Uniform continuity of a map of metric spaces: one $\delta$ serving every point

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $f : X \to Y$ be a function.

$f$ is uniformly continuous if for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that

$d_X(x,x') < \delta \;\Longrightarrow\; d_Y\big(f(x), f(x')\big) < \varepsilon \qquad \text{for all } x, x' \in X .$

The whole content is in the quantifier order. Continuity at a point $a$ allows $\delta$ to depend on $\varepsilon$ and on $a$ (Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$-$\delta$ form); uniform continuity demands one $\delta$ that works for every pair of points at once. Written with the quantifiers in order, continuity on $X$ is $\forall \varepsilon\, \forall a\, \exists \delta\, \forall x$ and uniform continuity is $\forall \varepsilon\, \exists \delta\, \forall a\, \forall x$; moving $\exists \delta$ to the left is the entire difference, and it is a strictly stronger condition.

Uniform continuity is a property of the triple $(d_X, d_Y, f)$. Both metrics are named, and neither may be replaced by a merely topologically equivalent one without changing the notion.

This definition was promised earlier and is now discharged. Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$-$\delta$ form records that uniform continuity is not defined there, and Topologically, uniformly and Lipschitz equivalent metrics on a set writes the condition out in full for the identity maps of two metrics on one set rather than naming it. With the definition above, uniform equivalence of $d$ and $d'$ says exactly that $\mathrm{id} : (X,d) \to (X,d')$ and $\mathrm{id} : (X,d') \to (X,d)$ are both uniformly continuous, which is how that condition is read from here on.

• Uniform continuity implies continuity, and the converse fails. The implication is immediate, since a $\delta$ serving every point serves each point; it is recorded with the rest of the hierarchy in Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent. The failure of the converse is witnessed by $x \mapsto 1/x$ on $(0,1)$ ($x \mapsto 1/x$ is continuous on $(0,1)$ and sends the Cauchy sequence $(1/(k+2))_{k \ge 0}$ to an unbounded one ↗).
• The condition is symmetric in $x$ and $x'$ and says nothing about a distinguished point, which is why it is stated with two free variables and no base point. In ball language it reads: for every $\varepsilon > 0$ there is $\delta > 0$ with $f[B_X(x,\delta)] \subseteq B_Y(f(x),\varepsilon)$ for every $x \in X$ simultaneously (Open ball, closed ball and sphere in a metric space).
• What uniform continuity buys. It transports Cauchy sequences (A uniformly continuous map sends Cauchy sequences to Cauchy sequences), which ordinary continuity does not, and that single property is what makes extension from a dense subspace possible (A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space) and what makes completion functorial enough to be unique (A completion is unique up to a unique isometry fixing the original space, and uniformly continuous maps into complete spaces extend through it).