Let $\left(\Omega_{1}, d_{1}\right)$ and $\left(\Omega_{2}, d_{2}\right)$ be metric spaces and let $f: \Omega_{1} \rightarrow \Omega_{2}$ be an arbitrary map. Denote by $U_{f}=\left\{x \in \Omega_{1}: f\right.$ is discontinuous at $\left.x\right\}$ the set of points of discontinuity of $f$. Show that $U_{f} \in \mathcal{B}\left(\Omega_{1}\right)$.
Hint: First show that for any $\varepsilon>0$ and $\delta>0$ the set
$$
U_{f}^{\delta, \varepsilon}:=\left\{x \in \Omega_{1}: \text { there are } y, z \in B_{\varepsilon}(x) \text { with } d_{2}(f(y), f(z))>\delta\right\}
$$
is open (where $\left.B_{\varepsilon}(x)=\left\{y \in \Omega_{1}: d_{1}(x, y)<\varepsilon\right\}\right) .$ Then construct $U_{f}$ from such $U_{f}^{\delta, \varepsilon}$