Remark 2

The definition of a uniformity by a pseudometric $f$ consists in taking as a fundamental system of entourages the inverse image under $f$ of the neighbourhood filter of 0 in the subspace $[0, +\infty]$ of $\overline{\mathbf{R}}$. Note that this procedure is quite analogous to that which allowed us to define the uniformities on a topological group (Chapter III, § 3, no. 1).

Let $f$ and $g$ be two pseudometrics on $X$. From Definition 2 it follows that the uniformity defined by $f$ is coarser than the uniformity defined by $g$ if and only if, for each $a > 0$ there exists $b > 0$ such that the relation $g(x, y) \leq b$ implies $f(x, y) \leq a$. A necessary and sufficient condition for $f$ and $g$ to be equivalent pseudometrics is that for each $a > 0$ there exists $b > 0$ such that $g(x, y) \leq b$ implies $f(x, y) \leq a$, and $f(x, y) \leq b$ implies $g(x, y) \leq a$.

In particular, if there exists a constant $k$ such that $f \leq kg$, the uniformity defined by $f$ is coarser than the uniformity defined by $g$.

Let $\varphi$ be a mapping of the interval $[0, +\infty]$ into itself, satisfying the following conditions: 1) $\varphi(0) = 0$, and $\varphi$ is continuous at 0; 2) $\varphi$ is increasing in $[0, +\infty]$ and is strictly increasing in a neighbourhood of 0; 3) for all $u \geq 0$ and $v \geq 0$, we have $\varphi(u + v) \leq \varphi(u) + \varphi(v)$. Then if $f$ is any pseudometric on a set $X$, the composition $g = \varphi \circ f$ is a pseudometric equivalent to $f$.

The reader may easily verify that we may, for example, take $\varphi$ to be any one of the following functions:

The last two examples show that there always exist bounded pseudometrics equivalent to any given pseudometric (finite or not).