Isolated Points and Topological Transitivity I'm looking for a hint on this question:
Suppose $X$ is a metric space with at least one isolated point and $T:X\to X$ is a topologically transitive dynamical system. Show that $X$ is necessarily finite, and $X=O(x)$ for any point $x\in X$.
Any help is welcome. Thanks in advance.
 A: Here I am using the given definition of "topologically transitive" that for any two nonempty open sets $U, V$ in $X$ there is a nonnegative integer $k$ such that $T^k(U) \cap V$ is nonempty.  (A different conclusion arises if we replace nonnegative with positive in this definition.) 

Let $x^*$ be the isolated point.  If you can show that the orbit of $x^*$ is finite, so $x^*$ eventually maps back to $x^*$, you can show that all points of $X$ are in that orbit and so the desired result is true.  
However, the result is false in general.  Here is a counter-example:
$$X = [0,1] \cup \{2\}$$
where "2" is the isolated point. Let $\{q_1, q_2, q_3, ...\}$ be a list of the rationals in $[0,1]$. 
Define $T:X\rightarrow X$ by
$$ T(x) = \left\{ \begin{array}{ll}
2 &\mbox{ if $x$ is an irrational in $[0,1]$} \\
q_1  & \mbox{ if $x=2$}\\
q_{k+1} & \mbox{ if $x=q_k$}  
\end{array}
\right.$$
In particular we have for any irrational $r \in [0,1]$: 
$$ r \rightarrow 2 \rightarrow q_1\rightarrow q_2\rightarrow q_3\rightarrow ...$$
We want to show that for every two nonempty open subsets $U, V$ of $X$, there is a $k\geq 0$ such that $f^k(U)\cap V$ is nonempty. Note that every open set that contains a point in $[0,1]$ contains an infinite number of irrationals and  rationals. 
Let $U$ and $V$ be open subsets of $X$.  


*

*If both $U$ and $V$ contain $2$ then $2 \in U=T^0(U)$ and $2 \in V$ and we are done. [This uses $k=0$ which we can do from the nonnegative definition. If we use a $k>0$ definition, then this step does not hold.] 

*If $V$ contains $2$ but $U$ does not, then $U$ must contain an element of $[0,1]$ and so it must contain an irrational in $[0,1]$, which maps to 2 and we are done. 

*If $U$ contains $2$ but $V$ does not, then $V$ must contain a rational in $[0,1]$. But 2 eventually maps to all rationals in $[0,1]$ so we are done. 

*If neither $U$ nor $V$ contains $2$, then they both contain an infinite number of rationals in $[0,1]$.   Let $q_k$ be any rational in $U$. Then $q_k$ eventually maps to all rationals $q_i$ such that $i \geq k$, and so it eventually maps to an element of $V$. 
A: Michael provides a counterexample to the statement:

Suppose $X$ is a metric space with at least one isolated point and $T\colon X\to X$ is a topologically transitive dynamical system where $T$ is not necessarily continuous. Show that $X$ is necessarily finite, and $X=O(x)$ for any point $x\in X$.

But judging by the context of the exercise I assume that $T$ is supposed to be continuous, since this is the standard assumption for a dynamical system when a topology is involved. So I will provide a proof of the statement if $T$ continuous. Moreover, I will use the following definition of transitivity

A topological dynamical system $T: X\to X$ is called topologically transitive if for every pair $U, V\subseteq X$ of open sets there exists a $k\in \mathbb{Z}_{\geq 0}$ such that $T^k(U) \cap V \neq \emptyset$.

As Micheal mentioned, it is enough to show that an isolated point is a periodic point. Let us do this now.
Let $x$ be an isolated point of $X$ and set
$$
\mathcal{X} = \{y\in X\mid \exists k\geq 0\colon T^k(y) = x\}.
$$
Note that by transitivity $\mathcal{X}$ is dense in $X$. Let $\mathcal{C}$ denote the set of all connected components of $X$ with non-empty interior and $\mathcal{D}$ the set of all connected components of $X$ with empty interior. Since each $C\in \mathcal{C}$ is open we can use density of $\mathcal{X}$ to find a $y_C\in C\cap \mathcal{X}$ so that there exists a $k\geq 0$ such that $T^k(y_C) = x$. Now $T$ is continuous, so $T^k$ is continuous and since continuous functions map connected sets on connected sets we must have that $T^k(C) = \{x\}$. Note that we can apply the same argument for any $D\in \mathcal{D}$ with $D\cap\mathcal{X}\neq \emptyset$. So we can partition $X = A\cup B$ where
$$
A = \{y\in X\mid \exists C\in\mathcal{C}\colon y\in C\text{ or }\exists D\in\mathcal{D}\colon D\cap\mathcal{X}\neq 0 \text{ and }y\in D\}
$$
and $B = A^c$. Observe that both $A$ and $B$ are $T$-invariant and $B$ has empty interior (by transitivity). Thus we have two dynamical systems $T|_A$ and $T|_B$ which we can study separately. By the above connected-component-argument we conclude that $x$ is in the orbit of every element in $A$. In particular, we can find a $k\geq 0$ for $T(x)$ such that $T^{k+1}(x) = x$, so $x$ is a periodic point of $T|_A$. Since $B$ has non-empty interior, it cannot be open, in particular for every $\epsilon > 0$ there must be an element in $A$ in an $\epsilon$-neighborhood of $B$, otherwise this $\epsilon$-neighborhood of $B$ is $B$ itself. So finiteness of $A$ implies that $B$ is empty. From this it follows that $X$ is indeed finite and $O(x) = X$ for all $x\in X$.
A: Here is a shorter proof, essentially along the ideas of Michael and Aaron, and some further details.
Note: In an earlier version I have used the a priori stronger version of topological transitivity (which I call below "$\mathbb{Z}_{\geq1}$-TT"; the corresponding argument is ($\ast$)); to rectify the definitions and implications I am adding further details.
For reference purposes, the content of the discussion below correspond to Exr.1.2.3 and Exr.1.2.4 from Grosse-Erdmann & Manguillot's Linear Chaos (p.25).

