Epsilon-delta uniform continuity

Statement

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces and let $f \colon X \to Y$ be a function. Then the following two conditions are equivalent:

1. For each real $\varepsilon > 0$ there is a real $\delta > 0$ such that $d_Y(f(x), f(y)) < \varepsilon$ for all $x, y \in X$ with $d_X(x, y) < \delta$.
2. For all sequences $(x_n)$ and $(y_n)$ in $X$ with $d_X(x_n, y_n) \to 0$, also $d_Y(f(x_n), f(y_n)) \to 0$.

The map $f$ is uniformly continuous when it satisfies these conditions.

Remarks

Condition 1 is the epsilon-delta definition of uniform continuity. Here $\delta$ depends only on $\varepsilon$, not on the points. Each uniformly continuous map is continuous. The converse fails: $x \mapsto x^2$ on $\mathbb{R}$ is continuous but not uniformly continuous. Take $x_n = n + 1/n$ and $y_n = n$; then $x_n^2 - y_n^2 > 2$. On a compact space, each continuous map to a metric space is uniformly continuous: this is the Heine–Cantor theorem. Each Lipschitz map is uniformly continuous.

Explanations

L1 human

For a uniformly continuous map, one input tolerance $\delta$ works in the whole space. The map $x \mapsto x^2$ gets steeper and steeper, so near large $x$ you need a smaller and smaller $\delta$, and no single $\delta$ works.