Showing that three statements are equivalent (Topology) $(a)$ Assume that $X$ is a compact metrizable space. Prove that the following conditions are equivalent:
$(i)$ $T$ is transitive.
$(ii)$ The set of transitive points for $T$ is dense in $X$.
$(iii)$ The set of transitive points for $T$ is non-empty.
Hint: Recall that every compact metrizable space has a countable basis (why?).
$(b)$ Does the equivalence in part $(a)$ still hold if instead of being compact and metrizable, the space $X$ is separable and completely metrizable?
A discrete-time (topological) dynamical system can be identified by a continuous map $T : X → X$ on a topological space $X$. A map $T$ is said to be transitive if for every pair of non-empty open sets $U, V ⊆ X$, there exists an integer $n > 0$ such that $T^n(U) ∩ V ≠ ∅$.
The (positive) orbit of a point $x ∈ X$ under $T$ is the set $O^+_T(x) :=$ {$T(x), T2(x), . . .$} .
A point $x ∈ X$ is said to be transitive for the map $T$ if its orbit $O^+_T(x)$ is dense in $X$.
My attempt:
$(i) => (ii):$ let $T: X→X$ be transitive, then for all $U, V ⊆ X$, there exists an integer $n > 0$ such that $T^n(U) ∩ V ≠ ∅$. But $O^+_T(x)$ $=$ {$T^n(U)$}$_{n>0}$ $=$ {$T(x), T^2(x),...$}, for $x \in X$, and $n>0$.
Hence $O^+_T(x) \cap V ≠ ∅$ and so $O^+_T(x)$ is dense.
$(ii) => (i):$ $O^+_T(x)$ is dense in $X$, so for all non empty $V ⊆ X$, $V \cap O^+_T(x)≠∅$. But $O^+_T(x)$ $=$ {$T^n(U)$}$_{n>0}$, for $x \in X$.
Hence there exists $n>0$ such that $V \cap T^n(U) ≠∅$.
$(i) => (iii):$ $T$ transitive implies that $U, V ⊆ X$, there exists an integer $n > 0$ such that $T^n(U) ∩ V ≠ ∅$. To get a contradiction, suppose $O^+_T(x) = \phi$, therefore, {$T^n(U)$}$_{n>0} = O^+_T(x) =\phi$. Then, $T^n(U) ∩ V = ∅$. Contradiction. So $O^+_T(x)≠ ∅$.
$(iii) =>(i):$ let $O^+_T(x)≠ ∅$. And let $V⊆ X$ be non empty. Thus, $O^+_T(x) \cap V≠ ∅$. Therefore, $T$ is transitive.
$(ii) => (iii):$ $O^+_T(x)$ is dense in $X$, so for all non empty $V⊆ X$, $O^+_T(x) \cap V ≠ ∅$
Hence $O^+_T(x)$ is non empty.
$(iii) =>(ii):$ $O^+_T(x)≠ ∅$. And let $V⊆ X$ be non empty. Thus, $O^+_T(x) \cap V≠ ∅$.
Hence, $O^+_T(x)$ is dense in $X$.
Wondering whether my attempt is correct or not, because I didn't use the given hint nor the fact that $X$ is compact metrizable. And what about part $(b)$, how to solve it? Thank you.
 A: I really can’t make sense of most of your argument. In most parts of it, for instance, you make assertions about $O_T^+(x)$ without ever specifying what $x$ is. In several places you seem to have confused $O_T^+(x)$ with the set of transitive points of $T$. There just isn’t enough substance there for me to turn it into a proof, so I’m simply going to show you one. I will show that (i), (ii), and (iii) are equivalent by showing that (i) implies (iii), (iii) implies (ii), and (ii) implies (i).
If $X$ is finite, it must be a discrete space, and it’s easy to check that each of the three conditions is equivalent to saying that $T$ is a permutation of $X$. Henceforth I will assume that $X$ is infinite.
(i) implies (iii):
Let $\mathscr{B}=\{B_n:n\in\Bbb N\}$ be a countable base for $X$, and let $T$ be transitive. Let $V_0=B_0$. Given a non-empty open set $V_n$ for some $n\in\Bbb Z^+$, there is a $k_{n+1}\in\Bbb Z^+$ such that $T^{k_{n+1}}[V_n]\cap B_{n+1}\ne\varnothing$. Then $V_n\cap\left(T^{k_{n+1}}\right)^{-1}[B_{n+1}]$ is a non-empty open set, so there is a $V_{n+1}\in\mathscr{B}$ such that $\operatorname{cl}V_{n+1}\subseteq V_n\cap\left(T^{k_{n+1}}\right)^{-1}[B_{n+1}]$ and hence $T^{k_{n+1}}[\operatorname{cl}V_{n+1}]\subseteq B_{n+1}$.
In this way we can produce open sets $V_n$ for $n\in\Bbb N$ such that for each $n\in\Bbb N$ we have $\operatorname{cl}V_{n+1}\subseteq V_n$ and $T^{k_{n+1}}[\operatorname{cl}V_{n+1}]\subseteq B_{n+1}$. Let $K=\bigcap_{n\ge 1}\operatorname{cl}V_n$; $X$ is compact, so $K\ne\varnothing$. Let $x\in K$; then for each $n\in\Bbb Z^+$ we have $x\in\operatorname{cl}V_n$, so $T^{k_n}(x)\in B_n$. And $x\in B_0$, so $x$ is a transitive point of $T$.
(iii) implies (ii):
Let $x\in X$ be a transitive point of $T$. Suppose that $x$ is an isolated point of $X$. Then $\{x\}$ is open, so there is an $n\in\Bbb Z^+$ such that $T^n(x)=x$. But then $\{T^k(x):k=1,\ldots,n\}$ is dense in $X$, so $X$ is finite, contradicting our assumption that $X$ is infinite. Thus, $x$ is not isolated.
Now suppose that $T^n(x)$ is isolated for some $n\in\Bbb Z^+$. $O_T^+(x)$ is dense in $X$, so there is a strictly increasing sequence $\langle k_i:i\in\Bbb Z^+\rangle$ in $\Bbb Z^+$ such that $\langle T^{k_i}:i\in\Bbb Z^+\rangle$ converges to $x$. It follows that $\langle T^{k_i+n}:i\in\Bbb Z^+\rangle$ converges to $T^n(x)$, and since $T^n(x)$ is isolated, this implies that $T^{k+n}(x)=T^n(x)$ for some $k\in\Bbb Z^+$. But then, just as in the preceding paragraph, $X$ is finite, contrary to our assumption, and we conclude that no point of $O_T^+(x)$ is isolated.
Let $n\in\Bbb Z^+$ be arbitrary, and let $y=T^n(x)$. Let $U$ be any non-empty open set in $X$. Then $V=U\setminus\{T^k(x):k=1,\ldots,n\}$ is a non-empty open set, so there is a $k\in\Bbb Z^+$ such that $T^k(x)\in V$. Clearly $k>n$, so $T^k(x)=T^{k-n}(y)\in O_T^+(y)$, and $y$ is a transitive point of $T$. Thus, every point of the dense set $O_T^+(x)$ is a transitive point of $T$.
(ii) implies (i):
Let $D\subseteq X$ be a dense set of transitive points of $T$, and let $U$ and $V$ be non-empty open sets in $X$. $U\cap D\ne\varnothing$, so let $x\in U\cap D$. Then $T^n(x)\in T^n[U]\cap V$ for some $n\in\Bbb Z^+$, so $T$ is transitive.
