\geq 1}(n_{k+1}-n_k)<\infty.\]

We say a subset A of \(\mathbb{N}_0\) is syndetic if the strictly increasing sequence of positive integers forming A is syndetic [1]. The complement of syndetic sets does not contain arbitrary long intervals.

We say an operator \(T \in B(X)\) is topologically ergodic if for any pair of nonempty open sets \(U \subseteq X\) and \(V \subseteq X\), \(N_T(U,V)\) is syndetic [1]. It is not hard to see that mixing operators are topologically ergodic and topologically ergodic operators are topologically transitive. Grosse-Erdmann and Peris proved in [6] that ergodic operators are weakly-mixing. An operator T is called weaklymixing if \(T \oplus T\) is topologically transitive. Remember that if X and Y are two Banach spaces, then \(X \oplus Y = \{(x,y); x \in X, y \in Y\}\) and if \(S \in B(X)\) and \(T \in\)B(Y), then the operator \(S \oplus T: X \oplus Y \to X \oplus Y\) is defined by \((S \oplus T)(x,y) =\)(Sx,Ty).

Madore and Martinez-Avendano introduced subspace-transitive operators in [7]. An operator T is called subspace-transitive with respect to a closed and non-trivial subspace M of X if for any non-empty relatively open sets \(U \subseteq M\) and \(V \subseteq M\), \(T^{-n}(U) \cap V\) contains a non-empty open subset of M for some \(n \in \mathbb{N}_0\). They also defined subspace-hypercyclic operators. For more information, see [8-10]. Also, in [11], one can read interesting properties of subspace-supercyclic operators.

Talebi and Moosapoor defined subspace-mixing operators in [12]. Let M be a closed and non-empty subspace of X. We say an operator \(T \in B(X)\) is M-mixing if for any relatively open sets \(U \subseteq M\) and \(V \subseteq M\) there exists \(N \in \mathbb{N}\) such that for any \(n \ge N\), \(T^n(U) \cap V \ne \emptyset\).

In this paper, we define subspace-ergodic operators and give examples of these operators. We show that by any given separable infinite-dimensional Banach space, we can construct subspace-ergodic operators. We demonstrate that an invertible operator T is subspace-ergodic if and only if \(T^{-1}\) is subspace-ergodic. We prove that the direct sum of two subspace-ergodic operators is subspace-ergodic. Also, we prove that if the direct sum of two operators is subspace-ergodic, then each of them is subspace-ergodic. Moreover, we investigate relations between subspace-ergodic operators and subspace-mixing operators. For example, we show that if T is subspace-mixing and invertible, then \(T^n\) and \(T^{-n}\) are subspace-ergodic for any \(n \in \mathbb{N}\).

2 Definitions and Some Results

First, we define the return set with respect to a subspace. As usual, when we talk about a subspace, it is considered a closed subspace. Also, the idea of this paper is given from subspace-hypercyclic and subspace-mixing operators. So, in the following definitions we will assume that M is a closed subspace.

Definition 2.1. Let \(T \in B(X)\) and let M be a closed and non-empty subspace of X. For \(A \subseteq M\) and \(B \subseteq M\), we define the return set from A to B with respect to M as follows:

\[N_T(A,B)_M = \{n \in \mathbb{N}_0; T^n(A) \cap B \neq \emptyset\}.\]

By Definition 2.1, if T is M-transitive, then \(N_T(U,V)_M\) is non-empty for any non-empty relatively open sets \(U \subseteq M\) and \(V \subseteq M\). Also, if T is an M-mixing operator, then \(N_T(U,V)_M\) is cofinite.

Now we define subspace-ergodic operators as follows:

Definition 2.2. Let \(T \in B(X)\) and let M be a closed and non-empty subspace of X. We say that T is M-ergodic or subspace-ergodic with respect to M if for any non-empty relatively open sets \(U \subseteq M\) and \(V \subseteq M\), the set \(N_T(U,V)_M\) is syndetic.

By definition, it is clear that any ergodic operator is subspace-ergodic since it is sufficient to consider M := X.

