Continuity via open sets

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 $a \in X$ and each $\varepsilon > 0$ there is $\delta > 0$ such that $d_X(x, a) < \delta$ implies $d_Y(f(x), f(a)) < \varepsilon$ for all $x \in X$.
2. For each open set $V$ of $Y$, the preimage $f^{-1}(V)$ is open in $X$.

Remarks

Condition 2 is the definition of a continuous map between topological spaces. Thus, for metric spaces, the epsilon-delta definition and the topological definition agree. Proof that 1 implies 2: let $a \in f^{-1}(V)$. Choose $\varepsilon$ with $B(f(a), \varepsilon) \subseteq V$, then $\delta$ from condition 1. Then $B(a, \delta) \subseteq f^{-1}(V)$. Images of open sets need not be open: the constant map $\mathbb{R} \to \mathbb{R}$, $x \mapsto 0$, is continuous, but the image of $\mathbb{R}$ is $\{0\}$.

Explanations

L1 human

The topological form of continuity uses only open sets and does not mention distances. This form lets us define continuity in spaces that have no metric at all.