\Phi_n(U)$, whenever $k>n$ and $V\in \mathscr{U}_k$ with $V\cap U\neq {\varnothing}$.

To prepare for the proof, we proceed with several observations.

\[proposition-Near-sel-v1:1\] Let $\varepsilon:X\to (0,+\infty)$ be a continuous function on a space $X$. Then there exists an open cover $\mathscr{V}$ of $X$ and a function $\delta:\mathscr{V}\to (0,+\infty)$ such that $\delta(V)\leq \varepsilon(p)$, for every $p\in V\in \mathscr{V}$.

Since $\varepsilon$ is continuous, each point $x\in X$ is contained in an open set $V_x$ such that $\varepsilon(p)>\frac{\varepsilon(x)}2=\delta_x$, for every $p\in V_x$. Take $\mathscr{V}=\{V_x:x\in X\}$, and for each $V\in \mathscr{V}$ set $\delta(V)=\delta_x$, where $\delta_x$ corresponds to some $x\in X$ with $V=V_{x}$.

Let $\mathscr{F}(Y)$ be the collection of all nonempty closed subsets of $Y$. A natural topology on $\mathscr{F}(Y)$ is the Hausdorff one $\tau_{H(\rho)}$ generated by the Hausdorff distance $H(\rho)$ associated to $\rho$. Recall that the *Hausdorff distance* is defined by $$\begin{aligned}
  \label{eq:Near-sel-v4:1}
  H(\rho)(S,T)&=\inf\big\{\varepsilon>0: S\subset
                \mathbf{O}_\varepsilon(T)\ \text{and}\ T\subset
                \mathbf{O}_\varepsilon(S)\big\}\\
              &=\sup\left\{\rho(S,y)+\rho(y,T): y\in S\cup
                T\right\},\qquad S,T\in \mathscr{F}(Y).\nonumber\end{aligned}$$ The Hausdorff topology $\tau_{H(\rho)}$ is always metrizable even though the distance $H(\rho)$ may assume infinite values. Moreover, given any nonempty set $S\subset Y$, we have that $\rho(y,S)=\rho\left(y,\overline{S}\right)$, for every $y\in Y$. So, we may consider $H(\rho)(S,T)$ defined on all nonempty subsets of $Y$. Then, according to and , a mapping $\varphi:X\operatorname{\leadsto}Y$ 