Topological weakly mixing implies totally transitive I'm currently reading the first chapter of Grosse-Erdmann & Peris Manguillot's book Linear Chaos, which chapter is about topological dynamical systems and I got stuck in an exercise which says that if $T$ is a dynamical system which is weakly mixing then $T$ is totally transitive:

Could someone give me a hint or a solution? Thanks in advance. I forgot to mention that a topological dynamical system is just a metric space $X$ with a continuous map $T : X \to X$ defined on it.
 A: Hint: The main tool to use is the Furstenberg Theorem (Thm.1.51 on p.20 of the text you are citing); that says that if a system is topologically weak mixing (TWM), then any Cartesian power of it is also TWM.

Let us fix some notations. Let $X$ be a topological space, $T:X\to X$ be a continuous self-map of $X$. For $n\in\mathbb{Z}_{\geq0}$: denote by
$$T^n=\underbrace{T\circ T\circ\cdots \circ T}_{n\text{ times}}:X\to X$$
the $n$-fold composition of $T$ with itself. For $q\in\mathbb{Z}_{\geq1}$ denote by
$$T^{(q)}=\underbrace{T\times T\times\cdots \times T}_{p\text{ times}}:X^q\to X^q$$
the $p$-fold Cartesian product of $T$ with itself, where $X^q=\underbrace{X\times X\times\cdots \times X}_{p\text{ times}}$.
Denote by $\mathcal{T}(X)^\ast$ the collection of all nonempty open subsets of $X$.
Definition: $T:X\to X$ is called:

*

*topologically transitive (TT) if $\forall U,V\in \mathcal{T}(X)^\ast,\exists n=n(T;U\to V)\in\mathbb{Z}_{\geq1}: T^n(U)\cap V\neq\emptyset$.

*totally topologically transitive (TTT) if $\forall p\in\mathbb{Z}_{\geq1}: T^p:X\to X$ is TT.

*topologically weak mixing (TWM) if $T^{(2)}:X^2\to X^2$ is TT.

*totally topologically weak mixing (TTWM) if $\forall p\in\mathbb{Z}_{\geq1}: T^p:X\to X$ is TWM.

*topologically strong mixing (TSM) if $\forall U,V\in \mathcal{T}(X)^\ast,\exists N=N(T;U\to V)\in\mathbb{Z}_{\geq1},\forall n\in\mathbb{Z}_{\geq N}: T^n(U)\cap V\neq\emptyset$.

*totally topologically strong mixing (TTSM) if $\forall p\in\mathbb{Z}_{\geq1}: T^p:X\to X$ is TSM.

Putting together the fact that any factor of a TWM system is TWM (Prop.1.48 on p.19) and the Furstenberg Theorem mentioned above we have:
Theorem: The following are equivalent:

*

*$T:X\to X$ is TWM,

*$\exists q\in\mathbb{Z}_{\geq1}:T^{(q)}:X^q\to X^q$ is TWM,

*$\forall q\in\mathbb{Z}_{\geq1}:T^{(q)}:X^q\to X^q$ is TWM.


Claim: If $T:X\to X$ is TWM, then  it is also TTWM, hence consequently TTT.
Proof of Claim: Say $T$ is TWM. Fix a number $p\in\mathbb{Z}_{\geq1}$ and open subsets $U,V,W,Z\in\mathcal{T}(X)^\ast$. We need to show that there is a number $Q\in\mathbb{Z}_{\geq1}$ such that
$$T^{Qp}(U)\cap W\neq\emptyset\text{ and }T^{Qp}(V)\cap Z\neq\emptyset.$$
By the Furstenberg Theorem above, since $T$ is TWM we have that $T^{(p)}$ is TWM, i.e. by definition $T^{(2p)}=(T^{(p)})^{(2)}:X^{2p}\to X^{2p}$ is TT. Define two open sets in $X^{2p}$ as follows:
$$A=U\times T^{-1}(U)\times \cdots \times T^{-p+1}(U)\times V \times T^{-1}(V)\times \cdots \times T^{-p+1}(V),$$
$$B = \underbrace{W\times W\times \cdots\times W}_{p \text{ many}}\times \underbrace{Z\times Z\times \cdots \times Z}_{p\text{ many}}.$$
Since $T^{(2p)}$ is TT, there is a number $m\in\mathbb{Z}_{\geq1}$ such that
$$\left(T^{(2p)}\right)^m(A)\cap B\neq \emptyset.$$
Note that we may assume that $m\geq p-1$ by applying the definition of TT to preimages of open sets ($\ast$). There are unique numbers $Q\in\mathbb{Z}_{\geq1}$ and $R\in\{0,1,...,p-1\}$ such that
$$m=Qp+R.$$
($Q$ and $R$ are uniquely determined by $m$ and $p$, but $m$ is not uniquely determined by the previous objects.)
Writing this out in terms of points, we have guaranteed that there exists $x_0,x_1,...,x_{p-1},y_0,y_1,...,y_{p-1}\in X $ such that
\begin{align*}
&u_0=x_0, u_1=T(x_1),..., u_{p-1}=T^{p-1}(x_{p-1})\in U\text{ and  }\\
&v_0=y_0, v_1=T(y_1),..., v_{p-1}=T^{p-1}(y_{p-1})\in V
\end{align*}
and
\begin{align*}
&T^{m}(u_0)=T^{m}(x_0), T^{m-1}(u_1)=T^{m}(x_1),..., T^{m-p+1}(u_{p-1})=T^{m}(x_{p-1})\in W\text{ and  }\\
&T^{m}(v_0)=T^{m}(y_0), T^{m-1}(v_1)=T^{m}(y_1),..., T^{m-p+1}(v_{p-1})=T^{m}(y_{p-1})\in Z.
\end{align*}
Thus we have
$$T^{pQ}(u_R)=T^{m-R}(u_R)=T^m(x_R)\in W \text{ and } T^{pQ}(v_R)=T^{m-R}(v_R)=T^m(y_R)\in Z,$$
that is,
$$T^{pQ}(u_R)\in T^{pQ}(U)\cap W\text{ and } T^{pQ}(v_R)\in  T^{pQ}(V)\cap Z.$$

Edit: As per request here is an argument for ($\ast$): Let $T:X\to X$ be TT,  $U,V\subseteq X$ be nonempty open. Let $N\in\mathbb{Z}_{\geq1}$. $T^{-N}(V)$ is nonempty open, so for some $n=n(U\to T^{-N}(V))\in\mathbb{Z}_{\geq1}:$ $T^n(U)\cap T^{-N}(V)\neq\emptyset$, whence $T^{N+n}(U)\cap V\neq\emptyset$.
A more detailed discussion about this matter can be found in my answer here: Isolated Points and Topological Transitivity . In particular, to be compatible with the conventions of the book in question, if $X$ is Hausdorff, topological transitivity with nonnegative return time is equivalent to topological transitivity with positive return time.
