Akin和Kolyada^[1]引入了Li-Yorke敏感的概念将Li-Yorke混沌和初值敏感联系起来,并且证明了弱混合动力系统是Li-Yorke敏感的, Moothathu^[16]介绍了一系列通过自然数子集表示的敏感,得到了syndetic敏感严格强于敏感,有限余敏感严格强于syndetic敏感.在文献[13]中,作者中讨论了关于一些敏感性和传递性的乘积系统,在文献[11, 12]中,刘等讨论了紧空间上的敏感, syndetic敏感, thickly syndetic敏感, Li-Yorke敏感.黄文等^[8]介绍了传递敏感的概念,并且得到了multi敏感,敏感紧和传递敏感关于极小系统是等价的.不仅如此,黄文等^{[6, 7]}证明了multi敏感, thick敏感, thickly syndetic敏感关于$ M $系统是等价的, multi敏感和thick敏感关于传递系统是等价的,李^[10]等介绍了超空间上$ (M(X), {T_M}) $上的multi-thick敏感, multi-syndetic敏感, multi-thickly-syndetic敏感. Godefroy^[4]证明了超循环的算子是敏感的,刘,崔,王等专家分别研究了Banach空间各种算子^[22-24],迭代动力系统可能是Hutchinson^[9]初次介绍和被Barnsley^[2]进一步发展. Ghane^[3]研究了迭代动力系统的敏感性,并且证明了如果迭代动力系统是强传递的非极小系统,那么此动力系统是敏感的,更多的关于敏感性,乘积系统,迭代动力系统, Frechet空间请参考文献[5, 14, 15, 17-21].但是,上面许多专家集中研究动力系统$ (X, T) $,其中$ X $是紧空间, $ T:X \to X $是连续满射.本文则集中研究动力系统$ (X, T) $,其中$ X $是Frechet空间, $ T:X \to X $是算子.本文主要得到了如下结果.

首先证明了以下命题等价: (1) $ (X, T) $是敏感的; (2) $ (X, T) $是multi敏感; (3) $ (X, T) $是multi-thick敏感; (4) $ (X, T) $是thick敏感.

不仅如此,还证明了如下命题$ (X, T) $是syndetic敏感当且仅当$ (X, T) $是thickly syndetic敏感.如果$ (X, T) $是syndetic传递,那么$ (X, T) $是syndetic敏感.如果$ (X, T) $是$ F $ -超循环,那么$ (X, T) $是thickly syndetic敏感.如果$ (X, T) $是syndetic传递,那么$ (X, T) $是传递敏感.最后证明了迭代动力系统也有类似的结果.

本部分先回忆一些动力系统的概念, $ {{\Bbb Z}^ + } $, $ {{\Bbb R}} $, $ {\Bbb C} $表示非负整数,实数,复数.

定义1.1  (1)集合$ S \subset {{\Bbb Z}_ + } $称为syndetic的是指它具有有界的间距,即存在$ M > 0 $,使得对任意的$ n \in {{{\Bbb Z}^ + }} $满足$ S \cap \{ n, n + 1, \cdots , n + M\} \ne \emptyset $;

(2)集合$ S \subset {{\Bbb Z}_ + } $称为thick的是指它包含了任意长的整数段,即存在序列$ {n_i} \to \infty $,使得$ S \supset \bigcup\limits _{i = 1}^\infty \{ {n_i}, {n_i} + 1, \cdots , {n_i} + i\} $;

(3)集合$ S \subset {{\Bbb Z}_ + } $称为thickly syndetic的是指对任意$ n \in {{\Bbb Z}^ + } $,存在一个syndetic集$ {S^n} = \{ s_1^n < s_2^n < \cdots \} $使得$ S \supset \bigcup\limits_{j = 1}^\infty {\{ s_j^n, s_j^n + 1, \cdots , s_j^n + n\} } $.

注1.1  分别记全体syndetic集, thick集, thickly syndetic集为$ {\cal F}_s $, $ {\cal F}_t $, $ {\cal F}_{ts} $.

定义1.2  (1)设$ p $是定义在$ {\Bbb K} $ ($ {{\Bbb R}} $或者$ {\Bbb C} $)上的向量空间$ X $非负实值函数,满足如下条件:

(ⅰ) $ p(x + y) \le p(x) + p(y) $, $ \forall x, y \in X $;

(ⅱ) $ p(\lambda x) = |\lambda |p(x) $, $ \forall x \in X, \alpha \in {\Bbb K} $;则称$ p $是$ X $上的一个半范数.

(2)半范数序列$ {({p_n})_n} $称为分离的,若对于任意的$ n \ge 1 $使得$ {p_n}(x) = 0 $,则$ x = 0 $.

(3)向量空间$ X $称为Frechet空间,若在其上定义一组分离的递增的半范数$ {({p_n})_n} $,并且在度量$ d(x, y) = \sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}\min (1, {p_n}(x - y))} $, $ x, y \in X $是完备的.

注1.2  (1)度量$ d $称为不变的,是指对任意的$ x, y, z \in X $, $ d(x, y) = d(x + z, y + z) $.

(2)令$ X $, $ Y $为Frechet空间,连续的线性函数$ T:X \to Y $称为算子.记全体这样的算子$ L(X, Y) $,如果$ Y = X $,则称$ T $为$ X $上的算子,记$ L(X) = L(X, X) $.

(3)令$ X $, $ Y $为Frechet空间,分别在其上定义一组分离的递增的半范数$ {({p_n})_n} $和$ {({q_n})_n} $,定义空间$ X \oplus Y = \{ (x, y):x \in X, y \in Y\} $,并且在其上定义一组分离的递增的半范数$ {r_n}(x, y) = {p_n}(x) + {q_n}(y) $,那么空间$ X \oplus Y = \{ (x, y):x \in X, y \in Y\} $为Frechet空间^[5].

注1.3  (1)令$ S:X \to X $和$ T:Y \to Y $分别为定义在Frechet空间$ X $和$ Y $上的算子,那么定义算子$ S \oplus T $为$ S \oplus T:X \oplus Y \to X \oplus Y $, $ (S \oplus T)(x, y) = (Sx, Ty) $, $ x \in X, y \in Y $.

