Proposition 2

Let D be a dense subset of X and A a subset of the space $\mathscr{C}_s(X)$. The following conditions are equivalent : (i) A is relatively compact in $\mathscr{C}_s(X)$. (ii) From every infinite sequence of elements of A, we can extract a sequence converging in $\mathscr{C}_s(X)$. (iii) Every infinite sequence of elements of A has a limit point in $\mathscr{C}_s(X)$. (iv) Let $(f_n)_{n \in \mathbf{N}}$ be a sequence of functions belonging to A and $(x_m)_{m \in \mathbf{N}}$ a sequence of points of D. If the iterated limits

exist, then they are equal. In addition, we have $\sup_{f \in A} |f(x)| < + \infty$ for all $x \in X$.

(i) $\Rightarrow$ (ii) : let $\overline{A}$ be the closure of A in $\mathscr{C}_s(X)$. Assume that A is compact, and consider a sequence of functions $f_n \in A$ (for $n \in \mathbf{N}$). Let $\phi$ be the continuous mapping $x \mapsto (f_n(x))_{n \in \mathbf{N}}$ from X into the metrizable space $K^\mathbf{N}$. The image $X'$ of X under $\phi$ is a compact metrizable space, since X is compact. Let E be the closed subspace of $\mathscr{C}_s(X)$ consisting of continuous functions $f$ on X such that the relation $\phi(x) = \phi(y)$ implies $f(x) = f(y)$ for every pair of points $x, y$ in X. By cor. 2 of GT, I, § 9, No. 4 and prop. 3 of GT, I, § 5, No. 2, the mapping $f' \mapsto f' \circ \phi$ is a homeomorphism $\phi^*$ from $\mathscr{C}_s(X')$ onto E. Hence the set $A' = (\phi^*)^{-1}(\overline{A})$ is compact in $\mathscr{C}_s(X')$, and it is clear that there exist elements $f'_n$ in $A'$ such that $\phi^*(f'_n) = f'_n \circ \phi$ is equal to $f_n$.

Since $X'$ is a compact metrizable space, there exists a countable dense subset D' in $X'$ (GT, IX, § 2, No. 8, prop. 12 and § 2, No. 9, prop. 16). Let $\mathscr{T}_1$ (resp. $\mathscr{T}_2$) be the topology on $A'$ induced by the topology of simple convergence on D' (resp. $X'$). Then $\mathscr{T}_1$ is metrizable, $\mathscr{T}_2$ is compact and finer than $\mathscr{T}_1$, hence $\mathscr{T}_1$ and $\mathscr{T}_2$ coincide; in other words, $A'$ is a compact metrizable subspace of $\mathscr{C}_s(X')$. Therefore, there exists a sequence $(f'_{n_k})$ extracted from $(f'_n)$ and converging to an element $f'$ of $\mathscr{C}_s(X')$. Therefore, the sequence $(f_{n_k})$ converges to $f = f' \circ \phi$ in $\mathscr{C}_s(X)$.

(ii) $\Rightarrow$ (iii) : this is clear.

(iii) $\Rightarrow$ (iv) : suppose that every infinite sequence of elements of A has a limit point in $\mathscr{C}_s(X)$. Let $x \in X$. The mapping $\phi_x : f \mapsto f(x)$ from A into K is continuous. Consequently, every infinite sequence in $\phi_x(A)$ has a limit point; since the field K (equal to $\mathbf{R}$ or $\mathbf{C}$) is metrizable, the set $\phi_x(A)$ is relatively compact in K, hence bounded. In other words, we have $\sup_{f \in A} |f(x)| < \infty$.

Let $f_n, x_m, \gamma$ and $\delta$ be as in (iv). Let $f$ be a limit point of the sequence $(f_n)$ in $\mathscr{C}_s(X)$, and let $x$ be a limit point of the sequence $(x_m)$ in the compact space X. For every $m$, the mapping $h \mapsto h(x_m)$ from $\mathscr{C}_s(X)$ into K is continuous. In view of the hypotheses, we have $f(x_m) = \lim_{n \to \infty} f_n(x_m)$, and hence $\gamma = \lim_{m \to \infty} f(x_m)$; since $f : X \to K$ is continuous and $x$ is a limit point of the sequence $(x_m)$, we get $\gamma = f(x)$. In an analogous way, we prove the equality $\delta = f(x)$, whence $\gamma = \delta$.

(iv) $\Rightarrow$ (i) : suppose that the set of numbers $f(x)$, as $f$ ranges over A, is bounded in K for all $x \in X$. This is equivalent to assuming that the closure $\overline{A}$ of A in the product space $K^X$ is compact (GT, I, § 9, No. 5). Suppose that A is not relatively compact in $\mathscr{C}_s(X)$. This means that there exists a function $u \in \overline{A}$ and a point $a \in X$ such that $u$ is not continuous at $a$. Hence there exists a real number $\varepsilon > 0$ such that in every neighbourhood U of $a$, there exists a point $x$ with $|u(x) - u(a)| \geq \varepsilon$.

We shall construct by induction a sequence $(x_n)_{n \in \mathbf{N}}$ of points in D and a sequence $(f_n)_{n \in \mathbf{N}}$ of elements of A, satisfying the following relations :

(1)$_m$

(2)$_m$

(3)$_{m,i}$

We take $x_0 = a$ with $f_0$ arbitrary in A (the set A is not empty, otherwise it will be relatively compact in $\mathscr{C}_s(X)$). Let $n \geq 1$ and $x_0, x_1, ..., x_{n-1}, f_0, f_1, ..., f_{n-1}$ satisfy relations (1)$_m$, (2)$_m$ for $1 \leq m < n$ and (3)$_{m,i}$ for $0 \leq m \leq i < n$. Since $u$ belongs to $\overline{A}$, there exists $f_n \in A$ satisfying (2)$_n$. Let $V_n$ be the set of all $x \in X$ such that $|f_m(x) - f_m(a)| \leq \frac{1}{n + 1}$ for $0 \leq m \leq n$. Since $f_n$ is continuous, $V_n$ is a neighbourhood of $a$; choose a point $x_n$ in $D \cap V_n$ such that $|u(x_n) - u(a)| \geq \varepsilon$, hence (1)$_n$ and (3)$_{m,n}$ are satisfied. Therefore, the construction can be continued.

Since $u(X)$ is a compact subset of K, there exists a sequence $(y_k)$ extracted from $(x_m)$ and such that the limit $\gamma = \lim_{k \to \infty} u(y_k)$ exists. By (2)$_m$, we have $u(x_i) = \lim_{n \to \infty} f_n(x_i)$ for all $i \in \mathbf{N}$, hence

(4)

On the other hand we have $f_n(a) = \lim_{i \to \infty} f_n(x_i)$ by (3)$_{m,i}$ hence $f_n(a) = \lim_{k \to \infty} f_n(y_k)$. Since $x_0 = a$, we deduce from (2)$_m$ that $\lim_{n \to \infty} f_n(a) = u(a)$. Consequently,

Finally, from (1)$_m$, we get $|\gamma - u(a)| \geq \varepsilon$, and so $\gamma \neq u(a)$. This contradicts assertion (iv); we have thus proved that (iv) implies (i).