Is the product of a proximal system with itself proximal? A topological dynamical system is a pair $(X,T)$ where $X$ is a compact metric space and $T$ is a continuous map from $X$ to itself. Two points $x,y\in X$ are said to to be proximal if for any $\epsilon>0$, there exists an positive integer $n$ such that $d(T^nx,T^ny)<\epsilon$. A topological dynamical system $(X,T)$ is called proximal if any two points $x,y\in X$ are proximal.
Now my question is: if a topological dynamical system $(X,T)$ is proximal, is the product system $(X\times X,T\times T)$ also proximal?
Note that to show that $(X\times X,T\times T)$ is proximal we need to prove that for two points $(x_1,x_2)$ and $(y_1,y_2)$ in $X\times X$ are proximal, i.e. for any $\epsilon>0$, there exists an positive integer $n$ such that $d((T^nx_1,T^nx_2),(T^ny_1,T^ny_2))<\epsilon$. Since $X$ is proximal, we know that $x_1$ and $y_1$ are proximal and $y_1$ and $y_2$ are proximal. So there exists an positive integer $n_1$ such that $$d(T^{n_1}x_1,T^{n_1}y_1)<\epsilon$$ and an positive integer $n_2$ such that $$d(T^{n_2}x_2,T^{n_2}y_2)<\epsilon.$$ But that is not enough, since to show that $X\times X$ is proximal we need to show there exists a uniform $n$, which means that we
need to find an $n$ such that $d(T^{n}x_1,T^{n}y_1)<\epsilon$ and $d(T^{n}x_2,T^{n}y_2)<\epsilon$.
 A: The answer is yes! It's a direct corollary of the following theorem - A dynamical system $(X,T)$ is proximal iff it has a fixed point which is the unique minimal subset of $X$. See - http://iopscience.iop.org/0951-7715/16/4/313 - E. Akin and S. Kolyada, Li-Yorke sensitivity, Nonlinearity, 16 (2003), 1421--1433  
A: Not a complete solution, but some thoughts that are too long to fit in a comment:
For some fixed $x,y,\epsilon$, we can consider the sequence $\{n_k\}_{x,y,\epsilon}$ of positive integers such that $d(T^{n_k}x,T^{n_k}y)<\epsilon$. Then this question is equivalent (via some kind of standard $\epsilon/2$ argument) to showing that $\{n_k\}_{x,y,\epsilon} \cap \{n_k\}_{z,w,\epsilon} \neq \emptyset$ for all $x,y,z,w,\epsilon$.
The compactness of $X$ constrains the sequences $\{n_k\}_{x,y,\epsilon}$ in two distinct ways that I can see. Neither is sufficient to get what we want (and they don't combine well) but they both seem suggestive to me.


*

*Since $X$ is compact, $T$ is uniformly continuous. Let $\omega$ be a uniform modulus of continuity for $T$. By setting $\epsilon=\omega^m(\epsilon_0)$ in the definition of proximality, we can find a sequence of $m$ consecutive integers in $\{n_k\}_{x,y,\epsilon_0}$; thus $\{n_k\}_{x,y,\epsilon}$ contains arbitrarily long runs of consecutive integers for all $x,y,\epsilon$. If we could constrain when these happened (if, say, proximality implied a modulus of continuity that wasn't too horrible), this might be useful...

*Let $U_{n,\epsilon}=\{(x,y) \in X \times X \, | \, d(T^n x, T^n y) < \epsilon\}$. Then $U$ is open (because $T$, and hence $T^n$, is continuous). Moreover, for any fixed $\epsilon$, the definition of proximality implies that the $U_{n,\epsilon}$ cover $X \times X$. So we can pass to a finite subcover. That is, for any $\epsilon$ there exists a finite set $\{m_1,\dots,m_k\}$ of exponents such that $d(T^{m_i}x,T^{m_i}y)<\epsilon$ for some $i<k$. Note that if we had $k=1$, we would be done, so in some sense we're only finitely far away from a solution...

A: I would argue yes.
First we will need to define a metric on $(X\times X$), I will chose the 2-norm. Call this metric d_p.
Hence you need to show that given a $\epsilon> 0,$ and any two points in the product space, $p_1=(x_1,y_1)$ and $p_2=(x_2,y_2)$,  $\exists n$ s.t. $d_p(T^n p_1,T^n p_2)<\epsilon$, where $T^np_1=(T^n x_1, T^n y_1)$ and so on.
But by definition, $d_p(p_1,p_2)=\sqrt{(d(x_1,x_2)^2+d(y_1,y_2)^2)}$
Now, all you need to do is chose $n_1$ and $n_2$ s.t. $d(T^{n_1}x_1,T^{n_1}x_2)<\epsilon/\sqrt 2$ and $d(T^{n_2}y_1,T^{n_2}y_2)<\epsilon/\sqrt 2 $
Then chose $n=max(n_1,n_2)$ and you are done.