Example 2.3. Let \(T \in B(X)\) be an ergodic operator and let I be the identity operator on X. Then \(T^p \oplus \alpha I\) is M-ergodic with respect to \(M := X \oplus \{0\}\) for any \(p \in \mathbb{N}\) and for any scalar \(\alpha\).

Proof. Let \(\alpha\) be a scalar. First, we show that \(T \oplus \alpha I\) is M-ergodic. Let \(U_1\) and \(V_1\) be non-empty relatively open subsets of M. Thus, there exist non-empty open subsets U and V of X such that \(U_1 = U \oplus \{0\}\) and \(V_1 = V \oplus \{0\}\). By hypothesis, T is an ergodic operator. Thus, the set \(\{n \in \mathbb{N}_0; T^n(U) \cap (V) \neq \emptyset\}\) is syndetic. Note that we have

\[(T \oplus \alpha I)^n(U \oplus \{0\}) \cap (V \oplus \{0\}) = (T^n(U) \oplus \{0\}) \cap (V \oplus \{0\})\]\[= (T^n(U) \cap V) \oplus (\{0\} \cap \{0\}) \quad (1)\]\[= (T^n(U) \cap V) \oplus \{0\}.\]

By Eq. (1), we deduce that \(\{n \in \mathbb{N}_0: (T \oplus \alpha I)^n(U_1) \cap (V_1) \neq \emptyset\}\) is syndetic too. Therefore, \(T \oplus \alpha I\) is M-ergodic. Now if T is an ergodic operator, then \(T^p\) is ergodic for any \(p \in \mathbb{N}\) [1, p.62]. Hence, similar to what was shown, \(T^p \oplus \alpha I\) is M-ergodic for any \(p \in \mathbb{N}\). On the other hand, \(\alpha\) is an arbitrary scalar. So, for any scalar \(\alpha\), \(T^p \oplus \alpha I\) is M-ergodic.

Similarly, \(\alpha I \oplus T^p\) is N-ergodic with respect to \(N := \{0\} \oplus X\) for any \(p \in \mathbb{N}\) and for any scalar \(\alpha\).

By the above examples, we can gain more subspace-ergodic operators from known ergodic operators. For instance, note to the following example:

Example 2.4. Let be a weighted shift on ² given by

\[T(x_1, x_2, x_3, x_4, \dots) = (2x_2, \frac{3}{2}x_3, \frac{4}{3}x_4, \dots).\]

As was proved in [1, Example 2.39], for any two non-empty open subsets and of 2 , we have (, ) is syndetic. Hence, is an ergodic operator and by Example 2.3, ⊕ is -ergodic with respect to : = ² ⊕ {0} for any ∈ ℕ and for any scalar .

As mentioned before, mixing operators are ergodic. Also, it is not hard to see that if is a mixing operator, then is mixing for any ∈ ℕ. So, by Example 2.3, we can create the following example:

Example 2.5. Let ∈ () be a mixing operator and let I be the identity operator on . Then ⊕ is -ergodic with respect to : = ⊕ {0} for any ∈ ℕ and for any scalar . For example, let = be the Rolewicz's operator on 2 , where is the backward shift on 2 and be a scalar with || > 1. As it said in the introduction, = is mixing. So, for any scalar α, the operator () ⊕ is M-ergodic with respect to : = ²⊕{0}. Also, it is easy to see that ⊕ () is -ergodic with respect to : = {0}⊕ 2 .

Corollary 2.6. Let be a separable and infinite-dimensional Banach space. Then there exists a subspace-ergodic operator on ⊕ .

Proof. By Theorem 1.1, there exists a mixing operator on . Then ⊕ is the desired operator by Example 2.5.

Now like subspace-hypercyclicity and subspace-supercyclicity, some questions arise for subspace-ergodicity as follows:

Question 1. Let be an invertible -ergodic operator. Can we conclude that −1 is also -ergodic?

Question 2. Let be an -ergodic operator. Is is M-ergodic for any ∈ ℕ?

Question 3. Let be a scalar with || = 1. Does the -ergodicity of imply -ergodicity of ?

