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 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 $d(T^nx,T^ny)>\varepsilon$.

Def2. $T$ is an expanding map if there exist a constant $c>1$ and a positive number $\varepsilon>0$, for any $x,y\in X$, if $d(x,y)<\varepsilon$, we have $d(Tx,Ty)>cd(x,y)$.

I once said that "an expanding map must be weakly expansive."(Must an expanding map be strongly expansive?). But just now, I find it's not easy to jump to this conclusion.

My quesitons are:

  1. Must an expanding map be weakly expansive? If not, does there exist a counterexample?

  2. Does there exist an example such that $T$ is an injection and an expansive map, but $T$ is not expanding?

  3. Does there exist such an example, $T$ is not an injection, but $T$ is strongly expansive?(ref to: Must an expanding map be strongly expansive?)


An (forward-)expansive map is weakly expanding (I assume that the space is cpct as well, this is a common assumption in topological dynamics).

Suppose the contrary. For every $\varepsilon$ there exists two different sequences of points $x_{\varepsilon},y_{\varepsilon}$ such that $\sup{d(T^{n}x_{\varepsilon},T^{n}y_{\varepsilon})}\leq \varepsilon$.

Pick $\varepsilon$ which is smaller than the epsilon indicated in the expansiveness assumption.

Then we have $d(Tx_{\varepsilon},Ty_{\varepsilon})>c\cdot d(x,y)$, in contradiction to the definition of the sequences. Notice we may assume that the supremum distance over the orbit occurs in the first time (hence $d(x,y)$), or at-least a very good approximation to it.

Remark - the only obstruction to this "continuous family of close orbits" is easily seen to be equivalent to the existence of isolated points, which do not exist by assumption.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.