First let us fix some definitions.
Let $X$ be a topological space, $T: X\to X$ be a continuous self-map of $X$. Denote by $\mathcal{T}(X)^\ast$ the collection of nonempty open subsets of $X$. For $Z\subseteq \mathbb{Z}_{\geq0}$, call $T$ topologically $Z$-transitive ($Z$-TT) if
$$\forall U,V\in\mathcal{T}(X)^\ast,\exists n=n(U\to V)\in Z: T^n(U)\cap V\neq \emptyset.$$
It's clear that if $Z_1\subseteq Z_2\subseteq\mathbb{Z}_{\geq0}$, then any $T$ that is $Z_1$-TT is also $Z_2$-TT.

Claim: Let $X$ be a Hausdorff ($\dagger$) topological space, $T:X\to X$ be continuous. Then the following are equivalent:

*

*$X$ is a finite set (with discrete topology) and $T$ is (set theoretically) transitive (hence $X$ is equal to the orbit of any point and $T$ is a homeomorphism).

*$X$ has an isolated point and $T$ is $\mathbb{Z}_{\geq1}$-TT.

*$X$ has an isolated point and $T$ is $\mathbb{Z}_{\geq0}$-TT.

Proof: (1$\implies$2) and (2$\implies$3) are clear.
(3$\implies$1) Let $x^\ast\in X$ be an isolated point. Then $\{x^\ast\}\subseteq X$ is both open and closed. Either $X=\{x^\ast\}$, in which case we are done, xor $X\setminus\{x^\ast\}$ is nonempty and open. Then by $\mathbb{Z}_{\geq0}$-TT there is an $m\in\mathbb{Z}_{\geq0}$ such that $T^m(X\setminus \{x^\ast\})\cap\{x^\ast\}\neq\emptyset$. These two open sets are disjoint initially, so $m\in\mathbb{Z}_{\geq1}$. Let $y\in X\setminus\{x^\ast\}$ be such that $T^m(y)=x^\ast$. $T^m$ is continuous, and $X\setminus\{x^\ast\}$ is open, so that there is a $U\in\mathcal{T}(X)^\ast$: $y\in U\subseteq X\setminus \{x^\ast\}$ and $T^m(U)=\{x^\ast\}$. Again by $\mathbb{Z}_{\geq0}$-TT there is an $n\in\mathbb{Z}_{\geq0}$ such that $T^n(\{x^\ast\})\cap U\neq\emptyset$, so that $\mathcal{O}_{x^\ast}(T)=\{T^n(x^\ast)\,|\, n\in\mathbb{Z}_{\geq0}\}$ is dense in $X$ and $T^n(x^\ast)\in U$, whence $T^p(x^\ast)=x^\ast$ for $p=m+n\in\mathbb{Z}_{\geq1}$, that is, $x^\ast$ is a periodic point of $T$, and so its orbit is finite, hence closed. Thus
$$\mathcal{O}_{x^\ast}(T)=\overline{\mathcal{O}_{x^\ast}(T)}=X.$$
($\ast$)(2$\implies$1) Let $x^\ast\in X$ be an isolated point. Then by $\mathbb{Z}_{\geq1}$-TT there is an $a\in\mathbb{Z}_{\geq1}$ such that $T^a(\{x^\ast\})\cap \{x^\ast\}\neq\emptyset$, so that $x^\ast$ is a $T$-periodic point, i.e. $\mathcal{O}_{x^\ast}(T)$ is finite, hence closed. Let $U\subseteq X$ be a nonempty open subset. Then again by $\mathbb{Z}_{\geq1}$-TT there is a $b\in\mathbb{Z}_{\geq1}$ such that $T^b(\{x^\ast\})\cap U\neq\emptyset$. Thus $\mathcal{O}_{x^\ast}(T)$ is dense in $X$. Thus
$$\mathcal{O}_{x^\ast}(T)=\overline{\mathcal{O}_{x^\ast}(T)}=X.$$
($\dagger$) Singletons being closed is enough, i.e. $T_1$ (see e.g. In $T_1$ space, all singleton sets are closed?).