Question 4. Let be an ergodic operator on . Is there a closed and non-trivial subspace of such that is -ergodic?

Question 5. Is the direct sum of two subspace-ergodic operators also subspaceergodic?

In this section, we answer Question 1 affirmatively. Also, we partially answer Question 2 and Question 3. In the next section, we answer Question 5.

In the next theorem, we show that the answer to Question 1 is positive.

Theorem 2.7. Let \(T \in B(X)\) be an invertible operator and let M be a closed and non-empty subspace of X. Then T is M-ergodic if and only if \(T^{-1}\) is M-ergodic.

Proof. Let \(U \subseteq M\) and \(V \subseteq M\) be two non-empty relatively open sets. Let \(n \in N_T(U,V)_M\). So, \(T^n(U) \cap V \neq \emptyset\). T is invertible and

\[T^{-n}(T^n(U)\cap V)\neq\emptyset.\]

Hence, \(T^{-n}(V) \cap U \neq \emptyset\). This means that \(n \in N_{T^{-1}}(V, U)_M\). So,

\[N_T(U,V)_M \subseteq N_{T^{-1}}(V,U)_M\].

Similarly, we have

\[N_{T^{-1}}(V,U)_M \subseteq N_T(U,V)_M.\]

Therefore, \(N_T(U, V)_M = N_{T^{-1}}(V, U)_M\).

Hence, \(N_T(U,V)_M\) is syndetic if and only if \(N_{T^{-1}}(V,U)_M\) is syndetic. Therefore, T is M-ergodic if and only if \(T^{-1}\) is M-ergodic

Now it is natural to note the powers of a subspace-ergodic operator. In the next theorem, we prove that if \(T^n\) is subspace-ergodic for some \(n \in \mathbb{N}\), then T is subspace-ergodic too.

Theorem 2.8. Let \(T \in B(X)\) and let M be a closed and non-empty subspace of X. If \(T^n\) is M-ergodic for some \(n \in \mathbb{N}\), then T is also M-ergodic.

Proof. Let \(U \subseteq M\) and \(V \subseteq M\) be two non-empty relatively open sets. By hypothesis, \(T^n\) is M-ergodic. So, \(N_{T^n}(U,V)_M\) is syndetic. But if \((T^n)^k(U) \cap V \neq \emptyset\), then \(T^{nk}(U) \cap V \neq \emptyset\). Hence, if \(k \in N_{T^n}(U,V)_M\), then \(k \in N_T(U,V)_M\). So,

\[N_{T^n}(U,V)_M \subseteq N_T(U,V)_M\].

Let \((m_k)_k\) be the elements of \(N_{T^n}(U,V)_M\) and let \((t_k)_k\) be the elements of \(N_T(U,V)_M\). Hence,

\[\sup_{1 \le k \le \infty} (m_{k+1} - m_k) \ge \sup_{1 \le k \le \infty} (t_{k+1} - t_k).\]

By definition, \(\sup_{1 \le k < \infty} (m_{k+1} - m_k) < \infty\) and so \(\sup_{1 \le k < \infty} (t_{k+1} - t_k)\) is less than infinity too. That means \(N_T(U,V)_M\) is syndetic and hence, T is an M-ergodic operator.

In the next theorem, we show that if T is an M-mixing operator, then \(T^n\) is M-ergodic for any \(n \in \mathbb{N}\) and therefore, we have a partial answer to Question 2.

Theorem 2.9. Let \(T \in B(X)\) and let M be a closed and non-empty subspace of X. If T is an M-mixing operator, then \(T^n\) is an M-ergodic operator for any \(n \in \mathbb{N}\)

Proof. It is immediately obtained by Definition 2.2 that T is M-ergodic. Now, let n > 1 be an arbitrary natural number. First, we show that \(T^n\) is M-mixing.

Suppose that \(U \subseteq M\) and \(V \subseteq M\) are non-empty relatively open sets. By hypothesis, there exists \(N \in \mathbb{N}\) such that \(T^k(U) \cap V\) is non-empty for any \(k \ge N\). On the other hand, for every \(k \in \mathbb{N}\) we have \(kn \ge k\). So, \(T^{kn}(U) \cap V \ne \emptyset\) for any \(k \ge N\). Hence,

\[(T^n)^k(U) \cap V \neq \emptyset\], for any \(k \ge N\).

