Lemma 1

Let $u$ be a continuous mapping of $A$ into $[ -i, +i ]$; then there is a continuous mapping $v$ of $X$ into $[ -i/3, +i/3 ]$, such that $|u(x) - v(x)| \leq 2/3$ for all $x \in A$.

Let $H$ be the set of all $x \in A$ such that $-i \leq u(x) \leq -i/3$, and let $K$ be the set of all $x \in A$ such that $i/3 \leq u(x) \leq i$; $H$ and $K$ are closed in $A$, and therefore in $X$, and do not intersect; hence by $(O_{V})$ there is a continuous mapping $v$ of $X$ into $[ -i/3, +i/3 ]$ which is equal to $-i/3$ on $H$ and to $+i/3$ on $K$. The mapping $v$ satisfies the conditions of the lemma.

We now define the functions $g_{n}$ by induction. Applying the lemma with $u = f$, we define $g_{0}$ to be a continuous mapping of $X$ into $[ -i/3, +i/3 ]$ such that $|f(x) - g_{0}(x)| \leq 2/3$ for all $x \in A$. Suppose now that a continuous mapping $g_{n}$ of $X$ into the interval

has been defined, such that $|f(x) - g_{n}(x)| \leq (\frac{2}{3})^{n+1}$ for all $x \in A$. Applying the lemma to the function $u(x) = (\frac{2}{3})^{n+1} (f(x) - g_{n}(x))$, we see that there exists a continuous mapping $h_{n+1}$ of $X$ into the interval

such that

for all $x \in A$; the induction is completed by taking $g_{n+1} = g_{n} + h_{n+1}$, since this function satisfies the inequality $|g_{n+1}(x)| \leq 1 - (2/3)^{n+2}$ for all $x \in X$, by virtue of the definition of $h_{n+1}$.

From this definition it follows that, for $m \geq p$ and $n \geq p$, we have

at each point $x \in X$; hence the sequence $(g_n(x))$ is a Cauchy sequence for each $x \in X$, and therefore converges to a point $g(x)$ of the interval $[ -1, +1 ]$; and since $f(x) - g_n(x)$ tends to 0 for all $x \in A$ as $n \to \infty$, $g$ is an extension of $f$ to $X$. It remains therefore only to show that $g$ is continuous on $X$.

Now let $x$ be any point of $X$; then, given any $\varepsilon > 0$, there exists an integer $n_0$ such that $|g_m(y) - g_n(y)| \leq \varepsilon$ for all $y \in X$ and all $m \geq n_0$ and all $n \geq n_0$; hence, letting $m$ tend to $+\infty$, we have

Let $V$ be a neighbourhood of $x$ such that $|g_n(y) - g_n(x)| \leq \varepsilon$ for all $y \in V$; then, for each $y \in V$ we shall have

which shows that $g$ is continuous at $x$, and completes the proof of Theorem 2. (The last part of the proof uses, in a particular case, the idea of uniform convergence, which we shall study in general in Chapter X, no. 1.)