A function can fail to be continuous for a reason visible at a single point. The graph may behave well on both sides, the formula may be simple, and every nearby value may be defined, yet the value assigned at one point may not match the values forced by its neighbours. Continuity at a point isolates this local question: if $x$ is forced close to $a$ inside the domain, is $f(x)$ forced close to $f(a)$? This page is a child of [Continuity](/page/Continuity). The parent concept usually asks whether a function is continuous throughout a domain. Here the unit of analysis is smaller. We focus on a fixed point $a$, learn how the epsilon-delta test works there, and then rebuild the global notion by checking every point. The definition looks modest, but it is the place where analysis learns to separate three ideas that are often blurred in intuition: the value $f(a)$, the nearby values $f(x)$, and the shape of the domain $E$ near $a$. A function may be continuous at an isolated point for a vacuous reason, discontinuous because of a jump, or continuous even when the domain has no interval around $a$. [example: A Removable Failure at One Point] Let $E = \mathbb R$, and define $f:E \to \mathbb R$ by \begin{align*} f(0)=1, \qquad f(x)=x \text{ for } x \ne 0. \end{align*} We show that $f$ is not continuous at $0$ by using the output tolerance $\varepsilon=1/2$. Let $\delta>0$ be arbitrary, set \begin{align*} \eta=\min\{\delta,1/2\}, \end{align*} and choose \begin{align*} x=\eta/2. \end{align*} Then $x \in E$, $x>0$, and since $\eta \le \delta$, \begin{align*} |x-0|=x=\eta/2<\eta \le \delta. \end{align*} Also $\eta \le 1/2$, so \begin{align*} 0<x=\eta/2 \le 1/4. \end{align*} Because $x \ne 0$, the defining rule gives $f(x)=x$, while $f(0)=1$. Hence \begin{align*} |f(x)-f(0)|=|x-1|. \end{align*} Since $0<x\le 1/4$, we have $x-1<0$, so $|x-1|=1-x$, and therefore \begin{align*} |f(x)-f(0)|=1-x \ge 1-\frac{1}{4}=\frac{3}{4}>\frac{1}{2}. \end{align*} Thus no matter how small $\delta$ is, there is a point of the domain within $\delta$ of $0$ whose function value is not within $1/2$ of $f(0)$, so $f$ is not continuous at $0$. The failure is caused only by the assigned value at $0$: if that value were changed to $0$, then the same formula would satisfy $|f(x)-f(0)|=|x|$, and choosing $\delta=\varepsilon$ would prove continuity at $0$. [/example] The example shows why pointwise continuity is not merely about the formula away from the point. The value at the point participates in the test. The nearby behaviour asks what $f(x)$ does as $x$ approaches $a$ through points of $E$, while the conclusion compares that behaviour to the already assigned value $f(a)$. ## Definition Before defining continuity at a point, we need the local language of being close inside a domain. The domain may be an interval, a punctured set, a finite set, or a more complicated subset of $\mathbb R$. The definition must therefore quantify only over those $x$ that actually belong to $E$. [definition: Continuity at a Point] Let $E \subset \mathbb R$, let $f: E \to \mathbb R$ be a function, and let $a \in E$. The function $f$ is continuous at $a$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that for every $x \in E$, \begin{align*} |x-a| < \delta \implies |f(x)-f(a)| < \varepsilon. \end{align*} [/definition] The quantifiers should be read in order. The tolerance $\varepsilon$ is chosen first by someone demanding a target accuracy in the output. The response is a radius $\delta$, which may depend on $\varepsilon$, on $a$, on $E$, and on $f$. Once $\delta$ is chosen, every point of the domain inside that radius must land inside the prescribed output window around $f(a)$. This page states the basic theory for real-valued functions on subsets of $\mathbb R$. Metric-space and topological versions use the same local logic with balls or neighbourhoods in place of absolute values, but those generalisations are not part of the defining convention used here. ## From Pointwise to Global Continuity Pointwise continuity tests one chosen point at a time, but many arguments need a statement about an entire domain. Without a separate global notion, a phrase like "[continuous function](/page/Continuous%20Function) on $E$" would be ambiguous: it could mean a formula behaves well at some points, or that the epsilon-delta test succeeds at every point of $E$. [definition: Continuity on a Set] Let $E \subset \mathbb R$ and let $f: E \to \mathbb R$ be a function. The function $f$ is continuous on $E$ if $f$ is continuous at every point $a \in E$. [/definition] Continuity on a set is therefore a