(2) $ (X, T) $称为线性动力系统,若$ X $是一个可分的Frechet空间, $ T:X \to X $是一个算子.或简称$ T $或者$ T:X \to X $是一个线性动力系统.从此刻起,如果没特别指出,所有的算子定义在一个可分的Frechet空间上.

(3)设$ A \subset {{\Bbb Z}^ + } $,定义$ \underline {dens} (A) = \mathop {\lim }\limits_{N \to \infty } \inf \frac{{card\{ 0 \le n \le N;n \in A\} }}{{N + 1}} $为$ A $的上密度.

(4)对于动力系统$ (X, T) $, $ x \in X $,以及$ U $, $ V \subset X $,定义$ N(x.U) = \{ n \in {{\Bbb Z}^ + };{T^n}x \in U\} $. $ A - A = \{ a - b \ge 0;a, b \in A\} $, $ N(U, V) = \{ n \in {{\Bbb Z}^ + };U \cap {T^{ - n}}V \ne \emptyset \} $$ = \{ n \in {{\Bbb Z}^ + };{T^n}U \cap V \ne \emptyset \} $.

定义1.3  (1) $ (X, T) $称为$ F $ -超循环的,若存在$ x \in X $使得对于任意的非空开集$ U \subset X $, $ \underline {dens} \{ n \in {{\Bbb Z}^ + }:{T^n}x \in U\} > 0 $.并且称$ x $是$ T $的$ F $ -超循环向量,记$ T $的所有$ F $ -超循环向量组成的集合为$ FHC(T) $.

(2) $ (X, T) $称为超循环的,若存在$ x \in X $使得$ orb(x, T) = \{ x, Tx, {T^2}x, \cdots \} $稠于$ X $.并且称$ x $是$ T $的超循环向量,记$ T $的所有超循环向量组成的集合为$ HC(T) $.

(3) $ (X, T) $称为传递的,是指对任意的非空开集$ U, V \subset X $有$ N(U, V) \ne \emptyset $.

(4) $ (X, T) $称为syndetic传递的,是指对任意的非空开集$ U, V \subset X $有$ N(U, V) $是syndetic集.

注1.4  (1) $ (X, T) $是超循环的当且仅当$ (X, T) $称为传递,并且$ HC(T) $是一个稠密的$ {G_\delta } $.

(2)设$ (X, T) $为动力系统,其中$ X $上的度量为$ d $,令$ {N_T}(U, \delta ) = \{ n \in {{\Bbb Z}^ + }; $存在$ {x_1}, {x_2} \in U $使得$ d({T^n}{x_1}, {T^n}{x_2}) > \delta \}, $$ B(x, r) = \{ y \in X;d(x, y) < r\} . $

定义1.4  设$ (X, T) $为线性动力系统,其中$ X $上的度量为$ d $,

(1) $ (X, T) $称为敏感的,是指存在$ \delta > 0 $ (敏感常数)使得对任意$ x \in X $和$ r > 0 $,能找到$ y \in B(x, r) $和$ n \in {{\Bbb Z}^+} $满足$ d({T^n}x, {T^n}y) > \delta $.

(2) $ (X, T) $称为syndetic (或者thick或者thickly syndetic)敏感的,是指存在$ \delta>0 $ (敏感常数)使得对任意非空开集$ U \subset X $满足$ {N_T}(U, \delta ) $是syndetic (或者thick或者thickly syndetic)集.

(3) $ (X, T) $称为multi敏感的,是指存在$ \delta >0 $ (敏感常数)使得对任意非空开集$ {U_1}, {U_2}, $$ \cdots , {U_k} \subset X $满足$ \bigcap\limits_{i = 1}^k {{N_T}({U_i}, \delta )} \ne \emptyset $.

(4) $ (X, T) $称为multi-syndetic (或者multi-thick或者multi-thickly syndetic)敏感的,是指存在$ \delta > 0 $ (敏感常数)使得对任意非空开集$ {U_1}, {U_2}, \cdots , {U_k} \subset X $满足$ \bigcap\limits_{i = 1}^k {{N_T}({U_i}, \delta )} $是syndetic (或者thick或者thickly syndetic)集.

(5) $ (X, T) $称为传递敏感的,是指存在$ \delta > 0 $ (敏感常数)使得对任意非空开集$ U, V, W \subset X $满足$ {N_T}(W, \delta ) \cap N(U, V) \ne \emptyset $.

定义1.5  设$ (X, d) $是Frechet空间和$ f_{\lambda} $, $ \lambda \in \Lambda $ (其中$ \Lambda $是有限非空数集)是$ X $上的算子,那么$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $称为迭代动力系统.

注1.5  记所有的无限序列$ \{\lambda_{i}\}_{i\in {{\Bbb Z}}^{+}} $,其中$ {\lambda _i} \in \Lambda $组成的集合为$ \Lambda^{{{\Bbb Z}}^{+}} $,例若$ \sigma \in {\Lambda ^{{{\Bbb Z}_ + }}} $,可记$ \sigma = \{\lambda_{0}, \lambda_{1}, \cdots \} $.对任意$ k\in {{\Bbb Z}}^{+} $,令$ {\cal F}^{k} = \{X; f_{\lambda_{k-1}}\circ \cdots\circ f_{\lambda_{0}}| \lambda_{0}, \cdots, \lambda_{k-1}\in \Lambda \} $, $ f_\sigma ^n(x) = {f_{{\lambda _{n - 1}}}} \circ \cdots \circ {f_{{\lambda _0}}}(x), x \in X $, $ f_\sigma ^{ - n}(U) = \{ x \in X;f_\sigma ^nx \in U\} $.

注1.6  对任意$ \delta > 0 $和任意非空开集$ U, V\subset X $,定义

$ N_{{\cal F}}(U, \delta) = \{n\in {{\Bbb Z}}^+; \sup\limits_{\sigma \in \Lambda^{{{\Bbb Z}}^{+}}}\mbox{diam}f_{\sigma}^{n}(U)>\delta \} $

和

$ N_{{\cal F}}(U, V) = \{n\in {{\Bbb Z}}^+; f_{\sigma}^{n}(U)\cap V\neq\emptyset\ \mbox{存在}\ \sigma \in \Lambda^{{{\Bbb Z}}^{+}}\}. $

一般在算子理论经常写作$ f_\sigma ^nx = f_\sigma ^n(x) $.