Therefore, \(T^n\) is M-mixing and hence, \(T^n\) is M-ergodic. But n > 1 is an arbitrary natural number. So, \(T^n\) is M-ergodic for any \(n \in \mathbb{N}\).

Lemma 2.10. Let \(T \in B(X)\) be an invertible operator and let M be a closed and non-empty subspace of X. Then T is M-mixing if and only if \(T^{-1}\) is M-mixing.

Proof. Let \(U \subseteq M\) and \(V \subseteq M\) be two non-empty relatively open sets. By hypothesis, T is an M-mixing operator. So, \(N_T(U,V)_M\) is cofinite. But

\[N_T(U,V)_M = N_{T^{-1}}(V,U)_M.\]

Hence, \(N_T(U, V)_M\) is cofinite if and only if \(N_{T^{-1}}(V, U)_M\) is cofinite. This means that T is M-mixing if and only if \(T^{-1}\) is M-mixing.

The next corollary is a direct result of Theorem 2.9 and Lemma 2.10.

Corollary 2.11. Let T be an invertible and M-mixing operator. Then for any \(n \in \mathbb{N}\), \(T^n\) and \(T^{-n}\) are M-ergodic.

Now we mention a theorem from [13] and by this we have a partial answer to Question 3.

Theorem 2.12. Let \(T \in B(X)\). Then T is M-mixing with respect to a non-empty and closed subspace M of X if and only if for any non-empty relatively open set \(U \subseteq M\) and any 0-neighborhood W in M there exists a positive integer N such that for any \(n \ge N\), \(T^n(U) \cap W \ne \emptyset\) and \(T^n(W) \cap U \ne \emptyset\) [13].

In other words, T is M-mixing if and only if \(N_T(U,W)_M\) and \(N_T(W,U)_M\) are cofinite, for any non-empty relatively open set \(U \subseteq M\) and any 0-neighborhood W in M.

Theorem 2.13. Let \(T \in B(X)\) be an M-mixing operator. Let \(\lambda\) be a scalar with \(|\lambda| = 1\). Then \(\lambda T\) is M-ergodic.

Proof. Let T be a subspace-mixing operator with respect to a closed and nonempty subspace M of X. We show that \(\lambda T\) is an M-mixing operator and hence it is M-ergodic. Let \(U \subseteq M\) be a relatively open set and let W be a 0neighborhood in M. By hypothesis, \(N_T(U,W)_M\) and \(N_T(W,U)_M\) are cofinite. We can find a balanced set \(W_1\), a neighborhood of zero in M such that \(W_1 \subseteq W\). Again by hypothesis, \(N_T(U,W_1)_M\) is cofinite. So, there exists \(N \in \mathbb{N}\) such that

\[T^{n}(U) \cap W_{1} \neq \emptyset \quad (n \ge N). \tag{2}\]

Let \(n \ge N\) be an arbitrary natural number. By Eq. (2), \(T^n(U) \cap W_1\) is non-empty. So, there exists \(x \in U\) such that \(T^n x \in W_1\). Since \(|\lambda^n| = 1\) and \(W_1\) is a balanced set.

\[\lambda^n T^n x \in \lambda^n W_1 \subseteq W_1\].

Hence, \(\lambda^n T^n x \in W_1\) and therefore, \((\lambda^n T^n)(U) \cap W_1 \neq \emptyset\). Since \(W_1 \subseteq W\), \((\lambda^n T^n)(U) \cap W \neq \emptyset\).

Therefore, \(N_{\lambda T}(U, W)_M\) is cofinite and similarly, \(N_{\lambda T}(W, U)_M\) is cofinite. So, by Theorem 2.12, \(\lambda T\) is M-mixing which completes the proof.

3 On the Direct Sum of Two Subspace-ergodic Operators

First, we show that the direct sum of two subspace-ergodic operators is subspace-ergodic. In fact, we show that the answer to Question 5 is positive. In this section, M and N always indicate closed and non-zero subspaces of X and Y respectively.

