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 $\epsilon&gt;0$, there exists an positive integer $n$ such that $d(T^nx,T^ny)&lt;\epsilon$. 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 $\epsilon&gt;0$, there exists an positive integer $n$ such that $d((T^nx_1,T^nx_2),(T^ny_1,T^ny_2))&lt;\epsilon$. 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 $$d(T^{n_1}x_1,T^{n_1}y_1)&lt;\epsilon$$ and an positive integer $n_2$ such that $$d(T^{n_2}x_2,T^{n_2}y_2)&lt;\epsilon.$$ 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 $d(T^{n}x_1,T^{n}y_1)&lt;\epsilon$ and $d(T^{n}x_2,T^{n}y_2)&lt;\epsilon$.
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
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 $\epsilon&gt; 0,$ 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. $d_p(T^n p_1,T^n p_2)&lt;\epsilon$, 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. $d(T^{n_1}x_1,T^{n_1}x_2)&lt;\epsilon/\sqrt 2$ and $d(T^{n_2}y_1,T^{n_2}y_2)&lt;\epsilon/\sqrt 2 $

Then chose $n=max(n_1,n_2)$ and you are done.