定义1.6  (1)迭代动力系统$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $称为传递的,是指对任意的非空开集$ U, V \subset X $满足$ {N_{{\cal F}}}(U, V) \ne \emptyset $.

(2)迭代动力系统$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $称为syndetic传递的,是指对任意的非空开集$ U, V \subset X $满足$ {N_{\cal F}}(U, V) $是syndetic集.

定义1.7  设$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $为迭代动力系统,其中$ X $上的度量为$ d $,

(1) $ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $称为敏感的,是指存在$ \delta > 0 $ (敏感常数)使得对任意$ x \in X $和$ r > 0 $,能找到$ y \in B(x, r) $，$ \sigma \in \Lambda^{{{\Bbb Z}}^{+}} $和$ n\in {{\Bbb Z}}^{+} $满足$ d(f_{\sigma}^{n}(x), f_{\sigma}^{n}(y))>\delta $.

(2) $ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $称为syndetic (或者thick或者thickly syndetic)敏感的,是指存在$ \delta > 0 $ (敏感常数)使得对任意非空开集$ U \subset X $满足$ N_{{\cal F}}(U, \delta) $是syndetic (或者thick或者thickly syndetic)集.

(3) $ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $称为multi敏感的,是指存在$ \delta > 0 $ (敏感常数)使得对任意非空开集$ {U_1}, {U_2}, \cdots , {U_k} \subset X $满足$ \bigcap\limits_{1\leq i\leq k}N_{{\cal F}}(U_{i}, \delta)\neq\emptyset $.

(4) $ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $称为multi-syndetic (或者multi-thick或者multi-thickly syndetic)敏感的,是指存在$ \delta > 0 $ (敏感常数)使得对任意非空开集$ {U_1}, {U_2}, \cdots , {U_k} \subset X $满足$ \bigcap\limits_{1\leq i\leq k}N_{{\cal F}}(U_{i}, \delta) $是syndetic (或者thick或者thickly syndetic)集.

(5) $ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $称为传递敏感的,是指存在$ \delta > 0 $ (敏感常数)使得对任意非空开集$ U, V, W \subset X $满足$ {N_{{\cal F}}}(U, \delta ) \cap {N_{{\cal F}}}(U, V) \ne \emptyset $.

命题2.1  $ (X, T) $是敏感的当且仅当$ (X, T) $是multi敏感的.

证  (必要性) 设$ (X, T) $是敏感的(敏感常数为$ \delta > 0 $),令$ d $为$ X $上的任意传递不变的度量,对任意的$ {r_1}, {r_2}, \cdots {r_k} > 0 $和$ {x_1}, {x_2}, \cdots, {x_k} \in X $,则$ B({x_1}, {r_1}), B({x_2}, {r_2}), \cdots , B({x_k}, {r_k}) $是$ X $上的非空开集,因为$ d $是传递不变的,所以

$ B({x_1}, {r_1}) = {x_1} + B(0, {r_1}), B({x_2}, {r_2}) = {x_2} + B(0, {r_2}), \cdots , B({x_k}, {r_k}) = {x_k} + B(0, {r_k}), $

又因存在$ B(0, l) \subset \bigcap\limits_{i = 1}^k {B(0, {r_i})} $,所以

(2.1) $ \begin{equation} \emptyset \ne {N_T}(B(0, l), \delta ) \subset {N_T}(\bigcap\limits_{i = 0}^k {B(0, {r_i}), \delta } ), \end{equation} $

对任意$ n \in {N_T}(B(0, l), \delta ) $,则存在$ {z_1}, {z_2} \in B(0, l) $满足$ d({T^n}{z_1}, {T^n}{z_2}) > \delta $,易知

$ {x_i} + {z_1}, {x_i} + {z_2} \in B({x_i}, {r_i}), {\quad} i = 1, \cdots , k, $

且

(2.2) $ \begin{equation} d({T^n}({x_i} + {z_1}), {T^n}({x_i} + {z_2})) = d({T^n}{x_i} + {T^n}{z_1}, {T^n}{x_i} + {T^n}{z_2}) = d({T^n}{x_1}, {T^n}{x_2}) > \delta. \end{equation} $

因此$ n \in \bigcap\limits_{i = 1}^k {{N_T}(B({x_i}, {r_i}), \delta )} $,所以$ (X, T) $是multi敏感的(敏感常数为$ \delta > 0 $).

(充分性) 显然的.

命题2.2  $ (X, T) $是multi敏感的当且仅当$ (X, T) $是multi-thick敏感的.

证  (必要性) 方法(1)设$ (X, T) $是multi敏感的(敏感常数为$ \delta > 0 $),令$ d $为$ X $上的任意传递不变的度量,对任意的$ {r_1}, {r_2}, \cdots {r_k} > 0 $和$ {x_1}, {x_2}, \cdots {x_k} \in X $,则$ B({x_1}, {r_1}), B({x_2}, {r_2}), \cdots , $$ B({x_k}, {r_k}) $是$ X $上的非空开集,剩下的证明类似命题3.2^[6].实际上因$ (X, T) $是multi敏感的(敏感常数为$ \delta > 0 $),对任意的$ N \in {{\Bbb Z}^ + } $,则

(2.3) $ \begin{equation} \bigcap\limits_{j = 0}^N {\bigcap\limits_{i = 1}^k {{N_T}({T^{ - j}}(B({x_i}, {r_i})), \delta )} } \ne \emptyset, \end{equation} $

因此取$ {n_N} \in \bigcap\limits_{j = 0}^N {\bigcap\limits_{i = 1}^k {{N_T}({T^{ - j}}(B({x_i}, {r_i})), \delta )} } $和$ {n_N} > N $,明显$ \{ {n_N}, {n_N} - 1, \cdots , {n_N} - N\} \subset \bigcap\limits_{i = 1}^k {{N_T}(B({x_i}, {r_i}), \delta )} $,因此$ (X, T) $是multi-thick敏感的(敏感常数为$ \delta > 0 $).

