Set of equicontinuity points

Question

Suppose that $ T: X \to X $ is a continuous onto map and $ \rho $ and $ \rho_{T} $ are metrics on that, which :
\begin{align*} &\rho_{T}(x,y) := \sup \lbrace \rho(T^{n}(x) , T^{n}(y)) : n \geq 0 \rbrace \\ &diam_{T}(A) := \sup \lbrace \rho_{T}(x,y): x,y \in A \rbrace ,\quad A \subseteq X \end{align*} Suppose that :
\begin{align*} Eq_{\varepsilon}(T) := \bigcup \lbrace U \subseteq X : U \text{ is open and}, diam_{T}(U) \leq \varepsilon \rbrace \end{align*}
I'm going to show that:
\begin{align*} T^{-1}(Eq_{\varepsilon}(T)) \subseteq Eq_{\varepsilon}(T) \end{align*} Notice that we know $Eq_{\varepsilon}(T)$ is a $G_{\delta}$ set.

Answer 1

Notice that $x\in E_{q_\epsilon}(T)$ if and only if there exists $U\subseteq X$ open such that $x\in U$ and $\text{diam}_T(U)\leq \epsilon$.

Let $x\in T^{-1}[E_{q_\epsilon}(T)]$, i.e, $T(x)\in E_{q_\epsilon}(T)$. Then there exists $U_0$ open containing $T(x)$ such that $\text{diam}_T(U_0)\leq \epsilon$. Which means that:

$$ \rho\Big(T^{n+1}(x),T^n(y) \Big)\leq \epsilon \quad \text{for all} \quad y\in U_0 \quad \text{and} \quad n\geq 0$$

Were searching for an open set $U'$ such that $x\in U'$ and $\text{diam}_T(U')\leq \epsilon$. I propose $U':=B_{\frac{\epsilon}{2}}(x)\cap T^{-1}[U_0]$. Since $T$ is continuous, $U'$ is open. We can also see that $x\in U'$.

It remains to show that $\text{diam}_T(U')\leq \epsilon$. Let $y,z\in U'$, and we'll show that:

$$ \rho\Big(T^{n}(y),T^n(z) \Big)\leq \epsilon \quad \text{for all} \quad n\geq 0$$

Since $T(y),T(z)\in U_0$, and $\text{diam}_T(U_0)\leq \epsilon$, we know that:

$$ \rho\Big(T^{n}(y),T^n(z) \Big)\leq \epsilon \quad \text{for all} \quad n\geq 1$$

Notice also that $y,z\in B_{\frac{\epsilon}{2}}(x)$, then by the triangle inequality:

$$ \rho(T^{0}(y),T^0(z))=\rho(z,y) \leq \rho(z,x)+\rho(x,y)\leq \frac{\epsilon}{2}+ \frac{\epsilon}{2} $$

And this shows that $x\in E_{q_\epsilon}(T)$.