Must an expanding map be weakly expansive?

Question

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?)

Answer 1

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.