In the previous section we have shown that $\mathbb{Z}_{\geq0}$-TT is equivalent to $\mathbb{Z}_{\geq1}$-TT for continuous self-maps of Hausdorff spaces with an isolated point. In fact they are equivalent even without the assumption of the existence of an isolated point. It's interesting to note that for the below argument Hausdorffness is really needed, as opposed to the above statement.
Claim: Let $X$ be a Hausdorff topological space, $T:X\to X$ be a continuous self-map of $X$. Then $T$ is $\mathbb{Z}_{\geq1}$-TT iff $T$ is $\mathbb{Z}_{\geq0}$-TT.
Proof: After the statement in the previous section we may assume $X$ has no isolated points, thus any nonempty open subset has at least two distinct points. It suffices to prove that $\mathbb{Z}_{\geq0}$-TT implies $\mathbb{Z}_{\geq1}$-TT.
Let $U,V\in\mathcal{T}(X)^\ast$. By $\mathbb{Z}_{\geq0}$-TT there is an $m\in\mathbb{Z}_{\geq0}$ such that $T^m(U)\cap V\neq\emptyset$, that is $W=U\cap T^{-m}(V)\in\mathcal{T}(X)^\ast$. Since $X$ has no isolated points there are $x,y\in W: x\neq y$. Since $X$ is Hausdorff, there are $R,S\in\mathcal{T}(X)^\ast$ such that
$$x\in R\subseteq W,\,\, y\in S \subseteq W,\,\, R\cap S=\emptyset.$$
By $\mathbb{Z}_{\geq0}$-TT, there is an $n\in\mathbb{Z}_{\geq0}$ such that $T^n(R)\cap S\neq\emptyset$. Initially $R$ and $S$ are disjoint, so $n\in\mathbb{Z}_{\geq1}$. Thus $\exists u\in R\subseteq W=U\cap T^{-m}(V): T^n(u)\in S\subseteq W=U \cap T^{-m}(V)$, that is, $u\in U, T^{p}(u)\in V$ for $p=m+n\in\mathbb{Z}_{\geq1}$, i.e. $T^p(U)\cap V\neq\emptyset$.

In fact, repackaging the previous argument as an induction step one can show something stronger:
Claim: Let $X$ be a Hausdorff topological space, $T:X\to X$ be a continuous self-map. Then $T$ is $\mathbb{Z}_{\geq0}$-TT iff
$$\forall U,V\in\mathcal{T}(X)^\ast: \#(\{n\in\mathbb{Z}_{\geq0}\,|\, T^n(U)\cap V\neq\emptyset\})=\infty.$$
In words, existence of a return time (say in $\mathbb{Z}_{\geq0}$) guarantees infinite returns. It is convenient to introduce some further notation; put
$$N^T(U\to V)=\{n\in\mathbb{Z}_{\geq0}\,|\, T^n(U)\cap V\neq\emptyset\}$$
and call it the set of return times from $U$ to $V$.
Proof: We may again assume $X$ has no isolated points. Let $U,V\in\mathcal{T}(X)^\ast$. By $\mathbb{Z}_{\geq0}$-TT there is an $m\in\mathbb{Z}_{\geq0}$ such that $T^m(U)\cap V\neq\emptyset$, that is $W=U\cap T^{-m}(V)\in\mathcal{T}(X)^\ast$. Note that $m+ N^T(W\to W)\subseteq N^T(U\to V)$ (where $m+ N^T(W\to W)=\{m+n\,|\, n\in N^T(W\to W)\}$). We'll show that $\#(N^T(W\to W))=\infty$.
As in the previous argument, there are $x_1,y_1\in W$ and $R_1,S_1\in\mathcal{T}(X)^\ast$ such that
$$x_1\in R_1\subseteq W,\,\, y_1\in S_1 \subseteq W,\,\, R_1\cap S_1=\emptyset.$$
By $\mathbb{Z}_{\geq0}$-TT, there is an $n_1\in\mathbb{Z}_{\geq0}$ such that $T^{n_1}(R_1)\cap S_1\neq\emptyset$. Initially $R_1$ and $S_1$ are disjoint, so $n_1\in\mathbb{Z}_{\geq1}$, whence $n_1\in N^T(W\to W)\cap \mathbb{Z}_{\geq1}$.
Put $W_1= W\cap T^{-n_1}(W)\in\mathcal{T}(X)^\ast$. Applying the previous paragraph to $W_1$ instead of $W$, there are $x_2,y_2\in W_1$ and $R_2,S_2\in\mathcal{T}(X)^\ast$ such that
$$x_2\in R_2\subseteq W_1,\,\, y_2\in S_2 \subseteq W_1,\,\, R_2\cap S_2=\emptyset.$$
By $\mathbb{Z}_{\geq0}$-TT, $\exists m_1\in N^T(W_1\to W_1)\cap \mathbb{Z}_{\geq1}$ and hence $n_2=n_1+m_1\in N^T(W\to W)\cap\mathbb{Z}_{>n_1}$. By induction ($W_2=W\cap T^{-n_2}(W), \cdots$ )  $N^T(W\to W)$ is thus infinite.