方法(2)设$ (X, T) $是multi敏感的(敏感常数为$ \delta > 0 $),令$ d $为$ X $上的任意传递不变的度量,对任意的$ {r_1}, {r_2}, \cdots {r_k} > 0 $和$ {x_1}, {x_2}, \cdots {x_k} \in X $,则$ B({x_1}, {r_1}), B({x_2}, {r_2}), \cdots , B({x_k}, {r_k}) $是$ X $上的非空开集,又因存在$ B(0, r) \subset \bigcap\limits_{i = 0}^k {B(0, {r_i})} $,对任意的$ N \in {{\Bbb Z}^ + } $, $ {T^{ - i}}(B(0, r)) $是$ X $中的非空开集, $ 0 \le i \le N $.因为$ (X, T) $是multi敏感的(敏感常数为$ \delta > 0 $),则$ \bigcap\limits_{i = 0}^N {{N_T}({T^{ - i}}(B(0, r)), \delta )} \ne \emptyset $,因此取$ {n_N} \in \bigcap\limits_{i = 0}^N {{N_T}({T^{ - i}}(B(0, r)), \delta )} $和$ {n_N} > N $,易知$ \{ {n_N}, {n_N} - 1, $$ \cdots , {n_N} - N\} \subset {N_T}(B(0, r), \delta ) $,因

(2.4) $ \begin{equation} \{ {n_N}, {n_N} - 1, \cdots , {n_N} - N\} \subset {N_T}(B(0, r), \delta ) \subset \bigcap\limits_{i = 1}^k {{N_T}(B(0, {r_i}), \delta )}, \end{equation} $

对任意的$ n \in \bigcap\limits_{i = 1}^k {{N_T}(B(0, {r_i}), \delta )} $,则存在$ z_1^i, z_2^i \in B(0, {r_i}) $满足$ d({T^n}z_1^i, {T^n}z_2^i) > \delta $,又$ {x_i} + z_1^i, $$ {x_i} + z_2^i \in B({x_i}, {r_i}) $和

(2.5) $ \begin{equation} d({T^n}({x_i} + z_1^i), {T^n}({x_i} + z_2^i)) = d({T^n}{x_i} + {T^n}z_1^i, {T^n}{x_i} + {T^n}z_2^i) = d({T^n}z_1^i, {T^n}z_2^i) > \delta, \end{equation} $

所以$ n \in \bigcap\limits_{i = 1}^k {{N_T}} (B({x_i}, {r_i}), \delta ) $,因此$ \{ {n_N}, {n_N} - 1, \cdots , {n_N} - N\} \subset \bigcap\limits_{i = 1}^k {{N_T}(B({x_i}, {r_i}), \delta )} $,故$ (X, T) $是multi-thick敏感的(敏感常数为$ \delta > 0 $).

(充分性) 显然的.

定理2.1  设$ (X, T) $为线性动力系统,则以下命题等价:

(1) $ (X, T) $是敏感的;

(2) $ (X, T) $是multi敏感的;

(3) $ (X, T) $是multi-thick敏感的;

(4) $ (X, T) $是thick敏感的.

证  $ (1) \Rightarrow (2) $见命题2.1; $ (2) \Rightarrow (3) $见命题2.2; $ (3) \Rightarrow (4) $, $ (4) \Rightarrow (1) $显然的.

推论2.1  $ (X, T) $是敏感的当且仅当$ (X, {T^n}) $是敏感的, $ n \in {{\Bbb Z}^ + } $.

证  (必要性) 由定理2.1知,若$ (X, T) $是敏感的，则$ (X, T) $是thick敏感,因此存在$ \delta > 0 $使得对任意非空开集$ U \subset X $, $ {N_T}(U, \delta ) $是thick集,所以对任意$ n \in {{\Bbb Z}^ + } $, $ \{ n, 2n, \cdots , ln, \cdots \} \cap {N_T}(U, \delta ) \ne \emptyset $,故$ (X, {T^n}) $是敏感的.

(充分性) 显然的.

命题2.3  $ (X \oplus Y, T \oplus S) $是敏感的当且仅当$ (X, T) $或者$ (Y, S) $是敏感,其中$ X $上的度量为$ {d_X} $和$ Y $上的度量为$ {d_Y} $, $ X \oplus Y $上的度量为$ {d_{X\oplus Y}} $

证  (必要性) 只需注意到

$ {d_X}({x_1}, {x_2}) + {d_Y}({y_1}, {y_2}) \ge \frac{1}{2}{d_{X \oplus Y}}(({x_1}, {x_2}), ({y_1}, {y_2})), \ ({x_1}, {y_1}), ({x_2}, {y_2}) \in X \oplus Y. $

(充分性) 显然的.

定理2.2  $ (X, T) $是syndetic敏感当且仅当$ (X, T) $是thickly syndetic敏感.

证  (必要性) 设$ (X, T) $是syndetic敏感(敏感常数为$ \delta > 0 $),令$ d $为$ X $上的任意传递不变的度量,对任意的$ {x_0} \in X $和$ {r_0} > 0 $,则$ B({x_0}, {r_0}) $是$ X $上的非空开集,且$ B({x_0}, {r_0}) = {x_0} + B(0, {r_0}) $,对任意的$ N \in {{\Bbb Z}^ + } $, $ {T^{ - i}}(B({x_0}, {r_0})) $为$ X $上的非空开集,则存在$ {x_i} \in X, {r_i} > 0 $使得

(2.6) $ \begin{equation} B({x_i}, {r_i}) = {x_i} + B(0, {r_i}) \subset {T^{ - i}}(B({x_0}, {r_0})), \end{equation} $

$ i = 1, 2, \cdots , N $,又存在$ B(0, l) \in \bigcap\limits_{i = 0}^N {B(0, {r_i})} $,因$ {N_T}(B(0, l), \delta ) $是syndetic集,则对任意的$ k \in {N_T}(B(0, l), \delta ) $,存在$ {y_1}, {y_2} \in B(0, l) $,使得$ d({T^k}{y_1}, {T^k}{y_2}) > \delta $,又$ {x_i} + {y_1}, {x_i} + {y_2} \in B({x_i}, {r_i}) $,和

(2.7) $ \begin{equation} d({T^k}({x_i} + {y_1}), {T^k}({x_i} + {y_2})) = d({T^k}{x_i} + {T^k}{y_1}, {T^k}{x_i} + {T^k}{y_2}) = d({T^k}{y_1}, {T^k}{y_2}) > \delta, \end{equation} $

所以$ k \in {N_T}(B({x_i}, {r_i}), \delta ) $, $ i = 0, 1, \cdots , N $,因此$ k \in \bigcap\limits_{i = 0}^N {{N_T}({T^{ - i}}B({x_0}, {r_0}), \delta )}, $取

$ {n_N} \in \bigcap\limits_{i = 0}^N {{N_T}({T^{ - i}}B({x_0}, {r_0}), \delta )} \ \mbox{和}\ {n_N} > N, $

明显$ \{ {n_N}, {n_N} - 1, \cdots , {n_N} - N\} \subset {N_T}(B({x_0}, {r_0}), \delta ) $,令

$ A = \{ m \in {{\Bbb Z}^ + };m \in {N_T}(B(0, l), \delta ), m > N\} = \{ s_1^N < s_2^N < \cdots \}, $

易知$ A $是syndetic集,又$ \bigcup\limits_{j = 1}^\infty {\{ s_j^N, s_j^N - 1, \cdots , s_j^N - N\} } \subset {N_T}(B({x_0}, {r_0}), \delta ) $,故$ (X, T) $是thickly syndetic敏感(敏感常数为$ \delta > 0 $).

(充分性) 显然的.

命题2.4  若$ (X, T) $是syndetic传递,那么$ (X, T) $是syndetic敏感.

证  令$ a \in X\backslash \{ 0\} $和$ d $为$ X $上的任意传递不变的度量, $ \delta = \frac{1}{4}d(a, 0) > 0 $,令$ U = B(a, \delta ) $,对任意的$ \varepsilon > 0 $和$ x \in X $,则$ B(x, \varepsilon ) = x + B(0, \varepsilon ) $和$ N(B(0, \varepsilon ), U) $是syndetic集,间距为$ {M_1} $,下证$ {N_T}(B(0, \varepsilon ), \delta ) $是syndetic集且间距为$ {M_1} $,令$ n \in {{\Bbb Z}^ + } $,取$ j \in \{ 1, 2, \cdots , {M_1}\} $和$ v \in B(0, \varepsilon ) $满足$ {T^{n + j}}v \in U $,则

(2.8) $ \begin{equation} d({T^{n + j}}v, {T^{n + j}}0) = d({T^{n + j}}v, 0) > 2\delta > \delta. \end{equation} $

因$ n \in {{\Bbb Z}^ + } $是任意的和$ j \le {M_1} $,故$ {N_T}(B(0, \varepsilon ), \delta ) $是syndetic集,则$ {N_T}(B(x, \varepsilon ), \delta ) $是syndetic集,实际上对任意$ n \in {N_T}(B(0, \varepsilon ), \delta ) $,则存在$ {a_1}, {a_2} \in B(0, \varepsilon ) $满足$ d({T^n}{a_1}, {T^n}{a_2}) > \delta $,明显, $ x + {a_1}, x + {a_2} \in B(x, \varepsilon ) $和

$ d({T^n}(x + {a_1}), {T^n}(x + {a_2})) = d({T^n}x + {T^n}{a_1}, {T^n}x + {T^n}{a_2}) = d({T^n}{a_1}, {T^n}{a_2}) > \delta, $

故$ n \in {N_T}(B(x, \varepsilon ), \delta ) $,因此$ {N_T}(B(x, \varepsilon ), \delta ) $是syndetic集,故$ (X, T) $是syndetic敏感.

推论2.2  若$ (X, T) $是syndetic传递,那么$ (X, T) $是传递敏感.

证  因为$ (X, T) $是syndetic传递,由定理2.2和命题2.4知$ (X, T) $是thickly syndetic敏感.明显$ (X, T) $是传递敏感.

推论2.3  若$ (X, T) $是$ F $ -超循环的,那么$ (X, T) $是thickly syndetic敏感.

证  若$ (X, T) $$ F $ -超循环的,当然$ (X, T) $是传递的,设$ U $, $ V $为$ X $的任意非空开集,则存在$ {n_0} \ge 0 $使得$ {T^{{n_0}}}U \cap V \ne \emptyset $,因为$ T $是连续的,则存在非空开集$ {U_0} \subset U $满足$ {T^{{n_0}}}{U_0} \subset V $,对任意的$ x \in FHC(T) $,则$ N(x, {U_0}) $上密度是正的,对任意$ m, n \in N(x, {U_0}) $, $ m \ge n $, $ {T^{{n_0} + m - n}}({T^n}x) = {T^{{n_0}}}({T^m}x) \in V $,因此

$ {n_0} + (N(x, {U_0}) - N(x, {U_0})) \subset N({U_0}, V) \subset N(U, V), $

由文献[18]知$ N(x, {U_0}) - N(x, {U_0}) $是syndetic集,则$ N(U, V) $是syndetic集.由定理2.2和命题2.4知因此$ (X, T) $是thickly syndetic敏感.

命题3.1  迭代动力系统$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $是敏感的当且仅当$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $是multi敏感的.

证  (必要性) 设$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $是敏感的(敏感常数为$ \delta > 0 $),令$ d $为$ X $上的任意传递不变的度量,对任意的$ {r_1}, {r_2}, \cdots {r_k} > 0 $和$ {x_1}, {x_2}, \cdots, {x_k} \in X $,则$ B({x_1}, {r_1}), B({x_2}, {r_2}), \cdots , $$ B({x_k}, {r_k}) $是$ X $上的非空开集,因为$ d $是传递不变的,所以

$ B({x_1}, {r_1}) = {x_1} + B(0, {r_1}), B({x_2}, {r_2}) = {x_2} + B(0, {r_2}), \cdots , B({x_k}, {r_k}) = {x_k} + B(0, {r_k}), $

