Let $(X, d)$ be a metric space and $T : X \to X$ be a continuous map that is topologically transitive. The history of topological transitivity is a long one and many authors opt for different and independent definitions. I have given two such definitions below:
Definition 1. For every pair of non-empty open sets $U$ and $V$ in $X$, there is a non-negative integer $n$ such that $f^n(U) \cap V \neq \emptyset$.
Definition 2. There is a point $x \in X$ such that the forward-orbit $\{T^n(x)\}_{n \geq 0}$ is dense in $X$.
My question concerns the hypotheses we can impose on $X$ to guarantee that it is surjection. From previous discussion on the forum, it is known that under Definition 1, $X$ being compact implies that $T$ is surjective. We also have examples (see here or here) of non-compact spaces $X$ with non-surjective topologically transitive maps $T$. My questions are mainly about the case of Definition 2. Specifically,
- Under Definition 2, does there exist a compact metric space $(X, d)$ with non-surjective topological transitive $T$?
- Under Definition 2, does there exist a complete metric space $(X, d)$ with non-surjective topological transitive $T$?
- Under Definition 1, does there exist a complete metric space $(X, d)$ with non-surjective topological transitive $T$?
If such a system exists for (1.), then it would be a system for (2.). The last question is a weakening of hypothesis about the result under Definition 1. Any observations, hints, and answers would be appreciated.
Note: Definition 1 and Definition 2 are equivalent when $(X, d)$ is complete, separable, and contains no isolated points.
Assume Definition 2. There exists a compact metric space $(X,d)$ with non-surjective topological transitive $T$. Let $X=\{0\}\cup\{2^{-n}:n\in\Bbb N\}$ be a subspace of $\Bbb R$ endowed with the usual metric and $T(x)=x/2$ for each $x\in X$.
On the other hand, if $T$ is a topologically transitive continuous map on a Hausdorff compact space $X$ without isolated points then there is a point $x\in X$ such that a set $X’=\{T^n(x)\}_{n \geq 1}$ is dense in $X$. Then the set $T(X)$ contains $X’$, so $T(X)$ is a dense. Since $T(X)$ is compact, it is closed in $X$. So $T(X)=X$.
Also there exists a completely metrizable space $X$ without isolated points and a topologically transitive non-surjective continuous map $T$ on $X$. Indeed, let $X_0$ be the unit circle $\{z\in\Bbb C:|z|=1\}$ endowed with the usual topology and $T$ be a rotation of $X$ into an angle $\phi$ such that $\phi/\pi$ is irrational. Pick any point $x\in X_0$ and put $X=X_0\setminus \{T^n(x)\}_{n<0}$. Then the space $X$ is a $G_\delta$ set of a completely metrizable space, so it is metrizable, $\{T^n(x)\}_{n \geq 1}$ is dense in $X$, but $x\not\in T(X)$. Remark that the map $T$ is topologically transitive also with respect to Definition 1.