A simple question about discrete dynamical system

Question

Given a self-mapping $f:X\to X$, $X$ is a Hausdorff space, and $f$ is continuous and topologically transitive, if $X$ is infinite, then $X$ contains no isolated points. Why? I don't know why? Please Help me!

Answer 1

Suppose that $X$ is Hausdorff, $f : X \to X$ is continuous and topologically transitive.

Assume there exists an isolated point $x \in X$. By definition, $\{x \}$ is a non-empty open subset, and since $f$ is continuous, $f^{-1}(\{ x \})$ is non-empty (see remark below) and open as well. By topological transitivity, there exists $n \ge 0$ such that $f^n(\{ x\}) \cap f^{-1}( \{x \}) \neq \emptyset$.

This means that $f^{n+1}(x) = x$. Since $n+1 \ge 1$, this implies that $x$ is a periodic point.

Let $k$ be the prime period of $x$. Since $X$ is Hausdorff, $X \setminus \{x, \ldots, f^{k-1}(x)\}$ is open. Since $f^m(\{x\}) \cap \left( X \setminus \{x, \ldots, f^{k-1}(x)\}\right)$ is empty for any $m \ge 0$, by topological transitivity, this implies that $X \setminus \{x, \ldots, f^{k-1}(x)\}$ is empty.

Hence $X = \{x, \ldots, f^{k-1}(x)\}$ is a finite set and consists of a single periodic orbit of $f$.

Remark : That $f^{-1}(\{x\})$ is non-empty is easy to see. Indeed, both $\{x\}$ and $X \setminus \{x\}$ are open subsets. If $X \setminus \{x\}$ was empty, then we would have $f(x) = x$. If $X \setminus \{x \}$ is non-empty, by topological transitivity, there exists $l \ge 0$ such that $f^l(X \setminus \{x \}) \cap \{x \} \neq \emptyset$. This implies $l \ge 1$ and that $f^{-l}(\{x\})$ is non-empty, so hence is $f^{-1}(\{x \})$.