Corollary 1

Let $X$ be a set and $Y$ a metric space. Then $\mathscr{B}_{\mathscr{S}}(X; Y)$ is closed in $\mathscr{F}_{\mathscr{S}}(X; Y)$. In particular, if $Y$ is complete then $\mathscr{B}_{\mathscr{S}}(X; Y)$ is complete with respect to the uniformity of $\mathscr{S}$-convergence.

For $\mathscr{B}_\mathfrak{S}(X; Y)$ is the inverse image of the subset $\prod_{A \in \mathfrak{S}} \mathscr{B}(A; Y)$ of the product $\prod_{A \in \mathfrak{S}} \mathscr{F}_u(X; Y)$ under the canonical mapping of $\mathscr{F}_\mathfrak{S}(X; Y)$ into $\prod_{A \in \mathfrak{S}} \mathscr{F}_u(A; Y)$; the first assertion therefore follows from § 1, no. 2, Remark 3, and the second follows from the first, if we take account of Theorem 1 of § 1, no. 5.