Theorem 1 (Urysohn)

Axiom $(O_v)$ is equivalent to the following:

$(O'_v)$ If $A$ and $B$ are any two disjoint closed subsets of $X$, then there exist two disjoint open sets $U, V$ such that $A \subset U$ and $B \subset V$.

It follows immediately that $(O_v)$ implies $(O'_v)$, for if $f$ is a continuous mapping of $X$ into $[0, 1]$ which is equal to $0$ on $A$ and to $1$ on $B$, then the open sets $\overline{f}^{-1}([0, 1/2[)$ and $\overline{f}^{-1}(]1/2, 1])$ contain $A$ and $B$ respectively and do not intersect.

To prove the converse, notice first that $(O'_v)$ is equivalent to the following axiom:

$(O''_v)$ Given any closed set $A$ and any open neighbourhood $V$ of $A$, there exists an open neighbourhood $W$ of $A$ such that $\overline{W} \subset V$.

If there is a continuous mapping $f : X \to [-1, +1]$ which is equal to $-1$ on $A$ and to $+1$ on $B$, and if we put $U(t) = \overline{f}^{-1}([-1, t[)$ for each $t \in [0, 1]$, then we have defined a family of open sets in $X$, indexed by $[0, 1]$, such that (i) $A \subset U(0)$, (ii) $B \subset \mathbf{C}U(1)$ and (iii) for each pair of real numbers $t, t'$ such that $0 \leq t < t' \leq 1$ we have

for $U(t)$ is contained in the closed set $\overline{f}^{-1}([-1, t])$. Conversely, suppose that we have defined a family $(U(t))$ of open sets $(0 \leq t \leq 1)$ with these three properties (i), (ii) and (iii). For each $x \in X$, put $g(x) = 1$ if $x \in \mathbf{C}U(1)$, and if $x \in U(1)$ let $g(x)$ be the greatest lower bound of the values of $t$ such that $x \in U(t)$. Clearly $0 \leq g(x) \leq 1$ for each $x \in X$, $g(x) = 0$ on $A$, $g(x) = 1$ on $B$; also $g$ is continuous on $X$, for if we put $g(x) = a$, we have $|g(y) - g(x)| \leq \varepsilon$ for all $y \in U(a + \varepsilon) \cap \overline{\mathbf{C}U}(a - \varepsilon)$, which is a neighbourhood of $x$ by (i) [with the conventions that $U(a + \varepsilon) = X$ if $a + \varepsilon > 1$, and $U(a - \varepsilon) = \varnothing$ if $a - \varepsilon < 0$].

Hence Theorem 1 will be proved if we can define a family $(U(t))$ of open sets satisfying conditions (i), (ii) and (iii) above; to do this we use Axiom $(\mathrm{O}'_V)$. Take $U(1) = \mathbf{C}B$; since $A \subset U(1)$ there exists an open set $U(o)$ such that $A \subset U(o)$ and $U(o) \subset U(1)$ by $(\mathrm{O}'_V)$. Suppose then that for each dyadic number $k/2^n$ ($k = 0, 1, \ldots, 2^n$) we have defined an open set $U(k/2^n)$, these sets being such that $\overline{U}(k/2^n) \subset U((k+1)/2^n)$ for $0 \leq k \leq 2^n - 1$. For each dyadic number

there exists by $(\mathrm{O}'_V)$ an open set $U((2k + 1)/2^{n+1})$ such that and Hence for each dyadic number $r$ such that $0 \leq r \leq 1$ we can define an open set $U(r)$, such that $A \subset U(o)$, $B \subset \mathbf{C}U(1)$, and for each pair of dyadic numbers $r, r'$ such that $0 \leq r < r' \leq 1$.

Now define, for each real number $t \in [0, 1]$,

by (2), this definition agrees with the preceding one for $t$ dyadic; also, if $0 \leq t < t' \leq 1$, then there exist two dyadic numbers $r, r'$ such that $t \leq r < r' \leq t'$; by (2) we have $U(t) \subset U(r)$, hence this proves (1) and therefore completes the proof.

Theorem 1 will enable us to show that two important categories of topological spaces are normal. In the first place: