Proposition 2

Let $X$ be a set and $Y$ a metric space. The set $\mathscr{B}(X; Y)$ of bounded mappings is both open and closed in the space $\mathscr{F}_u(X; Y)$.

If $u$ is bounded, then every mapping $v : X \to Y$ such that for all $x \in X$, we have $d(u(x), v(x)) \leq 1$ is bounded, because

hence $\mathscr{B}(X; Y)$ is open. On the other hand, if $u$ lies in the closure of $\mathscr{B}(X; Y)$ in $\mathscr{F}_u(X; Y)$, there is a mapping $u_0 \in \mathscr{B}(X; Y)$ such that $d(u(x), u_0(x)) \leq 1$ for all $x \in X$; hence $u$ is bounded.