Theorem 2 (Urysohn)

Axiom $(O_V)$ is equivalent to the following property: $(O''_V)$ Given any closed subset $A$ of $X$ and any continuous real-valued function $f$ (finite or not) defined on $A$, there exists an extension $g$ of $f$ to the whole space $X$, which is a continuous mapping of $X$ into $\overline{\mathbf{R}}$.

It is easy to see that $(O''_V)$ implies $(O_V)$; for if $B$ and $C$ are two disjoint closed subsets of $X$, then the function which is equal to 0 on $B$ and equal to $i$ on $C$ is defined and continuous on the closed set $B \cup C$, hence by $(O''_{V})$ has a continuous extension $f$ to $X$. If $g = \inf (f^{+}, 1)$, then $g$ is continuous on $X$, takes its values in $[0, 1]$ and is equal to $o$ on $B$ and to $i$ on $C$.

Let us show conversely that $(O_{V})$ implies $(O''_{V})$. Since $\overline{\mathbf{R}}$ and the interval $[ -i, +i ]$ are homeomorphic, we need consider only the case where the continuous mapping $f : A \to \overline{\mathbf{R}}$ takes its values in $[ -i, +i ]$. We shall construct an extension $g$ of $f$ to $X$ by forming a sequence $(g_{n})$ of continuous functions on $X$, such that the sequence $(g_{n}(x))$ converges for all $x \in X$ to a point of the interval $[ -i, +i ]$; this limit will, by definition, be the value of $g$ at $x$, and it will follow from the choice of the $g_{n}$ that the function $g$ satisfies the required conditions.

The definition of the $g_{n}$ rests on the following lemma: