Let $E$ be a bounded and complete metric space. $\Gamma(E)=\{T:E\to E\}$ such that $T$ is a contraction. We now add a distance to $\Gamma(E)$, namely $d_{sup}$ and we have $E$ with the classical distance. We define $\Phi:\Gamma(E)\to E$ such that $\Phi(T)=x_T$ with $x_T$ being the fixed point of $T$. How can I prove that $\Phi$ is continuous?
My attempt was to pick a neighbourhood of $\Phi(x)\in E$ and show that the preimage of said nbhd is also a neighbourhood of $x\in \Gamma(E)$, because the map $T$ is continuous due to it being a contraction. Tho I don't know how to express it formally. Thanks!
Fix $T \in \Gamma(E)$ with contraction constant $c_T < 1$, ie $d(Tx, Ty) \le c_T d(x, y)$. Then for $S \in \Gamma(E)$, $$\begin{align*}d(x_T, x_S) &= d(Tx_T, Sx_S)\\ & \le d(Tx_T, Tx_S) + d(Tx_S, Sx_S)\\ & \le c_T \cdot d(x_T, x_S) + d_{\sup}(T, S).\end{align*}$$ Rearranging, $d(x_T, x_S) \le \frac{1}{1-c_T} d_{\sup}(T, S)$, so $\Phi$ is continuous at $T$. Since $T$ was arbitrary, $\Phi$ is continuous.