Theorem 2

A topological space $X$ is uniformizable if and only if it satisfies the following axiom:

(OIV) Given any point $x_0 \in X$ and any neighbourhood $V$ of $x_0$, there exists a continuous real-valued function on $X$ which takes its values in $[0, 1]$, is equal to $0$ at $x_0$, and is equal to $1$ on $C_V$.

The condition is necessary. For if there is a uniformity on $X$ compatible with the topology of $X$, then by Theorem 1 this uniformity can be defined by a family $(f_i)$ of pseudometrics on $X$, and we may assume with no loss of generality that this family is saturated (no. 2). From the definition of the entourages of the uniformity defined by such a family of pseudometrics, there is a pseudometric $f_\alpha$ of the family $(f_i)$, and a number $a > 0$, such that $f_\alpha(x_0, x) \geq a$ for all $x \in C_V$. It follows that the function $g(x) = \inf \left( 1, \frac{1}{a} f_\alpha(x_0, x) \right)$ satisfies all the conditions laid down in (OIV).

The condition is sufficient. For let $\Phi$ be the set of all continuous mappings of $X$ into $[0, 1]$. Axiom (OIV) shows that the coarsest uniformity with respect to which all the functions belonging to $\Phi$ are uniformly continuous is compatible with the topology of $X$ (Chapter II, § 2, no. 3).