# Set of equicontinuity points

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.

• What exactly is the union over? There is an $\epsilon$ in the index, but it's not clear what union are you taking. – Keen-ameteur Sep 4 '19 at 14:56
• The union over all $U$ for which $diam_{T}(U) \leq \varepsilon$ – Reza Sep 4 '19 at 15:30
• And I assume you want to know how to prove what you are "going to show"? – Keen-ameteur Sep 4 '19 at 15:36
• I want to know how to prove that – Reza Sep 4 '19 at 15:52

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 $$x\in E_{q_\epsilon}(T)$$.