Epsilon-delta continuity in metric spaces

Statement

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces, let $f \colon X \to Y$ be a function and let $a \in X$. Then $f$ is continuous at $a$ if and only if for each real $\varepsilon > 0$ there is a real $\delta > 0$ such that $$d_X(x, a) < \delta \implies d_Y\bigl(f(x), f(a)\bigr) < \varepsilon \quad \text{for all } x \in X.$$

Remarks

In terms of open balls, the condition says: for each $\varepsilon > 0$ there is $\delta > 0$ with $f(B(a, \delta)) \subseteq B(f(a), \varepsilon)$. The number $\delta$ can depend on $\varepsilon$ and on $a$. When one $\delta$ works for all $a$, the map is uniformly continuous. For $X = Y = \mathbb{R}$, this is the epsilon-delta definition of continuity from calculus.

Explanations

L1 human

You can make $f(x)$ as close to $f(a)$ as you want by taking $x$ close enough to $a$. The tolerance $\varepsilon$ for the output comes first, and then you find a tolerance $\delta$ for the input.