又因存在$ B(0, l) \subset \bigcap\limits_{i = 1}^k {B(0, {r_i})} $,所以$ \emptyset \ne N_{{\cal F}}(B(0, l), \delta ) $,和$ N_{{\cal F}}(B(0, l), \delta ) \subset N_{{\cal F}}(\bigcap\limits_{i = 1}^k {B(0, {r_i})}, \delta ) $,对任意的$ n \in N_{{\cal F}}(B(0, l), \delta ) $,则存在$ {z_1}, {z_2} \in B(0, l) $, $ \sigma \in \Lambda^{{{\Bbb Z}}^{+}} $满足$ d({T_\sigma ^n}{z_1}, {T_\sigma ^n}{z_2}) > \delta $,易知$ {x_i} + {z_1}, {x_i} + {z_2} \in B({x_i}, {r_i}), i = 1, \cdots , k $,

(3.1) $ \begin{equation} d({T_\sigma ^n}({x_i} + {z_1}), {T_\sigma ^n}({x_i} + {z_2})) = d({T_\sigma ^n}{x_i} + {T_\sigma ^n}{z_1}, {T_\sigma ^n}{x_i} + {T_\sigma ^n}{z_2}) = d({T_\sigma ^n}{x_1}, {T_\sigma ^n}{x_2}) > \delta, \end{equation} $

因此$ n \in \bigcap\limits_{i = 1}^k {{N_{{\cal F}}}(B({x_i}, {r_i}), \delta )} $,所以$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $是multi敏感的(敏感常数为$ \delta > 0 $).

(充分性) 显然的.

命题3.2  迭代动力系统$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $是multi敏感的当且仅当$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $是multi-thick敏感的.

证  (必要性) 设$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $是multi敏感的(敏感常数为$ \delta > 0 $),令$ d $为$ X $上的任意传递不变的度量,对任意的$ {r_1}, {r_2}, \cdots {r_k} > 0 $和$ {x_1}, {x_2}, \cdots {x_k} \in X $,则$ B({x_1}, {r_1}), $$ B({x_2}, {r_2}), \cdots , B({x_k}, {r_k}) $是$ X $上的非空开集,又因存在$ B(0, r) \subset \bigcap\limits_{i = 1}^k {B(0, {r_i})} $,对任意的$ N \in {{\Bbb Z}^ + } $,则

$ \begin{eqnarray*} &&0 \in (B(0, r) \cap (\bigcap\limits_{{i_1} \in \Lambda } {{f_{{i_1}}}^{ - 1}(B(0, r))} ) \cap (\bigcap\limits_{{i_1}, {i_2} \in \Lambda } {{{({f_{{i_1}}} \circ {f_{{i_2}}})}^{ - 1}}(B(0, r))} ) \cap \cdots \\ && \cap (\bigcap\limits_{{i_{1, }}{i_2}, \cdots , {i_N} \in \Lambda } {{{({f_{{i_1}}} \circ {f_{{i_2}}} \circ \cdots \circ {f_{{i_N}}})}^{ - 1}}(B(0, r))} )) \ne \emptyset, \end{eqnarray*} $

则存在

$ \begin{eqnarray*} &&B(0, l) \subset (B(0, r) \cap (\bigcap\limits_{{i_1} \in \Lambda } {{f_{{i_1}}}^{ - 1}(B(0, r))} ) \cap (\bigcap\limits_{{i_1}, {i_2} \in \Lambda } {{{({f_{{i_1}}} \circ {f_{{i_2}}})}^{ - 1}}(B(0, r))} ) \cap \cdots \\ && \cap (\bigcap\limits_{{i_{1, }}{i_2}, \cdots , {i_N} \in \Lambda } {{{({f_{{i_1}}} \circ {f_{{i_2}}} \circ \cdots \circ {f_{{i_N}}})}^{ - 1}}(B(0, r))} )), \end{eqnarray*} $

因$ {N_{{\cal F}}}(B(0, l), \delta ) \ne \emptyset $,取$ {n_N} \in {N_{{\cal F}}}(B(0, l), \delta ) $和$ {n_N} > N $,则存在$ {y_1}, {y_2} \in B(0, l) $和$ \sigma \in \Lambda^{{{\Bbb Z}}^{+}} $满足$ d(T_\sigma ^{{n_N}}{y_1}, T_\sigma ^{{n_N}}{y_2}) > \delta $,同时

$ \begin{eqnarray*} &&B(0, l) \subset (B(0, r) \cap (\bigcap\limits_{{i_1} \in \Lambda } {{f_{{i_1}}}^{ - 1}(B(0, r))} ) \cap (\bigcap\limits_{{i_1}, {i_2} \in \Lambda } {{{({f_{{i_1}}} \circ {f_{{i_2}}})}^{ - 1}}(B(0, r))} ) \cap \cdots \\ && \cap (\bigcap\limits_{{i_{1, }}{i_2}, \cdots , {i_N} \in \Lambda } {{{({f_{{i_1}}} \circ {f_{{i_2}}} \circ \cdots \circ {f_{{i_N}}})}^{ - 1}}(B(0, r))} )), \end{eqnarray*} $

易知$ \{ {n_N}, {n_N} - 1, \cdots , {n_N} - N\} \subset {N_{{\cal F}}}(B(0, r), \delta ) $,又对任意$ n \in {N_{{\cal F}}}(B(0, r), \delta ) $,存在$ {z_1}, {z_2} \in B(0, r) $和$ {\sigma _1}\in \Lambda^{{{\Bbb Z}}^{+}} $满足$ d(T_{{\sigma _1}}^n{z_1}, T_{{\sigma _1}}^n{z_2}) > \delta $,因此$ {x_i} + {z_1}, {x_i} + {z_2} \in B({x_i}, {r_i}) $和

(3.2) $ \begin{equation} d(T_{{\sigma _1}}^n({x_i} + {z_1}), T_{{\sigma _1}}^n({x_i} + {z_2})) = d(T_{{\sigma _1}}^n{x_i} + T_{{\sigma _1}}^n{z_1}, T_{{\sigma _1}}^n{x_i} + T_{{\sigma _1}}^n{z_2}) = d(T_{{\sigma _1}}^n{z_1}, T_{{\sigma _1}}^n{z_2}) > \delta, \end{equation} $

所以$ n \in {N_{{\cal F}}}(B({x_i}, {r_i}), \delta ) $, $ i = 1, 2, \cdots , k $,故

(3.3) $ \begin{equation} \{ {n_N}, {n_N} - 1, \cdots , {n_N} - N\} \subset \bigcap\limits_{i = 1}^k {{N_{{\cal F}}}(B({x_i}, {r_i}), \delta )}, \end{equation} $

因此$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $是multi-thick敏感的(敏感常数为$ \delta > 0 $).

