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    Continuity

    For metric spaces \(X\), \(Y\) and \(f\colon X\to Y\), the following are equivalent: for every \(x\in X\) and \(\varepsilon>0\) there exists \(\delta>0\) with \(d(f(x),f(y))<\varepsilon\) whenever \(d(x,y)<\delta\); \(f(x_n)\to f(x)\) whenever \(x_n\to x\); the preimage of every open set is open; the preimage of every closed set is closed. If \(X\) is compact and \(f\) is continuous, then \(f\) is uniformly continuous, \(f(X)\) is compact, and for \(Y=\RR\), \(f\) attains its maximum and minimum. The functions \(x\mapsto x^2\) on \(\RR\) and \(x\mapsto1/x\) on \((0,1)\) are continuous and not uniformly continuous.

    Continuity and uniform continuity

    Let \(X\subseteq\RR\) and let \(f\colon X\to \RR\).

    • The function \(f\) is continuous on \(X\) if \[ \forall x_0\in X \quad \forall \varepsilon>0 \quad \exists \delta=\delta(\varepsilon, x_0)>0 \quad \forall y\in X: \quad \abs{x_0 - y} < \delta \implies \abs{f(x_0) - f(y)} < \varepsilon. \]

    • The function \(f\) is uniformly continuous on \(X\) if \[ \forall \varepsilon>0 \quad \exists \delta=\delta(\varepsilon)>0 \quad \forall x, y \in X: \quad \abs{x - y} < \delta \implies \abs{f(x) - f(y)} < \varepsilon. \]

    Substituting \(y = x_0 + h\), the two conditions read \[ \begin{aligned} &\forall x_0\in X \quad \forall \varepsilon>0 \quad \exists \delta(\varepsilon, x_0)>0 \quad \forall h \text{ with } x_0+h\in X: && \abs{h} < \delta \implies \abs{f(x_0) - f(x_0 + h)} < \varepsilon, \\ &\forall \varepsilon>0 \quad \exists \delta(\varepsilon)>0 \quad \forall x\in X \quad \forall h \text{ with } x+h\in X: && \abs{h} < \delta \implies \abs{f(x) - f(x + h)} < \varepsilon. \end{aligned} \] In the first, \(\delta\) may depend on the point \(x_0\); in the second, one \(\delta\) serves every point of \(X\).

    Equivalent characterizations of continuity

    On Euclidean subspaces, continuity is equivalently topological, epsilon--delta, sequential, or open-preimage continuity.Suggest a slogan

    Let \(E\subseteq\RR^n\) and let \(f\colon E\to\RR^m\). The following are equivalent:

    1. \(f\) is continuous on \(E\), with the subspace topology on \(E\).

    2. For every \(x\in E\) and every \(\varepsilon>0\) there exists \(\delta>0\) such that \(\norm{f(y)-f(x)}<\varepsilon\) for all \(y\in E\) with \(\norm{x-y}<\delta\).

    3. For every sequence \((x_n)_{n\geq1}\) in \(E\) converging to a point \(x\in E\), \(\lim_{n\to\infty}f(x_n)=f(x)\).

    4. For every open set \(G\subseteq\RR^m\), the set \(\inverseof{f}(G)\) is open in \(E\) with the subspace topology. [@Rud76].

    Continuous images of compact sets and the extreme value theorem

    Continuous maps preserve compactness, so continuous real-valued functions on compact sets attain their extrema.Suggest a slogan

    Let \(K\subseteq\RR^n\) be compact and let \(f\colon K\to\RR^m\) be continuous. Then \(f(K)\) is compact. In particular, if \(a<b\) and \(f\colon[a,b]\to\RR\) is continuous, then there exist \(p,q\in[a,b]\) such that \[ f(p)=\sup_{x\in[a,b]}f(x)\qquad\text{and}\qquad f(q)=\inf_{x\in[a,b]}f(x). \] [@Rud76].

    Continuous functions on compact sets are uniformly continuous

    Compactness upgrades continuity to uniform continuity.Suggest a slogan

    Let \(K\subseteq\RR^n\) be compact and let \(f\colon K\to\RR^m\) be continuous. Then \(f\) is uniformly continuous on \(K\). [@Rud76].

    Intermediate value theorem

    A continuous real function on an interval takes every value between its endpoint values.Suggest a slogan

    Let \(a<b\) and let \(f\colon[a,b]\to\RR\) be continuous with \(f(a)<f(b)\). For every \(y\in\RR\) with \(f(a)<y<f(b)\) there exists \(c\in(a,b)\) such that \(f(c)=y\). [@Rud76].

    Discontinuities of monotone functions

    Monotone functions have only countably many discontinuities, and every one is a jump.Suggest a slogan

    Let \(a<b\) and let \(f\colon(a,b)\to\RR\) be monotone. Then the set \(E\subseteq(a,b)\) of points at which \(f\) is discontinuous is at most countable. Moreover, every \(x\in E\) is a jump discontinuity: the one-sided limits \[ f(x-)\coloneqq\lim_{t\to x^-}f(t)\qquad\text{and}\qquad f(x+)\coloneqq\lim_{t\to x^+}f(t) \] both exist. [@Rud76].

    Heine--Cantor theorem

    A continuous map from a compact metric space to a metric space is uniformly continuous.Suggest a slogan

    Let \(X\) and \(Y\) be metric spaces with \(X\) compact. Every continuous function \(f\colon X\to Y\) is uniformly continuous [@Mun00]; [@Rud76].

    Let \(U\subseteq\RR^n\) and let \(f\colon U\to \RR\) be continuous. For every compact \(K \subseteq U\), the restriction \(\ro{f}{K}\) is uniformly continuous.