Continuity on Intervals and Sets

Continuity on a set is the precise hypothesis behind essentially every existence theorem in single-variable analysis. The Intermediate Value Theorem requires exactly "$f$ continuous on $[a,b]$" in the relative sense proven here — the endpoint values $f(a), f(b)$ must genuinely be the right one-sided limits, or the theorem's proof (via the least-upper-bound of the set of points where $f$ stays negative) breaks down. The Extreme Value Theorem similarly requires continuity on a closed, bounded interval; the closedness is used precisely so that endpoint continuity, as defined here via one-sided limits, is part of the hypothesis, ruling out functions like $f(x)=1/x$ on $(0,1]$ which blow up as the excluded endpoint is approached. Uniform Continuity takes this article's local, point-by-point definition and strengthens it to a single global $\delta$, a distinction essential to proving that continuous functions on compact sets are Riemann integrable. In numerical analysis, root-finding algorithms such as bisection depend on continuity on a closed interval including verified endpoint behavior. In probability theory, cumulative distribution functions are typically only right-continuous, a one-sided continuity notion structurally identical to the endpoint definition given here. In complex analysis and topology, the relatively-open-set characterization proven above generalizes verbatim to continuous maps between arbitrary topological spaces, making it the bridge between elementary $\varepsilon$–$\delta$ analysis and general topology.

:::quiz[Which is the correct condition for $f:[a,b]\to\mathbb{R}$ to be continuous at the endpoint $a$?]

• $\lim_{x\to a} f(x) = f(a)$, two-sided
• $\lim_{x\to a^+} f(x) = f(a)$ ✓
• $\lim_{x\to a^-} f(x) = f(a)$
• $f$ must be differentiable at $a$

:::quiz[Let $E=\{0\}\cup[1,2]$ and $f(0)=100$, $f(x)=x$ on $[1,2]$. Is $f$ continuous on $E$?]

• No, because $f(0)=100$ is far from nearby values
• No, because $E$ is not an interval
• Yes, because $0$ is an isolated point of $E$ so continuity there is automatic ✓
• Yes, but only because $100$ happens to be an integer

:::quiz[Why does the counterexample $E=[0,1]\cup[2,3]$ with $f=0$ on $[0,1]$, $f=1$ on $[2,3]$ NOT contradict the Intermediate Value Theorem?]

• It does contradict it; the theorem is simply false in this generality
• The IVT requires $E$ to be a single interval, and $[0,1]\cup[2,3]$ is not an interval ✓
• The IVT only applies to polynomials
• $f$ is not actually continuous on $E$ in this example

:::faq[Why can't we just use the ordinary two-sided limit definition at every point of $[a,b]$, including the endpoints?] Because at $x=a$, there are no points of the domain $[a,b]$ to the left of $a$, so the two-sided limit $\lim_{x\to a} f(x)$ is not defined at all — it would require values of $f$ at points less than $a$, which simply don't exist. The $\varepsilon$–$\delta$ definition, honestly interpreted, only ever quantifies over $x$ actually in the domain, so at $a$ the "$x\in E$" restriction silently forces $x > a$ (aside from $x=a$ itself), which is precisely the definition of the right-hand limit. This is not an ad hoc patch; it's what the general relative definition produces automatically once you take the "$x\in E$" clause seriously, as proved in the Corollary above.

:::faq[If $f$ is continuous on $(a,b)$, does that automatically make it continuous on $[a,b]$ once we define $f(a)$ and $f(b)$ suitably?] Only if the one-sided limits actually exist and you set the endpoint values equal to them. Continuity on $(a,b)$ says nothing by itself about whether $\lim_{x\to a^+}f(x)$ exists; for example $f(x)=\sin(1/x)$ is continuous on $(0,1)$ but has no limit as $x\to 0^+$, so no choice of $f(0)$ makes the extension continuous on $[0,1)$. When the one-sided limits do exist, defining $f(a):=\lim_{x\to a^+}f(x)$ and $f(b):=\lim_{x\to b^-}f(x)$ is exactly what's needed, by the Corollary proven above, to make the extended function continuous on the closed interval.

Ask the tutor: Why does the Extreme Value Theorem fail if the closed interval is replaced by an open one, even though the function is continuous on the whole open interval?

Think of continuity at $c$ as a promise: make $x$ close enough to $c$, and I will make $f(x)$ close to $f(c)$. The subtlety is the phrase "close enough" — close enough among which candidate $x$'s? If $f$ is only defined on $E$, it is meaningless to demand the promise hold for $x$ outside $E$; there is no $f(x)$ to control. So the promise is always tested using $x \in E$ only. This is not a weakening of continuity for the sake of convenience — it is the only logically coherent way to state the definition when the domain is not all of $\mathbb{R}$. At an interior point of an interval, $E$ surrounds $c$ symmetrically, so "close enough within $E$" is indistinguishable from "close enough in $\mathbb{R}$": the relative and absolute notions agree. At an endpoint like $a$ in $E=[a,b]$, the set $E$ only extends to the right of $a$; there is no left side to test, so the $\varepsilon$–$\delta$ condition automatically becomes a statement purely about the right-hand behavior — recovering exactly the one-sided limit condition $\lim_{x\to a^+}f(x)=f(a)$. A second, easily overlooked feature: continuity on a set is a pointwise property checked separately at each point, possibly with a different $\delta$ at each $c$ and each $\varepsilon$. This is what makes it different from — and weaker than — the single, uniform $\delta$ demanded by Uniform Continuity, a distinction the next figure is designed to make vivid alongside the endpoint picture.