(充分性) 显然的.

定理3.1  设$ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $为迭代动力系统,则以下命题等价:

(1) $ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $是敏感的;

(2) $ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $是multi敏感的;

(3) $ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $是multi-thick敏感的;

(4) $ {\cal F} = \{X; f_{\lambda}|\lambda \in\Lambda \} $是thick敏感的.

证  $ (1) \Rightarrow (2) $见命题3.1; $ (2) \Rightarrow (3) $见命题3.2; $ (3) \Rightarrow (4) $, $ (4) \Rightarrow (1) $显然的.

推论3.1  迭代动力系统$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是敏感的当且仅当$ {F_n} = \{ X;{f_{{\lambda _1}}} \circ {f_{{\lambda _2}}} \circ \cdots \circ {f_{{\lambda _n}}}|{\lambda _i} \in \Lambda, 1 \le i \le n\} $是敏感的, $ n \in {{\Bbb Z}^ + } $.

证  (必要性) 由定理3.1知,若$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是敏感的，则$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是thick敏感,因此存在$ \delta > 0 $使得对任意非空开集$ U \subset X $, $ {N_{{\cal F}}}(U, \delta ) $是thick集,所以对任意$ n \in {{\Bbb Z}^ + } $, $ {N_{{\cal F}}}(U, \delta ) \cap \{ n, 2n, \cdots , kn, \cdots \} \ne \emptyset $,故$ {N_{{{\cal F}_n}}}(U, \delta ) \ne \emptyset $,故$ {F_n} = \{ X;{f_{{\lambda _1}}} \circ {f_{{\lambda _2}}} \circ \cdots \circ {f_{{\lambda _n}}}|{\lambda _i} \in \Lambda, 1 \le i \le n\} $是敏感的.

(充分性) 显然的.

命题3.3  设$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $和$ {\cal G} = \{Y; g_{\gamma}| \gamma \in \Lambda_1\} $是两个迭代动力系统分别定义在Frechet空间$ X $和$ Y $,则$ {\cal F}\times {\cal G} $是敏感的当且仅当$ {\cal F} $或者$ {\cal G} $是敏感,其中$ X $上的度量为$ {d_X} $和$ Y $上的度量为$ {d_Y} $, $ X \oplus Y $上的度量为$ {d_{X\oplus Y}} $.

证  (必要性) 只需注意到

$ {d_X}({x_1}, {x_2}) + {d_Y}({y_1}, {y_2}) \ge \frac{1}{2}{d_{X \oplus Y}}(({x_1}, {x_2}), ({y_1}, {y_2})), \ ({x_1}, {y_1}), ({x_2}, {y_2}) \in X \oplus Y. $

(充分性) 显然的.

定理3.2  $ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是syndetic敏感当且仅当$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是thickly syndetic敏感.

证  (必要性) 设$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是syndetic敏感(敏感常数为$ \delta > 0 $),令$ d $为$ X $上的任意传递不变的度量,对任意的$ \varepsilon > 0 $和$ x \in X $,则$ B(x, \varepsilon ) $是$ X $上的非空开集,且$ B(x, \varepsilon ) = x + B(0, \varepsilon ) $,对任意的$ N \in {{\Bbb Z}^ + } $,有

$ \begin{eqnarray*} &&0 \in B(0, \varepsilon ) \cap (\bigcap\limits_{{\lambda _1} \in \Lambda } {f_{{\lambda _1}}^{ - 1}} (B(0, \varepsilon ))) \cap {(\bigcap\limits_{{\lambda _i} \in \Lambda, 1 \le i \le 2} {({f_{{\lambda _1}}} \circ {f_{{\lambda _2}}}} )^{ - 1}}(B(0, \varepsilon ))) \cdots \\ && \cap (\bigcap\limits_{{\lambda _i} \in \Lambda, 1 \le i \le N} {{{({f_{{\lambda _1}}} \circ {f_{{\lambda _2}}} \circ \cdots \circ {f_{{\lambda _N}}})}^{ - 1}}(B(0, \varepsilon ))} ), \end{eqnarray*} $

因此存在

$ \begin{eqnarray*} &&B(0, r) \subset B(0, \varepsilon ) \cap (\bigcap\limits_{{\lambda _1} \in \Lambda } {f_{{\lambda _1}}^{ - 1}} (B(0, \varepsilon ))) \cap {(\bigcap\limits_{{\lambda _i} \in \Lambda, 1 \le i \le 2} {({f_{{\lambda _1}}} \circ {f_{{\lambda _2}}}} )^{ - 1}}(B(0, \varepsilon ))) \cdots \\ && \cap (\bigcap\limits_{{\lambda _i} \in \Lambda, 1 \le i \le N} {{{({f_{{\lambda _1}}} \circ {f_{{\lambda _2}}} \circ \cdots \circ {f_{{\lambda _N}}})}^{ - 1}}(B(0, \varepsilon ))} ), \end{eqnarray*} $

取$ {n_N} \in {N_{{\cal F}}}(B(0, r), \delta ) $和$ {n_N} > N $,因为

$ \begin{eqnarray*} &&B(0, r) \subset B(0, \varepsilon ) \cap (\bigcap\limits_{{\lambda _1} \in \Lambda } {f_{{\lambda _1}}^{ - 1}} (B (0, \varepsilon ))) \cap {(\bigcap\limits_{{\lambda _i} \in \Lambda, 1 \le i \le 2} {({f_{{\lambda _1}}} \circ {f_{{\lambda _2}}}} )^{ - 1}}(B(0, \varepsilon ))) \cdots \\ && \cap (\bigcap\limits_{{\lambda _i} \in \Lambda, 1 \le i \le N} {{{({f_{{\lambda _1}}} \circ {f_{{\lambda _2}}} \circ \cdots \circ {f_{{\lambda _N}}})}^{ - 1}}(B(0, \varepsilon ))} ), \end{eqnarray*} $

