Let $u, v$ be two elements of $\mathscr{F}_\mathscr{S}(X; Y)$ such that $(u, v) \in W(A, V)$ for every entourage $V$ of $Y$ and every $A \in \mathscr{S}$. Since $Y$ is Hausdorff it follows that $u$ and $v$ coincide on every set $A \in \mathscr{S}$, and since $\mathscr{S}$ covers $X$ we must have $u = v$.

Let $V$ be an entourage of $Y$ and let $A_i \in \mathscr{S}_i (i = 1, 2)$; then it follows immediately from the definitions that $W(A_1 \times A_2, V)$ is identified with $W(A_1, W(A_2, V))$ by the canonical bijection, and the result is immediate.

b) Let $X, X'$ be two sets; let $\mathscr{S}$ (resp. $\mathscr{S}'$) be a set of subsets of $X$ (resp. $X'$); let $Y$ be a uniform space; and let $g : X' \to X$ be a mapping such that, for each $A' \in \mathscr{S}'$, $g(A')$ is contained in a finite union of sets of $\mathscr{S}$. Then the mapping $u \to u \circ g$ of $\mathscr{F}_\mathscr{S}(X, Y)$ into $\mathscr{F}_{\mathscr{S}'}(X'; Y)$ is uniformly continuous.

These propositions are immediate consequences of the description of a fundamental system of entourages for the uniformity of $\mathscr{S}$-convergence given in no. 2, Remark 1; the details of the proofs are left to the reader. Proposition 4 implies, in particular:

Since the structure of pointwise convergence in $B$ is coarser than that of $\mathscr{S}$-convergence, it is enough to show that for each $A \in \mathscr{S}$ and each closed entourage $V$ of $Y$, $W(A, V)$ is closed in $B$ with respect to the topology of pointwise convergence (Chapter II, § 3, no. 3, Proposition 7). Now $W(A, V)$ is the intersection of the inverse images of $V$ under the mappings $(u, v) \to (u(x), v(x))$ as $x$ runs through $A$; these mappings are continuous with respect to the topology of pointwise convergence (no. 2, Remark 6), and the result follows.

This follows immediately from Proposition 5.

For every Cauchy filter with respect to $\mathscr{G}_2$-convergence is also a Cauchy filter with respect to $\mathscr{G}_1$-convergence, and we may apply Corollary 1.

Let $\Phi$ be a Cauchy filter on $\overline{H}$, and define a mapping $v : X \to Y$ as follows. If $x \in B$, $\Phi(x)$ is a Cauchy filter on $\overline{H(x)}$ (no. 2, Remark 6), hence by hypothesis it has at least one limit point; take $v(x)$ to be one of these limits. If $x \notin B$, take $v(x)$ to be any point of $Y$. With this definition of $v$, it is clear that $\Phi$ converges pointwise to $v$ in $B$, and $v$ is therefore a limit of $\Phi$ in $\mathscr{F}_{\mathscr{G}}(X; Y)$ by Proposition 5.

In particular, if $Y$ is complete, the hypothesis of Corollary 3 of Proposition 5 is satisfied for every $H \subset \mathscr{F}(X; Y)$; hence:

For if $u, v$ are continuous mappings of $X$ into $Y$, the mapping $x \mapsto (u(x), v(x))$ of $X$ into $Y \times Y$ is continuous, and the hypothesis that $(u(x), v(x)) \in V$ for all $x \in A$ therefore implies that $(u(x), v(x)) \in \overline{V} = V$ for all $x \in \overline{A}$ (Chapter I, § 2, no. 1, Theorem 1).

If $\bar{\mathscr{G}}$ denotes the set of closures in $X$ of the sets of $\mathscr{G}$, Proposition 6 shows that, on $\mathscr{C}(X; Y)$, the structures of $\mathscr{G}$-convergence and $\bar{\mathscr{G}}$-convergence are identical.

G-convergence in spaces of continuous mappings

For if $(u, v)$ belongs to all the sets $W(A, V)$, where $A \in \mathfrak{S}$ and V is an entourage of Y, the hypothesis that Y is Hausdorff tells us that $u(x) = v(x)$ for all $x \in B$; if $u$ and $v$ are continuous, then $u = v$ by the principle of extension of identities (Chapter I, § 8, no. 1, Proposition 2, Corollary 1).

In particular, on $C(X; Y)$, the topology of pointwise convergence in a dense subset of X is Hausdorff.