Theorem 3.1. Let \(S \in B(X)\) be an M-ergodic operator and let \(T \in B(Y)\) be an N-ergodic operator. Then, \(S \oplus T\) is \(M \oplus \{0\}\)-ergodic and \(\{0\} \oplus N\)-ergodic.

Especially, \(T \oplus T\) is \(N \oplus \{0\}\)-ergodic and \(\{0\} \oplus N\)-ergodic.

Proof. Let \(U \subseteq M\) and \(V \subseteq M\) be non-empty relatively open sets. Then,

\[(S \oplus T)^n(U \oplus \{0\}) \cap (V \oplus \{0\}) = (S^n \oplus T^n)(U \oplus \{0\}) \cap (V \oplus \{0\})\]\[= (S^n(U) \cap V) \oplus (T^n(\{0\}) \cap \{0\})\]\[= (S^n(U) \cap V) \oplus \{0\}.\]

So,

\[N_{S \oplus T} ((U \oplus \{0\}), (V \oplus \{0\}))_{M \oplus \{0\}} = N_S(U, V)_M.\] (3)

By hypothesis, \(N_S(U,V)_M\) is syndetic. So, by Eq. (3), \(N_{S\oplus T}(U\oplus\{0\}), (V\oplus\{0\}))_{M\oplus\{0\}}\) is syndetic. This means that \(S\oplus T\) is \(M\oplus\{0\}\)-ergodic. Similarly, \(S\oplus T\) is \(\{0\}\oplus N\)-ergodic.

By Theorem 3.1 and Theorem 2.7 we can conclude the following corollary:

Corollary 3.2. Let \(S \in B(X)\) be an M-ergodic operator and let \(T \in B(Y)\) be an N-ergodic operator. If S and T are invertible operators, then \((S \oplus T)^{-1}\) is \(M \oplus \{0\}\)-ergodic and \(\{0\} \oplus N\)-ergodic.

If \(S \in B(X)\) and \(T \in B(Y)\) are topologically ergodic operators, then \(S^p \oplus T^q\) is topologically ergodic on \(X \oplus Y\) for any \(p, q \in \mathbb{N}\) [1, p. 173].

Now the question arises if this is also true for subspace-ergodic operators? We partially answer this question in the next theorem.

Theorem 3.3. Let ∈ () be an -ergodic operator and let ∈ () be an -ergodic operator. Then

• (i) if is an M-mixing operator, then ⊕ is ⊕ -ergodic for any ∈ ℕ,
• (ii) if is an N-mixing operator, then ⊕ is ⊕ -ergodic for any ∈ ℕ.

Proof. We prove part (i) and the proof of part (ii) is similar. First, we prove that ⊕ is ⊕ -ergodic. Let 1,² ⊆ and 1,² ⊆ be non-empty relatively open sets. By hypothesis, is an -mixing operator. So, there exists a natural number such that for any ≥ ,

\[S^n(U_1)\cap (U_2)\neq \emptyset.\]

So,

\[{n \in \mathbb{N}; n \geq p} \subseteq N_S(U_1, U_2)_M.\]

By hypothesis, is an -ergodic operator and so (1,² ) is syndetic. On the other hand,

\[(S \oplus T)^n(U_1 \oplus V_1) \cap (U_2 \oplus V_2) = \left(S^n(U_1) \cap (U_2)\right) \oplus \left(T^n(V_1) \cap (V_2)\right).\]

Hence,

\[\begin{split} N_{S \oplus T}((U_1 \oplus V_1), (U_2 \oplus V_2))_{M \oplus N} &= N_S(U_1, U_2)_M \cap N_T(V_1, V_2)_N \\ & \supseteq \{n \in \mathbb{N}; n \geq p\} \cap N_T(V_1, V_2)_N \\ & = \{n \in \mathbb{N}; n \geq p \text{ and } n \in N_T(V_1, V_2)_N\}. \end{split}\]

Since (1,² ) is syndetic, we can deduce that ⊕((¹ ⊕ 1), (² ⊕ 2))⊕ is syndetic. Now note that if is -mixing, then is -mixing for any ∈ ℕ which completes the proof.