显然, $ \{ {n_N}, {n_N} - 1, \cdots , {n_N} - N\} \subset {N_{{\cal F}}}(B(0, \varepsilon ), \delta ) $,对任意的$ n \in {N_{{\cal F}}}(B(0, \varepsilon ), \delta ) $,存在$ {x_1}, {x_2} \in B(0, \varepsilon ) $和$ \sigma \in \Lambda^{{{\Bbb Z}}^{+}} $满足$ d(f_\sigma ^n{x_1}, f_\sigma ^n{x_2}) > \delta $,则$ x + {x_1}, x + {x_2} \in B(x, \varepsilon ) $和

(3.4) $ \begin{equation} d(f_\sigma ^n(x + {x_1}), f_\sigma ^n(x + {x_2})) = d(f_\sigma ^nx + f_\sigma ^n{x_1}, f_\sigma ^nx + f_\sigma ^n{x_2}) = d(f_\sigma ^n{x_1}, f_\sigma ^n{x_2}) > \delta, \end{equation} $

故$ n \in {N_{{\cal F}}}(B(x, \varepsilon ), \delta ) $,令

(3.5) $ \begin{equation} A = \{ m \in {{\Bbb Z}^ + };m \in {N_{{\cal F}}}(B(0, r ), \delta ), m > N\} = \{ s_1^N < s_2^N < \cdots \}, \end{equation} $

则$ \bigcup\limits_{j = 1}^\infty {\{ s_j^N, s_j^N - 1, \cdots , s_j^N - N\} } \subset {N_{{\cal F}}}(B(x, \varepsilon ), \delta ) $,故$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是thickly syndetic敏感(敏感常数为$ \delta > 0 $).

(充分性) 显然的.

引理3.1 若迭代动力系统$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是传递的,则集合$ A \subset X $是一个稠密的$ {G_\delta } $ -集,其中对任意$ x \in A $, $ \bigcup\limits_{\sigma \in {\Lambda ^{{{\Bbb Z}}^ {+} }}} {\bigcup\limits_{n = 0}^\infty {f_\sigma ^nx} } $稠于$ X $.

证  因为$ X $是可分的,所以$ X $存在可数稠子集$ \{ {y_j};j \ge 1\} $,且$ B({y_j}, \frac{1}{m}) $, $ m, j \ge 1 $,是$ X $的一组可数基$ {({U_k})_{k \ge 1}} $,因此, $ x \in A $当且仅当对任意$ k \ge 1 $,存在$ n \ge 0 $和$ \sigma \in {\Lambda ^{{{\Bbb Z}}^ {+ }}} $满足$ f_\sigma ^nx \in {U_k} $.也就是说$ A = \bigcap\limits_{k = 1}^\infty {\bigcup\limits_{\sigma \in {\Lambda ^{{{\Bbb Z}^ + }}}} {\bigcup\limits_{n = 0}^\infty {f_\sigma ^{ - n}({U_k})} } } $,因$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是传递的,因此$ \bigcup\limits_{\sigma \in {\Lambda ^{{{\Bbb Z}}^ {+ }}}} {\bigcup\limits_{n = 0}^\infty {f_\sigma ^{-n}({U_k})} } $, $ k \ge 1 $,开稠于$ X $,由Baire定理知$ A $是一个稠密的$ {G_\delta } $ -集.

命题3.4  若迭代动力系统$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是syndetic传递,那么$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是syndetic敏感.

证  令$ a \in X\backslash \{ 0\} $和$ d $为$ X $上的任意传递不变的度量, $ \delta = \frac{1}{4}d(a, 0) > 0 $,令$ U = B(a, \delta ) $,对任意的$ \varepsilon > 0 $和$ x \in X $,则$ B(x, \varepsilon ) = x + B(0, \varepsilon ) $和$ {N_{{\cal F}}}(B(0, \varepsilon ), U) $是syndetic集,间距为$ {M_1} $,下证$ {N_{{\cal F}}}(B(0, \varepsilon ), \delta ) $是syndetic集且间距为$ {M_1} $,令$ n \in {{\Bbb Z}^ + } $,取$ j \in \{ 1, 2, \cdots , {M_1}\} $和$ \sigma \in {\Lambda ^{{{\Bbb Z}_ + }}} $, $ a \in B(0, \varepsilon ) $满足$ f_\sigma ^{n + j}a \in U $,则

(3.6) $ \begin{equation} d(f_\sigma ^{n + j}a, f_\sigma ^{n + j}0) = d(f_\sigma ^{n + j}a, 0) > 2\delta > \delta, \end{equation} $

因$ n \in {{\Bbb Z}^ + } $是任意的和$ j \le {M_1} $,因此$ {N_{\cal F}}(B(0, \varepsilon ), \delta ) $是syndetic集,则$ {N_{{\cal F}}}(B(x, \varepsilon ), \delta ) $是syndetic集,实际上对任意$ n \in {N_{{\cal F}}}(B(0, \varepsilon ), \delta ) $,则存在$ {x_1}, {x_2} \in B(0, \varepsilon ) $, $ \sigma \in {\Lambda ^{{{\Bbb Z}^ + }}} $满足$ d(f_\sigma ^n{x_1}, f_\sigma ^n{x_2}) > \delta $,明显$ x + {x_1}, x + {x_2} \in B(x, \varepsilon ) $,且

(3.7) $ \begin{equation} d(f_\sigma ^n(x + {x_1}), f_\sigma ^n(x + {x_2})) = d(f_\sigma ^nx + f_\sigma ^n{x_1}, f_\sigma ^nx + f_\sigma ^n{x_2}) = d(f_\sigma ^n{x_1}, f_\sigma ^n{x_2}) > \delta, \end{equation} $

所以$ n \in {N_{{\cal F}}}(B(x, \varepsilon ), \delta ) $,因此$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是syndetic敏感.

推论3.2 (1)若$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是syndetic传递的,则$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是thickly syndetic敏感.

(2)若$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是syndetic传递的,则$ {\cal F} = \{X; f_{\lambda}| \lambda \in \Lambda\} $是传递敏感.

证  (1)由命题3.4和定理3.2, (2)由推论3.2(1)可得.