:::svg[Endpoint continuity of $f(x)=\sqrt{x}$ on $[0,4]$ tested at $c=0$ using only points of $E=[0,4]$]

:::corollary[Reduction to endpoints on $[a,b]$] Let $f : [a,b] \to \mathbb{R}$ with $a < b$. Then $f$ is continuous on $[a,b]$ if and only if: (i) $f$ is continuous at every $c \in (a,b)$ in the ordinary two-sided sense; (ii) $\displaystyle\lim_{x \to a^+} f(x) = f(a)$; and (iii) $\displaystyle\lim_{x \to b^-} f(x) = f(b)$.

Example 1 — a genuinely one-sided endpoint check. Let

Show $f$ is continuous on $[1,3]$, using the endpoint corollary. Interior points $(1,3)$. For $c \in (1,3)$, $f(x) = \sin(x-1)/(x-1)$ is a quotient of continuous functions (sine composed with the continuous map $x \mapsto x-1$, divided by $x-1$) on the set $(1,3)$ where the denominator $x - 1 \ne 0$. By the algebra-of-continuity theorem (parts 2, 3, 4, applied on the open set $(1,3)$ where ordinary and relative continuity coincide), $f$ is continuous at every such $c$. Left endpoint $c=1$. We must check $\lim_{x\to 1^+} f(x) = f(1) = 1$. Substituting $u = x-1 \to 0^+$,

using the standard one-sided (and in fact two-sided) limit $\lim_{u\to 0}\frac{\sin u}{u}=1$. This matches $f(1)=1$, so $f$ is right-continuous at $1$. Right endpoint $c=3$. Since $3 \in (1,3]$ is not an endpoint requiring a separate one-sided argument beyond ordinary continuity — wait: $3$ is the right endpoint of $[1,3]$, but $f$ is already given by the same closed-form expression at $x=3$ (no case split there), and that expression is continuous at $3$ as a quotient of continuous functions with nonzero denominator $3-1=2\ne 0$; two-sided continuity at an endpoint always implies the required one-sided continuity, since the one-sided limit is a special case of the two-sided one. Concretely $\lim_{x\to 3^-} f(x) = f(3) = \sin(2)/2$ by continuity of the formula itself. Hence by the Corollary, $f$ is continuous on $[1,3]$.

Example 2 — continuity on a set that is neither open nor closed, and a genuinely removable behavior only at one non-endpoint. Let $E = [0,1) \cup \{2\}$, and define $f: E \to \mathbb{R}$ by

Show $f$ is continuous on $E$. Points in $[0,1)$. Fix $c \in [0,1)$. Choose $r = \min(c, 1-c)$ if $c > 0$; if $c=0$, we treat it as a one-sided endpoint of $E$ on the left, since no points of $E$ lie left of $0$. In either case, for $x \in E$ near $c$ (within a small enough radius), $x$ lies in $[0,1)$, not at the isolated point $2$ — because $2$ is at distance $\ge 1$ from any $c\in[0,1)$, we may always choose $\delta \le 1 - c$ (or, at $c=0$, $\delta\le 1$) to exclude $x=2$ from consideration entirely. On $[0,1)$, $f(x) = x^2$ is a product of the continuous identity function with itself (Theorem, part 2), hence continuous at each $c \in [0,1)$ relative to $[0,1)$; at $c=0$ this uses the one-sided (right) limit $\lim_{x\to 0^+} x^2 = 0 = f(0)$, matching the endpoint case of the definition since $0$ is the left edge of $[0,1)\subset E$. Explicitly at $c=0$: given $\varepsilon>0$, take $\delta = \min(1,\sqrt\varepsilon)$. If $x\in E$ and $|x-0|<\delta$, then since $\delta \le 1$, $x \ne 2$ (as $|2-0|=2 > \delta$), forcing $x \in [0,1)$, and $|f(x)-f(0)| = x^2 < \delta^2 \le \varepsilon$. The isolated point $c=2$. Since $\{2\}$ is isolated in $E$ (indeed $(1.5,2.5)\cap E = \{2\}$), continuity there is automatic: given any $\varepsilon>0$, take $\delta = 0.5$; then $x\in E, |x-2|<0.5$ forces $x=2$, so $|f(x)-f(2)|=0<\varepsilon$. Hence $f$ is continuous on all of $E = [0,1)\cup\{2\}$ — despite the enormous, discontinuous-looking jump from values near $1$ (approaching $1$) to the isolated value $5$ at $x=2$. This illustrates that continuity on $E$ only ever compares $f$ at points that are actually close within $E$; since no point of $[0,1)$ is close to $2$ inside $E$, no contradiction arises. If we had instead defined $E' = [0,2]$ with the same formulas, $f$ would fail to be continuous at $c=1$-adjacent points approaching from the right of $1$ within $E'$ — but $1 \notin E$ in our actual example, so this concern does not arise here.