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Let $(X,d)$ is a metric space, $T:X\rightarrow X$ is a continuous self-map.

Def1. $T$ is strongly sensitive if there exist $\epsilon>0$, for any $x\in X$ and any positive number $\delta>0$, we can find a point $y\in B(x,\delta)$ and $n\in\mathbb{N}$, such that $d(T^nx,T^ny)>\epsilon$.

Def2. $T$ is (weakly) sensitive if there exist $\epsilon>0$, for any $x\in X$ and any positive number $\delta>0$, we can find a point $y\in B(x,\delta)$ and $n\in\mathbb{N}\cup\{0\}$, such that $d(T^nx,T^ny)>\epsilon$.

Of course, Def1 implies Def2, my quesiton is Def2 implies Def 1, too? If not, does there exist a counterexample?

James Chan
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1 Answers1

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I believe Def 2 does imply Def 1.

Suppose $T$ is (weakly) sensitive.

Then, there exists an $\epsilon > 0$ such that [rest of Def 2].

Let $x \in X$ and $\delta > 0$ be chosen arbitrarily.

Pick a number $0 < \delta' < \min\{\epsilon,\delta\}$.

Then by our choice of $\epsilon$, there exists a $y \in B(x,\delta')$ and $n \in \mathbb{N}\cup\{0\}$ such that $d(T^nx,T^ny) > \epsilon$.

If $n = 0$, then $d(x,y) = d(T^0x,T^0y) > \epsilon \ge \min\{\epsilon,\delta\} > \delta'$, a contradiction since $y \in B(x,\delta')$.

Thus, $n \in \mathbb{N}$. Also, since $\delta' < \min\{\epsilon,\delta\} \le \delta$, we have $y \in B(x,\delta') \subseteq B(x,\delta)$.

Hence, $\epsilon > 0$ is such that for any $x \in X$ and any $\delta > 0$, there exists a $y \in B(x,\delta)$ and $n \in \mathbb{N}$ such that $d(T^nx,T^ny) > \epsilon$.

Therefore, $T$ is strongly sensitive.

JimmyK4542
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