A function can be perfectly well-defined by a simple rule and still destroy the shape of a space. The map that sends every rational number to $0$ and every irrational number to $1$ respects neither nearness nor passage to limits: a sequence of rationals and irrationals can approach the same real number while the function values jump between two separated points. Continuity is the topological condition that prevents this kind of tearing. It asks that whenever the target declares a set of acceptable outputs to be open, the corresponding set of inputs was already open in the source.

The point of the topological definition is that it separates the idea of continuity from formulas, derivatives, distances, and coordinates. On metric spaces it recovers the familiar $\varepsilon$-$\delta$ condition, but it also applies to quotient spaces, manifolds, function spaces, ordered spaces, and spaces built by gluing. The central question is not whether a graph can be drawn without lifting a pencil. The central question is: which maps respect the open-set structure that records observable local behaviour?

[example: A Jump Detected by an Open Set] Let $f:\mathbb R\to\mathbb R$ be given by \begin{align*} f(x) &= \begin{cases} 0, & x<0,\\ 1, & x\ge 0. \end{cases} \end{align*} We test continuity by pulling back the open interval $(1/2,3/2)$ in the codomain. For $x\in\mathbb R$, \begin{align*} x\in f^{-1}((1/2,3/2)) &\iff f(x)\in (1/2,3/2)\\ &\iff f(x)=1\\ &\iff x\ge 0. \end{align*} The middle equivalence holds because the only possible values of $f$ are $0$ and $1$, with \begin{align*} 0\notin (1/2,3/2), \qquad 1\in (1/2,3/2). \end{align*} Therefore \begin{align*} f^{-1}((1/2,3/2))&=[0,\infty). \end{align*} This set is not open in the usual topology on $\mathbb R$: if it were open, then since $0\in[0,\infty)$ there would be some $\varepsilon>0$ with \begin{align*} (-\varepsilon,\varepsilon)\subset [0,\infty). \end{align*} But $-\varepsilon/2\in(-\varepsilon,\varepsilon)$ and $-\varepsilon/2<0$, so $-\varepsilon/2\notin[0,\infty)$, a contradiction. Thus one [open set](/page/Open%20Set) in the target detects the jump at $0$, showing why continuity is tested by preimages of open sets. [/example]

The same test also explains why topology is the right language. A formula may look harmless, but the topology decides what counts as a small change. Changing the topology on the domain or codomain can turn the same underlying set-theoretic map from continuous to discontinuous, or from discontinuous to continuous. Continuous functions therefore belong to topology as much as to analysis.

Before defining a continuous function, we need the ambient objects on which continuity is measured. The essential data are not lengths or angles, but the collections of subsets declared open. Those collections encode what it means to test a point locally.

[definition: Topological Space] A [topological space](/page/Topological%20Space) is a pair $(X, \tau_X)$, where $X$ is a set and $\tau_X \subset \mathcal P(X)$ satisfies: \begin{align*} &\varnothing \in \tau_X, \qquad X \in \tau_X, \\ &\text{if } \mathcal U \subset \tau_X, \text{ then } \bigcup_{U \, \in \, \mathcal U} U \in \tau_X, \\ &\text{if } U_1,\ldots,U_n \in \tau_X, \text{ then } \bigcap_{i=1}^n U_i \in \tau_X. \end{align*} The elements of $\tau_X$ are called open sets. [/definition]

A topological space is therefore a set equipped with a chosen language of local tests. Once that language is fixed on the source and target, continuity asks whether a function respects those tests by pulling them back.

## Definition

A function between topological spaces should preserve what can be detected locally in the target. If an observer in $Y$ can restrict attention to an open region $V$, then a continuous map should make the corresponding condition on $X$ locally detectable as well. This leads to the preimage condition.

[definition: Continuous Function] Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces. A function $f: X \to Y$ is continuous if for every $V \in \tau_Y$, the preimage \begin{align*} f^{-1}(V) &= \{x \in X : f(x) \in V\} \end{align*} belongs to $\tau_X$. [/definition]

This definition pulls information backwards. It does not ask that images of open sets be open, because a map can collapse a whole interval to one point and still preserve every target-side test. The fastest way to separate the correct direction from the tempting wrong direction is to examine constant maps.

[example: Constant Maps] Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces, fix $y_0\in Y$, and define $c:X\to Y$ by $c(x)=y_0$ for every $x\in X$. We show that $c$ is continuous by checking the preimage of an arbitrary open set $V\in\tau_Y$. For each $x\in X$, \begin{align*} x\in c^{-1}(V) &\iff c(x)\in V \\ &\iff y_0\in V. \end{align*} If $y_0\in V$, then the last condition is true for every $x\in X$, so \begin{align*} c^{-1}(V)&=X. \end{align*} If $y_0\notin V$, then the last condition is false for every $x\in X$, so \begin{align*} c^{-1}(V)&=\varnothing. \end{align*} Thus \begin{align*} c^{-1}(V) &= \begin{cases} X, & y_0\in V,\\ \varnothing, & y_0\notin V. \end{cases} \end{align*} By the definition of a topological space, $X\in\tau_X$ and $\varnothing\in\tau_X$, so $c^{-1}(V)\in\tau_X$ for every $V\in\tau_Y$. Hence $c$ is continuous. This also shows why continuity is not about sending open sets to open sets: if $U\subset X$ is nonempty, choose $u\in U$, and then \begin{align*} c(U) &=\{c(x):x\in U\}\\ &=\{y_0\}, \end{align*} because $c(u)=y_0$ gives $y_0\in c(U)$ and every value $c(x)$ equals $y_0$. The singleton $\{y_0\}$ need not be open in $Y$. [/example]

Open sets are the native tests for continuity, but many arguments detect failure by trapping points in complementary constraints. The formal language of closed sets belongs just after the primary definition, because it gives the first alternative test for the same concept.

## Open and Closed Tests

Many continuity arguments are easier to run backwards through closed constraints than through open regions. To use that complementary language without ambiguity, we first name closed sets.

[definition: Closed Set] Let $(X,\tau_X)$ be a topological space. A subset $F \subset X$ is closed if $X \setminus F \in \tau_X$. [/definition]

The open-set definition is often inconvenient when the available information is phrased as membership in a closed constraint, such as a diagonal, a compact subset of $\mathbb R$, or the zero set of a function. The next theorem supplies the closed-set test needed in those situations, and it is also the version that connects most directly to compactness and Hausdorff separation.

[quotetheorem:1010]

The theorem says that continuity preserves both kinds of topological tests: open neighbourhood tests and closed obstruction tests. It also prepares the ground for compactness and convergence, where closed subsets often carry the usable information.

## Local Continuity and Neighbourhoods