Bamerni and Kilicman showed in [14] that the direct sum of two subspacemixing operators is also subspace-mixing. Now we extend their statement as follows:

Theorem 3.4. Let ∈ () be an -mixing operator and let ∈ () be an -mixing operator. Then, ( ⊕ ) is ⊕ -mixing for any ∈ ℕ.

Moreover, if and are invertible, then ( ⊕ ) − is also ⊕ -mixing for any ∈ ℕ. Especially, ( ⊕ ) and ( ⊕ ) − are ⊕ -ergodic for any ∈ ℕ.

Proof. Let 1,² ⊆ and 1,² ⊆ be non-empty relatively open sets. is an -mixing operator. So, there exists a natural number ¹ such that for any ≥ 1,

\[S^{n}(U_{1}) \cap (U_{2}) \neq \emptyset. \tag{4}\]

On the other hand, T is an N-mixing operator. So, there exists a natural number \(N_2\) such that for any \(n \ge N_2\),

\[T^n(V_1) \cap (V_2) \neq \emptyset. \tag{5}\]

Let \(p = max\{N_1, N_2\}\). So, by Eq. (4) and Eq. (5), for any \(n \ge p\) we have,

\[S^n(U_1) \cap (U_2) \neq \emptyset\] and \(T^n(V_1) \cap (V_2) \neq \emptyset\).

Hence, for any \(n \ge p\),

\[(S \oplus T)^n(U_1 \oplus V_1) \cap (U_2 \oplus V_2) = (S^n(U_1) \cap U_2) \oplus (T^n(V_1) \cap V_2) \neq \emptyset.\]

So, \(N_{(S \oplus T)^n}((U_1 \oplus V_1), (U_2 \oplus V_2))_{M \oplus N}\) is cofinite and hence, \((S \oplus T)^n\) is \(M \oplus N\)-mixing.

The proof of the rest of the theorem is an easy consequence of Lemma 2.10 and Theorem 2.9.

Finally, we prove that subspace-ergodicity of the direct sum of two operators, indicates subspace-ergodicity of each of them.

Theorem 3.5. Let \(S \in B(X)\) and \(T \in B(Y)\). If \(S \oplus T\) is an \(M \oplus N\)-ergodic operator, then S is an M-ergodic operator and T is an N-ergodic operator.

Especially, if \(T \oplus T\) is \(N \oplus N\) ergodic, then T is N-ergodic.

Proof. Let \(U_1 \subseteq M\) and \(U_2 \subseteq M\) be non-empty relatively open sets. We prove that \(N_S(U_1, U_2)_M\) is syndetic.

Suppose that \(V_1 \subseteq N\) and \(V_2 \subseteq N\) be non-empty relatively open sets. By hypothesis, \(S \oplus T\) is \(M \oplus N\)-ergodic. So, \(N_{S \oplus T}((U_1 \oplus V_1), (U_2 \oplus V_2))_{M \oplus N}\) is syndetic. Let \(n \in N_{S \oplus T}((U_1 \oplus V_1), (U_2 \oplus V_2))_{M \oplus N}\). So,

\[(S \oplus T)^n (U_1 \oplus V_1) \cap (U_2 \oplus V_2) \neq \emptyset.\]

And hence,

\[(S^n(U_1) \cap (U_2)) \oplus (T^n(V_1) \cap (V_2)) \neq \emptyset.\]

Therefore, \(S^n(U_1) \cap U_2\) must be non-empty and hence, \(n \in N_S(U_1, U_2)_M\). This means

\[N_{S \oplus T}((U_1 \oplus V_1), (U_2 \oplus V_2))_{M \oplus N} \subseteq N_S(U_1, U_2)_M.\]

Since \(N_{S \oplus T}((U_1 \oplus V_1), (U_2 \oplus V_2))_{M \oplus N}\) is syndetic, \(N_S(U_1, U_2)_M\) is syndetic. Similarly, T is also N-ergodic.