universal statement over points. In practice, many proofs first establish continuity at an arbitrary point and then conclude continuity on the whole domain. Conversely, many discontinuity arguments locate a single point where the pointwise test fails. ## Domain Dependence A point can be uninteresting from the perspective of nearby behaviour if the domain has no other nearby points. We need a name for such points because they create automatic examples of pointwise continuity and warn us that continuity depends on the domain, not only on a formula. [definition: Isolated Point] Let $E \subset \mathbb R$ and let $a \in E$. The point $a$ is an isolated point of $E$ if there exists $r > 0$ such that \begin{align*} E \cap (a-r, a+r) = \{a\}. \end{align*} [/definition] At an isolated point, the continuity condition has no neighbouring domain points to test once $\delta$ is small enough. The next theorem records the resulting automatic continuity, which is essential when working with finite sets, sequences regarded as subspaces of $\mathbb R$, and piecewise domains. [quotetheorem:8493] This theorem is often the first warning that continuity is a property of a function together with its domain. The same formula may behave differently when its domain is changed, because the possible approaches to $a$ have changed. ## Quantifiers and Local Control The epsilon-delta definition is a game of local control. It does not ask that $f(x)$ equal $f(a)$ near $a$, nor that $f$ vary linearly, nor that there be a visible interval in the graph. It asks for a guarantee: every desired output accuracy can be enforced by a sufficiently small input restriction. ### Relative Neighbourhoods To avoid repeating the phrase "points sufficiently close to $a$ inside $E$", it is useful to name the local neighbourhood of $a$ relative to the domain. This language also connects the real-line definition to metric and topological versions of continuity. [definition: Relative Neighbourhood in $E$] Let $E \subset \mathbb R$ and let $a \in E$. A subset $N \subset E$ is a relative neighbourhood of $a$ in $E$ if there exists $r > 0$ such that \begin{align*} E \cap (a-r, a+r) \subset N. \end{align*} [/definition] With this language, continuity at $a$ says that the preimage under $f$ of every sufficiently small interval around $f(a)$ contains some relative neighbourhood of $a$. This phrasing is often the bridge to topology, where intervals are replaced by open sets. ### Basic Examples A first useful test case is a constant function. It demonstrates that the definition is about controlling variation, and in the constant case there is no variation to control. [example: Constant Functions at a Point] Let $E \subset \mathbb R$, let $c \in \mathbb R$, and define $f:E \to \mathbb R$ by $f(x)=c$ for every $x \in E$. Fix $a \in E$. We show that $f$ is continuous at $a$. Let $\varepsilon>0$ be given, and choose any $\delta>0$. If $x \in E$ and $|x-a|<\delta$, then the defining rule gives $f(x)=c$ and $f(a)=c$, so \begin{align*} |f(x)-f(a)|=|c-c|=|0|=0. \end{align*} Since $\varepsilon>0$, we have \begin{align*} 0<\varepsilon. \end{align*} Therefore \begin{align*} |f(x)-f(a)|<\varepsilon. \end{align*} Thus the epsilon-delta condition holds at the arbitrary point $a \in E$, so $f$ is continuous at every point of $E$. The reason is that the output error is identically zero, independent of how close $x$ is to $a$. [/example] The constant example is the easiest successful strategy: make the output difference identically zero. More often, the output difference is bounded by a simple expression involving $|x-a|$. Linear functions give the model computation. [example: A Linear Function] Let $m,b \in \mathbb R$, and define $f:\mathbb R \to \mathbb R$ by $f(x)=mx+b$. Fix $a \in \mathbb R$. We prove continuity at $a$ by controlling $|f(x)-f(a)|$ in terms of $|x-a|$. For any $x \in \mathbb R$, \begin{align*} f(x)-f(a)=(mx+b)-(ma+b). \end{align*} The constant terms cancel, so \begin{align*} (mx+b)-(ma+b)=mx-ma. \end{align*} Factoring out $m$ gives \begin{align*} mx-ma=m(x-a). \end{align*} Therefore, by multiplicativity of absolute value, \begin{align*} |f(x)-f(a)|=|m(x-a)|=|m|\,|x-a|. \end{align*} Let $\varepsilon>0$ be given. If $m=0$, choose any $\delta>0$. Then for every $x \in \mathbb R$ with $|x-a|<\delta$, \begin{align*} |f(x)-f(a)|=|0|\,|x-a|=0<\varepsilon. \end{align*} If $m \ne 0$, then $|m|>0$, so choose \begin{align*} \delta=\frac{\varepsilon}{|m|}. \end{align*} This $\delta$ is positive because both $\varepsilon$ and $|m|$ are positive. If $x \in \mathbb R$ and $|x-a|<\delta$, then \begin{align*} |f(x)-f(a)|=|m|\,|x-a|<|m|\delta. \end{align*} Substituting the chosen value of $\delta$ gives \begin{align*} |m|\delta=|m|\frac{\varepsilon}{|m|}=\varepsilon. \end{align*} Hence \begin{align*} |f(x)-f(a)|<\varepsilon. \end{align*} Thus the epsilon-delta condition holds at the arbitrary point $a \in \mathbb R$, so every affine function is continuous at every point of $\mathbb R$. [/example] ### Local Lipschitz Control The computation reveals a common proof pattern. To prove continuity at $a$, estimate $|f(x)-f(a)|$ by an expression that tends to $0$ as $x$ approaches $a$. The next theorem isolates the most common version of this pattern, where the output error is controlled by a constant multiple of the input error. [quotetheorem:10150] Local Lipschitz control is stronger than continuity, but it is common in computations. Differentiability, polynomial algebra, and many estimates in analysis produce exactly this kind of bound near a fixed point. ## Limits and the Value at the Point A limit describes what nearby values do as $x$ approaches $a$. Continuity at $a$ adds the requirement that this limiting behaviour agree with the value already assigned to the function at $a$. That distinction explains removable discontinuities and many examples from calculus. The next definition is not a replacement for continuity; it isolates the nearby behaviour so that continuity can be described as a matching condition. Since the domain may omit $a$ in limit problems, the definition uses points of $E$ different from $a$. [definition: Limit of a Function at a Point] Let $E \subset \mathbb R$, let $a \in \mathbb R$, let $L \in \mathbb R$, and let $f: E \to \mathbb R$ be a function. We say that $f(x)$ tends to $L$ as $x$ tends to $a$ through $E$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that for every $x \in E$, \begin{align*} 0 < |x-a| < \delta \implies |f(x)-L| < \varepsilon. \end{align*} [/definition] The difference between this definition and continuity at $a$ is the condition $0 < |x-a|$. A limit ignores the value at $a$; continuity tests it. The next theorem is needed to make this relationship exact in the setting where there is real approach behaviour to measure: away from isolated points, pointwise continuity is precisely the agreement between the punctured limit and the assigned value. [quotetheorem:10151] The non-isolated hypothesis is included for conceptual substance: it keeps the limit statement from becoming vacuous as an account of nearby behaviour. With the punctured-neighbourhood definition above, the same formal equivalence may still hold at isolated points, but there it says little beyond the fact that no nontrivial approach points exist. [example: Repairing a Removable Discontinuity] Let $E=\mathbb R$, and define $g:E\to\mathbb R$ by \begin{align*} g(0)=1,\qquad g(x)=\frac{\sin x}{x}\text{ for }x\ne 0. \end{align*} We show directly that $g$ is continuous at $0$. We use the standard sine estimate \begin{align*} |\sin x-x|\le \frac{|x|^3}{6} \end{align*} for all real $x$, which follows from [Taylor's theorem](/theorems/827) with remainder applied to $\sin x$ at $0$. Let $\varepsilon>0$ be given, and choose \begin{align*} \delta=\min\{1,\sqrt{6\varepsilon}\}. \end{align*} Then $\delta>0$. Suppose $x\in E$ and $|x-0|<\delta$. If $x=0$, then \begin{align*} |g(x)-g(0)|=|g(0)-g(0)|=0<\varepsilon. \end{align*} If $x\ne 0$, then $g(x)=\sin x/x$ and $g(0)=1$, so \begin{align*} |g(x)-g(0)|=\left|\frac{\sin x}{x}-1\right|. \end{align*} Combining the two terms over the denominator $x$ gives \begin{align*} \left|\frac{\sin x}{x}-1\right|=\left|\frac{\sin x-x}{x}\right|. \end{align*} Since $x\ne 0$, absolute values divide as \begin{align*} \left|\frac{\sin x-x}{x}\right|=\frac{|\sin x-x|}{|x|}. \end{align*} Using the sine estimate, \begin{align*} \frac{|\sin x-x|}{|x|}\le \frac{|x|^3/6}{|x|}. \end{align*} Because $|x|>0$, the right-hand side reduces to \begin{align*} \frac{|x|^3/6}{|x|}=\frac{|x|^2}{6}. \end{align*} From $|x|<\delta$ we get \begin{align*} \frac{|x|^2}{6}<\frac{\delta^2}{6}. \end{align*} Since $\delta\le \sqrt{6\varepsilon}$, we have \begin{align*} \frac{\delta^2}{6}\le \frac{(\sqrt{6\varepsilon})^2}{6}=\varepsilon. \end{align*} Therefore \begin{align*} |g(x)-g(0)|<\varepsilon. \end{align*} Thus $g$ is continuous at $0$. The assigned value $g(0)=1$ repairs the punctured formula because the nearby values $\sin x/x$ approach exactly $1$ as $x$ approaches $0$. [/example] The same punctured formula would fail if we assigned any value other than $1$ at $0$. The local behaviour of the surrounding points selects the only