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Continuity as Preimages of Open Sets

The epsilon–delta definition of continuity secretly says nothing about numbers — it says that pulling back an open target keeps you inside an open source. We extract that statement, make it the definition, and prove it matches the old one. Then the basics of continuity become one-line consequences of the open-set axioms.

Rereading epsilon–delta

Recall the epsilon–delta definition: f is continuous at a if for every epsilon > 0 there is delta > 0 so that |x - a| < delta forces |f(x) - f(a)| < epsilon. Read the conclusion as: whenever x lands in the ball of radius delta around a, f(x) lands in the ball of radius epsilon around f(a). In open-ball language, the preimage of every open ball around f(a) contains an open ball around a. Since open sets are exactly unions of open balls, this is the same as saying the preimage of any open set is open.

The topological definition

Let X and Y be topological spaces. A map f : X → Y is a continuous map if for every open set V ⊆ Y, the preimage f⁻¹(V) is open in X. That is the entire definition — no points, no epsilons, no distances. It is global rather than pointwise, but it captures exactly the same maps in the metric setting, as the next proof shows.

Theorem. For f : R -> R, the two definitions agree:
  (A) epsilon-delta continuous at every point a;
  (B) preimage of every open set is open.

(A) => (B):
  Let V ⊆ R be open. Take any a ∈ f^{-1}(V), so f(a) ∈ V.
  Since V is open, some ball (f(a)-epsilon, f(a)+epsilon) ⊆ V.
  By (A) there is delta > 0 with |x-a| < delta => |f(x)-f(a)| < epsilon,
      i.e. f maps (a-delta, a+delta) into (f(a)-epsilon, f(a)+epsilon) ⊆ V.
  Hence (a-delta, a+delta) ⊆ f^{-1}(V).
  Every point of f^{-1}(V) has such a ball around it, so f^{-1}(V) is open.

(B) => (A):
  Fix a and epsilon > 0. The set V = (f(a)-epsilon, f(a)+epsilon) is open,
  so by (B) the set U = f^{-1}(V) is open and contains a.
  Openness of U gives delta > 0 with (a-delta, a+delta) ⊆ U,
  meaning |x-a| < delta => f(x) ∈ V => |f(x)-f(a)| < epsilon.
  That is exactly the epsilon-delta condition at a.            ∎
The epsilon–delta and preimage definitions of continuity are equivalent — both directions, explicitly.

Why preimages, and a free theorem

The payoff is that hard theorems become trivial. Take the composition of continuous maps is continuous. With epsilon–delta this needs two nested choices of delta; with preimages it is one line. If f : X → Y and g : Y → Z are continuous and W ⊆ Z is open, then g⁻¹(W) is open in Y, so f⁻¹(g⁻¹(W)) is open in X. But f⁻¹(g⁻¹(W)) = (g∘f)⁻¹(W), so the preimage of W under g∘f is open. Done. An equivalent test is that f is continuous iff the preimage of every closed set is closed — just take complements.