0

Let $(X,d)$ is a metric space and $X$ has no isolated points, $T:X\rightarrow X$ is a continuous self-map.

Def1. $T$ is strongly expansive if there exist $\varepsilon>0$, for any $x,y\in X$, $x\neq y$, we can find a number $n\in\mathbb{N}$, such that $d(T^nx,T^ny)>\varepsilon$.

Def2. $T$ is (weakly) expansive if there exist $\varepsilon>0$, for any $x,y\in X$, $x\neq y$, we can find a number $n\in\mathbb{N}\cup\{0\}$, such that $d(T^nx,T^ny)>\varepsilon$.

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

James Chan
  • 333

1 Answers1

2

If $T$ is injective, the two properties are equivalent.Suppose that $\epsilon$ witnesses the weak expansiveness of $T$. Let $x$ and $y$ be distinct points of $X$. If $d(Tx,Ty)>\epsilon$, we're done, so suppose that $0<d(Tx,Ty)\le\epsilon$. By hypothesis there is an $n\ge 0$ such that $d(T^{n+1}x,T^{n+1}y)=d(T^nTx,T^nTy)>\epsilon$, and clearly $n+1\ge 1$, so $T$ is strongly expansive.

If $T$ is not injective, however, they are not. Let $X=\{x\in\Bbb R:|x|\ge1\}$ with the usual metric. Let $Tx=2|x|$ for $x\in X$. Let $x,y\in X$ with $x\ne y$. If $|x|=|y|$, then $d(x,y)>1$. Otherwise, $d(Tx,Ty)>0$, and for sufficiently large $n$ we must have $d(T^nx,T^ny)>1$. Thus, $T$ is weakly expansive. However, $d(T^n(1),T^n(-1))=0$ for all $n\ge 1$, so $T$ is not strongly expansive.

Brian M. Scott
  • 616,228
  • If $Tx\neq Ty$? And suppose $d(Tx,Ty)<\epsilon$, by hypothesis? What hypothesis? I can't undersand this. – James Chan Dec 31 '15 at 11:40
  • @James: The hypothesis stated at the beginning, that $T$ is weakly expansive. However, there was an oversight in my original answer, since I implicitly assumed that the map was injective; I've repaired that error. – Brian M. Scott Dec 31 '15 at 12:09
  • Venture to ask a question, does there exist an example, such that $T$ is weakly expansive but not strong expansive, and $X$ is a compact metric space? Of course, $X$ has no isolated points. – James Chan Dec 31 '15 at 13:01