value compatible with continuity. [example: A Jump Discontinuity] Define $H:\mathbb R\to\mathbb R$ by $H(x)=0$ for $x<0$ and $H(x)=1$ for $x\ge 0$. We show that $H$ is not continuous at $0$ by taking the output tolerance $\varepsilon=1/2$. Let $\delta>0$ be arbitrary, and choose \begin{align*} x=-\frac{\delta}{2}. \end{align*} Then $x\in\mathbb R$ and $x<0$, so the defining rule gives \begin{align*} H(x)=0. \end{align*} Also $0\ge 0$, so \begin{align*} H(0)=1. \end{align*} The chosen point lies within the input window because \begin{align*} |x-0|=\left|-\frac{\delta}{2}\right|=\frac{\delta}{2}<\delta. \end{align*} But its output is not within $1/2$ of $H(0)$, since \begin{align*} |H(x)-H(0)|=|0-1|=|-1|=1\ge \frac{1}{2}. \end{align*} Thus for every $\delta>0$ there is a point $x\in\mathbb R$ with $|x-0|<\delta$ but $|H(x)-H(0)|\ge 1/2$, so $H$ is discontinuous at $0$. The failure is a jump: points immediately to the left of $0$ have value $0$, while the value at $0$ is $1$. [/example] A jump cannot be repaired by changing one value if the two sides force different limiting behaviours. This distinguishes removable failures from genuine one-sided disagreement. ## Sequential Tests Epsilon-delta proofs give direct control, but sequences often expose failure more quickly. A function is continuous at $a$ exactly when it respects every sequence from the domain converging to $a$. This turns a local quantifier statement into a convergence-preservation statement. To state this cleanly, we recall the local notion of convergence inside a subset of the real line. The sequence must stay in the domain because the function is only defined there. [definition: Sequence Converging to a Point Through a Set] Let $E \subset \mathbb R$ and let $a \in \mathbb R$. A sequence $(x_n)_{n=1}^{\infty}$ in $E$ converges to $a$ through $E$ if \begin{align*} \lim_{n \to \infty} x_n = a. \end{align*} [/definition] This definition has no continuity in it; it only records the allowed approaches. Once the approaches are named, we need a theorem that converts the epsilon-delta test into convergence preservation for every such approach. [quotetheorem:268] The sequential characterisation is especially useful for disproving continuity. To show failure at $a$, it suffices to find one sequence $x_n \in E$ with $x_n \to a$ for which $f(x_n)$ does not converge to $f(a)$. [example: Oscillation Near the Point] Define $f:\mathbb R\to\mathbb R$ by $f(0)=0$ and $f(x)=\sin(1/x)$ for $x\ne 0$. We use two sequences approaching $0$ whose images stay at different values. For $n\in\mathbb N$, set \begin{align*} x_n=\frac{1}{\frac{\pi}{2}+2\pi n}. \end{align*} Since $\frac{\pi}{2}+2\pi n\ge 2\pi n$, we have \begin{align*} 0<x_n\le \frac{1}{2\pi n}. \end{align*} Thus $x_n\to 0$. Also $x_n\ne 0$, so \begin{align*} f(x_n)=\sin\left(\frac{1}{x_n}\right). \end{align*} By the definition of $x_n$, \begin{align*} \frac{1}{x_n}=\frac{\pi}{2}+2\pi n. \end{align*} Using the $2\pi$-periodicity of sine, \begin{align*} f(x_n)=\sin\left(\frac{\pi}{2}+2\pi n\right)=\sin\left(\frac{\pi}{2}\right)=1. \end{align*} Now set \begin{align*} y_n=\frac{1}{\frac{3\pi}{2}+2\pi n}. \end{align*} Since $\frac{3\pi}{2}+2\pi n\ge 2\pi n$, we have \begin{align*} 0<y_n\le \frac{1}{2\pi n}, \end{align*} so $y_n\to 0$. Again $y_n\ne 0$, and \begin{align*} \frac{1}{y_n}=\frac{3\pi}{2}+2\pi n. \end{align*} Therefore \begin{align*} f(y_n)=\sin\left(\frac{3\pi}{2}+2\pi n\right)=\sin\left(\frac{3\pi}{2}\right)=-1. \end{align*} The sequence $(x_n)$ converges to $0$, but $(f(x_n))$ is the constant sequence $1$, so it converges to $1$ rather than to $f(0)=0$. By *[Sequential Characterisation of Continuity](/theorems/268) at a Point*, $f$ is not continuous at $0$. The second sequence $(y_n)$ shows the oscillation explicitly: points also approach $0$ along which the function values are constantly $-1$. [/example] This example shows a failure different from a jump. The problem is not a single wrong assigned value, since the surrounding values do not settle toward any one number. The next theorem records the positive version of the same idea, which is the practical rule used when substituting limits into continuous functions. [quotetheorem:10152] The theorem is the operational form of pointwise continuity. It is what justifies substitutions such as replacing $x_n$ by its limit inside a continuous expression. ## Algebra of Continuous Functions at a Point Continuity at a point is stable under the usual algebraic operations. This is why complicated elementary functions can be built from simpler continuous pieces rather than proved from