Let $u_0 : X \to Y$ lie in the closure of H in $F_u(X; Y)$. For each symmetric entourage V of Y, there is a mapping $u \in H$ such that $(u_0(x), u(x)) \in V$ for all $x \in X$; on the other hand, by hypothesis there is a set $M \in \mathfrak{T}$ such that $(u(x), u(x')) \in V$ whenever $x$ and $x'$ are in M. Since $(u_0(x), u(x)) \in V$ and $(u_0(x'), u(x')) \in V$, it follows that $(u_0(x), u_0(x')) \in V$ whenever $x$ and $x'$ are in M, and therefore $u_0(\mathfrak{T})$ is a Cauchy filter base on Y.

If V is the neighbourhood filter of $x_0$ in X, $u(x_0)$ is a cluster point of $u(V)$; hence $u$ is continuous at $x_0$ if and only if $u(V)$ is a Cauchy filter base on Y (Chapter II, § 3, no. 2, Proposition 5, Corollary 2).

For v lies in the closure of the set of the $u_\lambda$ in $F_u(X; Y)$, and therefore $v(\mathfrak{T})$ is a Cauchy filter base on Y by virtue of Proposition 8; this shows that v has a limit y with respect to $\mathfrak{T}$ because Y is complete. Let

X' = X \cup \{ \omega \} be the topological space associated with the filter $\mathscr{F}$ (Chapter I, § 6, no. 5), and extend $u_\lambda$ (resp. $v$) to a mapping $\bar{u}_\lambda$ (resp. $\bar{v}$) of $X'$ into $Y$ by putting $\bar{u}_\lambda(\omega) = y_\lambda$ [resp. $\bar{v}(\omega) = y$]. Then the mappings $\bar{u}_\lambda, \bar{v}$ are continuous on $X'$, and $\bar{u}_\lambda$ converges uniformly in $X$ to $\bar{v}$ with respect to $\mathscr{G}$; since $X$ is dense in $X'$, the Corollary to Proposition 6 shows that $\bar{u}_\lambda$ converges uniformly in $X'$ to $\bar{v}$, and in particular that $y = \lim_{\mathscr{G}} y_\lambda$.

For each $x \in X$, the set of mappings of $X$ into $Y$ which are continuous at $x$ is closed in $F_u(X; Y)$ (Proposition 8, Corollary 1); hence the intersection $C(X; Y)$ of these sets is also closed.

This result may be expressed in the form that the uniform limit of continuous functions is continuous.

For, by Theorem 2, $C_u(X; Y)$ is a closed uniform subspace of the uniform space $F_u(X; Y)$, which is complete by Theorem 1 of no. 5.

Suppose that $u$ lies in the closure of $\tilde{C}_\mathscr{S}(X; Y)$ in $F_\mathscr{S}(X; Y)$; then (no. 2), for each $A \in \mathscr{S}$, $u|A$ lies in the closure of $C(A; Y)$ in $F_u(A; Y)$, and is therefore continuous by Theorem 2.

By virtue of Corollary 2 it is enough to show that, if we take $\mathscr{S}$ to be the set of compact subsets of $X$, we have $\tilde{C}_\mathscr{S}(X; Y) = C(X; Y)$ in both cases under consideration. This is clear if $X$ is locally compact. If $X$ is metrizable, and $u : X \to Y$ is a mapping whose restriction to every compact subset of $X$ is continuous, then for each $x \in X$ and each sequence $(z_n)$ of points of $X$ which converges to $x$, we have $u(x) = \lim_{n \to \infty} u(z_n)$, and therefore $u$ is continuous at $x$ (Chapter IX, § 2, no. 6, Proposition 10).

Note that the argument above applies whenever every point of $x$ has a countable fundamental system of neighbourhoods.

Let $f_0 : X \to Y$ be a continuous mapping, let $x_0$ be a point of $X$ and let $V$ be an entourage of $Y$. The set $T$ of continuous mappings $f : X \to Y$ such that $(f(x), f(x_0)) \in V$ for all $x \in X$ is a neighbourhood of $f_0$ in $\mathscr{C}_u(X; Y)$. On the other hand, since $f_0$ is continuous, there is a neighbourhood $U$ of $x_0$ in $X$ such that $(f_0(x), f_0(x_0)) \in V$ for all $x \in U$. Consequently we have $(f(x), f(x_0)) \in V$ whenever $(f, x) \in T \times U$, and the result is proved.