Continuity in Topology — Mathematics Knowledge System

Overview

Continuity is usually first learned via the ε-δ definition on ℝ: f is continuous at x if points close to x map to points close to f(x). Topology strips this down to its essence, defining continuity purely in terms of open sets, with no distance required: a function f: X → Y between topological spaces is continuous if the preimage of every open set in Y is open in X. This single, elegant condition recovers the ε-δ definition exactly when X and Y are metric spaces (in particular ℝ), and extends continuity to arbitrary topological spaces where no notion of distance exists at all.

Intuition

Think of continuity as a promise about 'not tearing' the space: wherever you look in the target space Y and mark out an open region, the set of points in X that land in that region should itself be an unbroken, open region — no isolated boundary points sneaking in. If pulling back an open 'patch' from Y ever produced a jagged, non-open set in X, that would be exactly where the function tears the space apart, since some points just outside the preimage would map arbitrarily close to points inside it without themselves landing in the patch. Homeomorphisms — the topological analogue of 'the same shape' — are exactly the continuous bijections whose inverse is also continuous, so neither direction is allowed to tear or glue anything.

Formal Definition

Definition

Let X and Y be topological spaces. A function f: X → Y is continuous if:

\forall V \subseteq Y \text{ open}, \quad f^{-1}(V) = \{ x \in X : f(x) \in V \} \text{ is open in } X

The preimage of every open set of Y must be an open set of X

Open-set definition of continuity

f \text{ continuous at } x_0 \iff \forall \varepsilon>0\, \exists \delta>0: d_X(x,x_0)<\delta \implies d_Y(f(x),f(x_0))<\varepsilon

When X, Y are metric spaces, choosing V to be an ε-ball around f(x₀) recovers exactly the classical ε-δ definition of continuity at a point

Metric (ε-δ) special case

f: X \to Y \text{ is a homeomorphism} \iff f \text{ is a continuous bijection with continuous inverse } f^{-1}

A bicontinuous bijection — the topological notion of 'the same space,' since it preserves exactly the open-set structure in both directions

Homeomorphism (reference)

Notation

Notation	Meaning
$f^{-1}(V)$	The preimage of V under f: all x ∈ X with f(x) ∈ V
$C(X,Y)$	The set of all continuous functions from X to Y

Properties

Composition of continuous functions

f: X\to Y,\ g: Y \to Z \text{ continuous} \implies g \circ f: X \to Z \text{ continuous}

Example: Preimages compose: (g∘f)⁻¹(W) = f⁻¹(g⁻¹(W)), a preimage of a preimage of an open set, hence open.

Constant functions are continuous

f(x) = c \text{ for all } x \implies f \text{ is continuous}

Example: The preimage of any open V is either ∅ or all of X, and both are always open.

Closed-set characterization

f \text{ is continuous} \iff f^{-1}(C) \text{ is closed in } X \text{ for every closed } C \subseteq Y.

Condition: An equivalent dual formulation, since preimages commute with complements.

Topology-dependence

\text{Continuity depends on the topologies chosen on } X \text{ and } Y, \text{ not just the underlying function.}

Example: The identity map from (X, discrete) to (X, indiscrete) is always continuous, but the reverse identity map generally is not.

Worked Examples

Take a basic open set V = (a,b) in the codomain.
$f^{-1}((a,b)) = \left\{ x : a < 2x+1 < b \right\}$
Solve the inequality for x, which produces another open interval.
$f^{-1}((a,b)) = \left( \frac{a-1}{2}, \frac{b-1}{2} \right)$
Since preimages of basic open intervals are open intervals (hence open), and open sets are unions of such intervals, f is continuous.
$f^{-1}(\text{open}) \text{ is open}$

Answer: f(x) = 2x+1 is continuous, since preimages of open intervals are open intervals.

Practice Problems

Difficulty 5/10

A function f: X → Y between topological spaces is continuous if and only if:

f maps open sets to open setsthe preimage of every open set of Y is open in Xf is a bijectionf is bounded

Difficulty 6/10

Give an example of a continuous function whose image of an open set is not open.

Difficulty 7/10

Prove that if f: X → Y and g: Y → Z are continuous, then g∘f: X → Z is continuous.

Quiz

The topological (open-set) definition of continuity generalizes which classical definition?

The ε-δ definition of continuity in metric spacesThe definition of a derivativeThe definition of a limit of a sequence onlyThe definition of compactness

A homeomorphism is best described as:

Any continuous functionA continuous bijection whose inverse is also continuousAny bijection between topological spacesA function that preserves distances exactly

Summary

f: X → Y is continuous iff the preimage of every open set of Y is open in X — a definition using only open sets, no distance.
For metric spaces (including ℝ), this open-set definition is exactly equivalent to the classical ε-δ definition of continuity.
A homeomorphism is a continuous bijection with continuous inverse — the topological notion of two spaces being 'the same,' since it preserves open sets in both directions.

References

WebsiteWikipedia — Continuous function