scratch every time. The first operation to isolate is composition. It answers a natural question: if $g$ sends points near $a$ to points near $g(a)$, and $f$ behaves continuously near $g(a)$, does $f \circ g$ behave continuously near $a$? [definition: Composition of Functions] Let $E,F \subset \mathbb R$, let $g: E \to F$, and let $f: F \to \mathbb R$. The composition $f \circ g: E \to \mathbb R$ is the function defined by \begin{align*} (f \circ g)(x) = f(g(x)) \quad \text{for every } x \in E. \end{align*} [/definition] Composition is where the local nature of continuity matters. The main obstruction is that the tolerance for the outer function $f$ is imposed near $g(a)$, not directly near $a$. Continuity of $g$ is what converts closeness to $a$ into closeness to $g(a)$, allowing the two local controls to be chained. [quotetheorem:10153] This theorem lets us assemble continuous functions modularly. Polynomial and rational examples also require closure under sums, products, and quotients, so we need the following algebra rules to avoid reproving epsilon-delta estimates for every expression. [quotetheorem:10154] The quotient clause includes a local nonzero conclusion. Without it, division might not even define a function near the point. Continuity of $g$ at $a$ and the condition $g(a) \ne 0$ prevent nearby denominator values from crossing zero after the neighbourhood is made small enough. [example: A Rational Function at a Point] Define $r: \mathbb R \setminus \{1\} \to \mathbb R$ by \begin{align*} r(x)=\frac{x^2+1}{x-1}. \end{align*} We verify continuity at $a=2$ directly. First, \begin{align*} r(2)=\frac{2^2+1}{2-1}=\frac{5}{1}=5. \end{align*} For $x\ne 1$, \begin{align*} r(x)-r(2)=\frac{x^2+1}{x-1}-5. \end{align*} Putting the two terms over the common denominator $x-1$ gives \begin{align*} \frac{x^2+1}{x-1}-5=\frac{x^2+1-5(x-1)}{x-1}. \end{align*} Expanding the numerator, \begin{align*} x^2+1-5(x-1)=x^2+1-5x+5=x^2-5x+6. \end{align*} Factoring, \begin{align*} x^2-5x+6=(x-2)(x-3). \end{align*} Hence \begin{align*} |r(x)-r(2)|=\left|\frac{(x-2)(x-3)}{x-1}\right|=\frac{|x-2|\,|x-3|}{|x-1|}. \end{align*} Let $\varepsilon>0$ be given, and choose \begin{align*} \delta=\min\left\{\frac{1}{2},\frac{\varepsilon}{3}\right\}. \end{align*} If $x\in \mathbb R\setminus\{1\}$ and $|x-2|<\delta$, then $|x-2|<1/2$. Therefore \begin{align*} |x-1|=|(x-2)+1|\ge 1-|x-2|>1-\frac{1}{2}=\frac{1}{2}. \end{align*} Also, \begin{align*} |x-3|=|(x-2)-1|\le |x-2|+1<\frac{1}{2}+1=\frac{3}{2}. \end{align*} Substituting these bounds into the expression for $|r(x)-r(2)|$ gives \begin{align*} |r(x)-r(2)|<\frac{|x-2|(3/2)}{1/2}=3|x-2|. \end{align*} Since $|x-2|<\delta\le \varepsilon/3$, \begin{align*} 3|x-2|<3\cdot \frac{\varepsilon}{3}=\varepsilon. \end{align*} Thus $|r(x)-r(2)|<\varepsilon$, so $r$ is continuous at $2$. At $a=1$ there is no pointwise continuity question for this function, because $1\notin \mathbb R\setminus\{1\}$; continuity at a point is only defined for points in the domain. [/example] The last sentence is important: a function cannot be continuous or discontinuous at a point outside its domain. Many apparent singularities are not pointwise failures until a value is assigned at the missing point. ## Domains, One-Sided Behaviour, and Subspace Effects ### Endpoints and Missing Sides Continuity at a point only tests points that belong to the domain. On a half-interval, an endpoint has only one side available. On a sparse set, sequences may approach in restricted ways. This is why the same formula can have different continuity properties on different domains. The endpoint phenomenon is common enough to name. It lets us describe continuity on intervals such as $[0,1]$ without pretending that points less than $0$ are part of the problem. [definition: Right Continuity at a Point] Let $E \subset \mathbb R$, let $f: E \to \mathbb R$, and let $a \in E$. The function $f$ is right continuous at $a$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that for every $x \in E$, \begin{align*} 0 \le x-a < \delta \implies |f(x)-f(a)| < \varepsilon. \end{align*} [/definition] A one-sided endpoint can also face the opposite direction. To treat right endpoints of intervals and left-hand limits with the same precision, we need the left-sided analogue of the preceding condition. [definition: Left Continuity at a Point] Let $E \subset \mathbb R$, let $f: E \to \mathbb R$, and let $a \in E$. The function $f$ is left continuous at $a$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that for every $x \in E$, \begin{align*} 0 \le a-x < \delta \implies |f(x)-f(a)| < \varepsilon. \end{align*} [/definition] Left and right continuity become a pair of tests when the domain has points on both sides of $a$. The possible obstruction is that a full neighbourhood of $a$ contains points approaching from both directions, so controlling only one side may miss half of the domain near $a$. When both directional controls are available, the smaller of the two input radii controls the whole neighbourhood. [quotetheorem:10155] The theorem is most useful when $E$ contains points on both sides of $a$, but the statement remains valid in restricted domains. It tells us that the full neighbourhood test can be assembled from the two directional tests. [example: Endpoint Continuity on a Closed Interval] Define $f:[0,1]\to\mathbb R$ by $f(x)=\sqrt{x}$. We show that $f$ is continuous at the endpoint $0$, where the only domain points tested are points $x\in[0,1]$. Let $\varepsilon>0$ be given, and choose \begin{align*} \delta=\varepsilon^2. \end{align*} Since $\varepsilon>0$, we have $\delta>0$. Suppose $x\in[0,1]$ and $|x-0|<\delta$. Because $x\in[0,1]$, we have $x\ge 0$, so \begin{align*} |x-0|=|x|=x. \end{align*} Thus \begin{align*} 0\le x<\delta=\varepsilon^2. \end{align*} Also $f(0)=\sqrt{0}=0$, and therefore \begin{align*} |f(x)-f(0)|=|\sqrt{x}-0|=|\sqrt{x}|. \end{align*} Since $x\ge 0$, the number $\sqrt{x}$ is nonnegative, so \begin{align*} |\sqrt{x}|=\sqrt{x}. \end{align*} From $0\le x<\varepsilon^2$ and $\varepsilon>0$, taking square roots gives \begin{align*} \sqrt{x}<\sqrt{\varepsilon^2}=\varepsilon. \end{align*} Hence \begin{align*} |f(x)-f(0)|<\varepsilon. \end{align*} Therefore $f$ is continuous at $0$ as a function on $[0,1]$. The point is an endpoint effect: the continuity test only uses points of the domain, so no values to the left of $0$ are required. [/example] The domain is not a technical footnote here. It determines which nearby points are eligible for the continuity test. ### Sparse Domains A second domain effect occurs when the domain is dense or sparse. The next example shows that restricting a discontinuous-looking formula to a smaller domain can change the question completely. [example: A Function Continuous on a Sparse Domain] Let $E=\{0\}\cup\{1/n:n\in\mathbb N\}$, and define $f:E\to\mathbb R$ by $f(x)=x$. First, $0$ is not isolated in $E$: if $r>0$, choose $n\in\mathbb N$ with $n>1/r$. Then \begin{align*} 0<\frac{1}{n}<r \end{align*} so $1/n\in E\cap(-r,r)$ and $1/n\ne 0$. We prove continuity at $0$. Since $f(0)=0$, for every $x\in E$ we have \begin{align*} |f(x)-f(0)|=|x-0|=|x|. \end{align*} Let $\varepsilon>0$ be given, and choose \begin{align*} \delta=\varepsilon. \end{align*} If $x\in E$ and $|x-0|<\delta$, then \begin{align*} |f(x)-f(0)|=|x|=|x-0|<\delta=\varepsilon. \end{align*} Thus $f$ is continuous at $0$. Now fix $n\in\mathbb N$. We show that $1/n$ is isolated in $E$. Choose \begin{align*} r_n=\frac{1}{2n(n+1)}. \end{align*} The distance from $1/n$ to $0$ is \begin{align*} \left|\frac{1}{n}-0\right|=\frac{1}{n}>\frac{1}{2n(n+1)}=r_n. \end{align*} If $m\in\mathbb N$ and $m\ne n$, then the closest reciprocal to $1/n$ occurs at $1/(n+1)$ on the right-hand side of the sequence, and in all cases \begin{align*} \left|\frac{1}{m}-\frac{1}{n}\right|\ge \frac{1}{n(n+1)}>r_n. \end{align*} Therefore \begin{align*} E\cap\left(\frac{1}{n}-r_n,\frac{1}{n}+r_n\right)=\left\{\frac{1}{n}\right\}. \end{align*} So $1/n$ is an isolated point of $E$. Finally, let $\varepsilon>0$ be given and choose $\delta=r_n$. If $x\in E$ and \begin{align*} \left|x-\frac{1}{n}\right|<\delta, \end{align*} then the isolation just proved forces $x=1/n$. Hence \begin{align*} \left|f(x)-f\left(\frac{1}{n}\right)\right|=\left|\frac{1}{n}-\frac{1}{n}\right|=0<\varepsilon. \end{align*} Thus $f$ is continuous at every point $1/n$ as well as at $0$, so $f$ is continuous on the sparse domain $E$. [/example] This example is a compact model for subspace thinking. Continuity at a point is always relative to the available approaches in the domain. ## Topological Reformulation The epsilon-delta definition belongs to metric spaces, but the idea behind it is topological: inverse images of neighbourhoods of $f(a)$ should contain neighbourhoods of $a$. The pointwise version is the local form of the open-set definition of continuity. To state the reformulation, we need the preimage of a set. This construction turns output conditions into input conditions, which is exactly what the epsilon-delta definition does. [definition: Preimage] Let $A$ and $B$ be sets, let $f: A \to B$ be a function, and let $S \subset B$. The preimage of $S$ under $f$ is \begin{align*} f^{-1}(S) = \{x \in A : f(x) \in S\}. \end{align*} [/definition] Continuity at $a$ can now be phrased without naming absolute values explicitly. The issue is whether every allowed output neighbourhood around $f(a)$ pulls back to an input condition that already contains some relative neighbourhood of $a$. This is the same local control as epsilon-delta, but expressed through preimages rather than inequalities. [quotetheorem:10156] This theorem is the pointwise ancestor of the global topological statement: a function is continuous when preimages of open sets are open in the appropriate relative topology. The pointwise version is often more flexible in calculations, while the open-set version is better for structural arguments. [example: The Neighbourhood Test for $x^2$ at $3$] Define $f:\mathbb R\to\mathbb R$ by $f(x)=x^2$, and let $a=3$. For the output neighbourhood $V=(8,10)$, we have \begin{align*} f(a)=f(3)=3^2=9 \end{align*} and hence $f(a)\in V$ because $8<9<10$. We compute an input neighbourhood around $3$ whose image lies in $V$. Since \begin{align*} (\sqrt{8})^2=8<9=3^2<10=(\sqrt{10})^2 \end{align*} and all three numbers are positive, we have $\sqrt{8}<3<\sqrt{10}$. Therefore \begin{align*} 3-\sqrt{8}>0 \end{align*} and \begin{align*} \sqrt{10}-3>0. \end{align*} Choose \begin{align*} r=\min\{3-\sqrt{8},\sqrt{10}-3\}. \end{align*} Then $r>0$, $r\le 3-\sqrt{8}$, and $r\le \sqrt{10}-3$. Now suppose $x\in(3-r,3+r)$. Then \begin{align*} 3-r<x<3+r. \end{align*} From $r\le 3-\sqrt{8}$, we get \begin{align*} 3-r\ge \sqrt{8}, \end{align*} so $x>\sqrt{8}$. From $r\le \sqrt{10}-3$, we get \begin{align*} 3+r\le \sqrt{10}, \end{align*} so $x<\sqrt{10}$. Hence \begin{align*} \sqrt{8}<x<\sqrt{10}. \end{align*} The lower inequality gives \begin{align*} x^2-8=x^2-(\sqrt{8})^2=(x-\sqrt{8})(x+\sqrt{8})>0, \end{align*} so $x^2>8$. The upper inequality gives \begin{align*} 10-x^2=(\sqrt{10})^2-x^2=(\sqrt{10}-x)(\sqrt{10}+x)>0, \end{align*} so $x^2<10$. Therefore $x^2\in(8,10)$, which means $x\in f^{-1}(V)$. Thus \begin{align*} (3-r,3+r)\subset f^{-1}(V). \end{align*} The output interval $(8,10)$ around $f(3)=9$ therefore pulls back to an input interval around $3$, exactly as the neighbourhood formulation of continuity predicts. [/example] The example is not meant as an efficient proof that $x^2$ is continuous. It shows how the geometric idea of pulling back neighbourhoods matches the epsilon-delta calculation. ## Failure Modes and Diagnostics A failed continuity test has a precise structure. There is some output tolerance $\varepsilon > 0$ such that no matter how small the input window is, a domain point inside that window escapes the output window around $f(a)$. This negation is often the cleanest way to diagnose discontinuity. It is useful to record the negated form as a definition-like diagnostic, but it is not a new kind of continuity. It is the formal failure of the original pointwise condition. [definition: Discontinuity at a Point] Let $E \subset \mathbb R$, let $f: E \to \mathbb R$ be a function, and let $a \in E$. The function $f$ is discontinuous at $a$ if there exists $\varepsilon > 0$ such that for every $\delta > 0$ there exists $x \in E$ with \begin{align*} |x-a| < \delta \quad \text{and} \quad |f(x)-f(a)| \ge \varepsilon. \end{align*} [/definition] This formulation is useful because it tells us exactly what evidence a counterexample must provide. Directly producing a new escaping point for every possible $δ$ can be awkward, especially when the bad behaviour is oscillatory or dense. The next diagnostic question is whether those infinitely many local failures can be organized into a single witness. A sequence approaching $a$ is designed for exactly this purpose: it tests the function along points that get arbitrarily close to $a$, while recording whether the corresponding values actually approach $f(a)$. [quotetheorem:10157] The witness sequence is often easier to find than a direct epsilon. Oscillation, jumps, and incompatible dense subsets all become visible through carefully chosen sequences. [example: Dirichlet-Type Failure at Every Point] Define $f:\mathbb R\to\mathbb R$ by \begin{align*} f(x)=1 \text{ for } x\in\mathbb Q,\qquad f(x)=0 \text{ for } x\in\mathbb R\setminus\mathbb Q. \end{align*} We show that $f$ is discontinuous at an arbitrary point $a\in\mathbb R$ by using the output tolerance $\varepsilon=1/2$. First suppose $a\in\mathbb Q$. Let $\delta>0$ be given. Choose $n\in\mathbb N$ such that \begin{align*} \frac{\sqrt{2}}{n}<\delta. \end{align*} Set \begin{align*} x=a+\frac{\sqrt{2}}{n}. \end{align*} Since $a$ is rational and $\sqrt{2}/n$ is irrational, $x$ is irrational. Hence $f(x)=0$, while $f(a)=1$. Also \begin{align*} |x-a|=\left|a+\frac{\sqrt{2}}{n}-a\right|=\left|\frac{\sqrt{2}}{n}\right|=\frac{\sqrt{2}}{n}<\delta. \end{align*} The output difference is \begin{align*} |f(x)-f(a)|=|0-1|=|-1|=1\ge \frac{1}{2}. \end{align*} Now suppose $a\notin\mathbb Q$. Let $\delta>0$ be given. Choose $n\in\mathbb N$ such that \begin{align*} \frac{1}{n}<\delta. \end{align*} Let $k=\lfloor na\rfloor+1$ and set \begin{align*} x=\frac{k}{n}. \end{align*} Then $x\in\mathbb Q$. Because $\lfloor na\rfloor\le na<\lfloor na\rfloor+1=k$, we have \begin{align*} 0<k-na\le 1. \end{align*} Dividing by the positive number $n$ gives \begin{align*} 0<\frac{k}{n}-a\le \frac{1}{n}<\delta. \end{align*} Thus \begin{align*} |x-a|=\left|\frac{k}{n}-a\right|=\frac{k}{n}-a<\delta. \end{align*} Since $x$ is rational and $a$ is irrational, $f(x)=1$ and $f(a)=0$, so \begin{align*} |f(x)-f(a)|=|1-0|=|1|=1\ge \frac{1}{2}. \end{align*} In both cases, every input window around $a$ contains a point whose output is at distance at least $1/2$ from $f(a)$. Therefore $f$ is discontinuous at $a$, and since $a\in\mathbb R$ was arbitrary, $f$ is discontinuous at every point of $\mathbb R$. [/example] This example is the sharpest reminder that continuity is about all nearby domain points, not just a convenient family of approaches. Dense interlacing of two behaviours prevents local control at every scale. Not every discontinuity is so severe. Some failures are removable, some are jumps, and some are oscillatory. The pointwise framework does not require a taxonomy, but it gives a common test for all of them. [remark: Common Discontinuity Patterns] Let $E \subset \mathbb R$, let $f: E \to \mathbb R$, and let $a \in E$ be a non-isolated point. A discontinuity at $a$ may occur because the punctured limit exists but differs from $f(a)$, because different one-sided limits disagree, or because nearby values do not approach any single limiting value. [/remark] The remark is a guide for diagnosis rather than a classification theorem. In a first analysis course, these three patterns account for most examples and help the reader decide what evidence to look for. ## Beyond and Connected Topics Pointwise continuity is the local atom from which global continuity is built. The parent page [Continuity](/page/Continuity) studies what happens when the same condition is imposed at every point of a domain, including the open-set viewpoint and standard closure properties. [Uniform continuity](/page/Uniform%20Continuity) changes the order of dependence in the epsilon-delta statement. For continuity at a point, $\delta$ may depend on $a$. For uniform continuity on a set, one $\delta$ must work for all points at once. This distinction becomes decisive on non-compact domains and leads naturally to compactness arguments in [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Differentiability is a stronger local condition. If a real function is differentiable at $a$, then it is continuous at $a$, but continuity does not imply differentiability. The absolute value function at $0$ is the standard example: it is continuous at $0$ but has no derivative there. In metric spaces, the same pointwise idea is written using the metric $d$ instead of absolute value. In topological spaces, neighbourhoods replace metric balls entirely. This abstraction preserves the local logic while allowing continuity to apply to spaces of functions, manifolds, and other non-Euclidean objects. In complex analysis, continuity at a point remains a prerequisite for stronger notions such as [complex differentiability](/page/Complex%20Differentiability) and holomorphicity. The page [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis) develops this direction, where local control interacts with [power series](/page/Power%20Series) and contour integrals. In analysis of functions, pointwise continuity interacts with sequences, compactness, [uniform convergence](/page/Uniform%20Convergence), and integration. The notes [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions) are a natural continuation after the basic pointwise theory is comfortable. ## References Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Androma, [Continuity](/page/Continuity). Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). Androma, [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). T. M. Apostol, *Mathematical Analysis* (1974). W. Rudin, *Principles of Mathematical Analysis* (1976). S. Abbott, *Understanding Analysis* (2015).