Continuity

Attempt these from memory before reading.

• Define continuity at a point, using quantifiers.
• What are the three types of discontinuity?
• State the intermediate value theorem. What does “closed, bounded” buy?
• What is the difference between continuity and uniform continuity?
• Is there a function that is continuous everywhere but differentiable nowhere?

Check your answers against the text. The last prompt is the one most people answer wrongly — and the answer is the most important result on this page.

The intuitive definition — “you can draw it without lifting your pen” — is nearly right, and it is worth knowing exactly how it is wrong. Three functions destroy it.

Dirichlet’s function $D(x) = 1$ if $x \in \mathbb{Q}$, $0$ otherwise, is discontinuous everywhere. Its graph cannot be drawn at all.

Thomae’s function $t(x) = 1/q$ if $x = p/q$ in lowest terms, $0$ if $x$ is irrational, is continuous at every irrational and discontinuous at every rational. Its graph looks like a firework; the pen is lifted infinitely often.

Weierstrass’ function $W(x) = \sum_{n=0}^{\infty} a^n \cos(b^n \pi x)$ (with $0 < a < 1$, $b$ odd, $ab > 1 + 3\pi/2$) is continuous everywhere and differentiable nowhere. It can be drawn without lifting the pen, and yet it has no tangent anywhere.

So “drawn without lifting the pen” is neither necessary (Dirichlet) nor sufficient (Weierstrass) for what continuity actually means. The $\varepsilon$-$\delta$ definition is not pedantry; it is the only definition that survives these examples.

Let $f : D \to \mathbb{R}$ where $D \subseteq \mathbb{R}$. We say $\lim_{x \to a} f(x) = L$ if for every $\varepsilon > 0$ There exists $\delta > 0$ such that

Definition. $f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$. In epsilon-delta form: For every $\varepsilon > 0$ There exists $\delta > 0$ such that

Remark. A function is continuous on a set $E$ if it is continuous at every point of $E$. A function is globally continuous (or “continuous”) if it is continuous on its entire domain.

Definition. $f$ is discontinuous at $a$ if it is not continuous at $a$. Discontinuities are Classified as:

• Removable: $\lim_{x \to a} f(x)$ exists but does not equal $f(a)$ (or $f(a)$ is undefined).
• Jump: $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$ both exist but are unequal.
• Essential (or infinite/oscillatory): At least one one-sided limit does not exist.

Proposition 4.3. Polynomials are continuous on $\mathbb{R}$. Rational functions $p(x)/q(x)$ are Continuous wherever $q(x) \neq 0$. The functions $\sin x$, $\cos x$, $e^x$, $\ln x$ are continuous on their domains.

Theorem 4.1 (Algebra of Continuous Functions). If $f$ and $g$ are continuous at $a$ Then $f+g$ $f-g$, $fg$ And (where defined) $f/g$ are continuous at $a$.

Theorem 4.2. Compositions of continuous functions are continuous: if $f$ is continuous at $a$ and $g$ is continuous at $f(a)$ Then $g \circ f$ is continuous at $a$.

The epsilon-delta definitions can be reformulated in terms of sequences, which is often more Convenient for …/1-number-and-algebra/3_proof-and-logics.

Proposition 4.2a (Sequential Criterion for Limits). $\lim_{x \to c} f(x) = L$ if and only if For every sequence $(x_n)$ with $x_n \to c$ and $x_n \neq c$ for all $n$ We have $f(x_n) \to L$.

Proof. ($\Rightarrow$) Let $\varepsilon > 0$. Choose $\delta > 0$ from the $\varepsilon$-$\delta$ definition. Since $x_n \to c$ There exists $N$ with $|x_n - c| \lt \delta$ for $n \geq N$. Then $|f(x_n) - L| \lt \varepsilon$ for $n \geq N$.

($\Leftarrow$) Suppose the $\varepsilon$-$\delta$ condition fails. Then there exists $\varepsilon > 0$ such That for every $n \in \mathbb{N}$ There exists $x_n$ with $0 \lt |x_n - c| \lt 1/n$ but $|f(x_n) - L| \geq \varepsilon$. Then $x_n \to c$ but $f(x_n) \not\to L$ Contradicting the hypothesis. $\blacksquare$

Corollary 4.2b. $f$ is continuous at $c$ if and only if for every sequence $(x_n)$ with $x_n \to c$ We have $f(x_n) \to f(c)$.

This is especially useful for proving that a function is not continuous: find one sequence Converging to $c$ whose image does not converge to $f(c)$.

Theorem 4.3 (IVT). If $f : [a,b] \to \mathbb{R}$ is continuous and $f(a) \lt y \lt f(b)$ (or $f(b) \lt y \lt f(a)$), then there exists $c \in (a,b)$ such that $f(c) = y$.

Proof. Assume $f(a) \lt y \lt f(b)$. Let $S = \{x \in [a,b] : f(x) \lt y\}$. Since $a \in S$ $S$ is non-empty and bounded above by $b$. Let $c = \sup(S)$. We show $f(c) = y$.

If $f(c) \lt y$ Then by continuity at $c$ There exists $\delta > 0$ such that $f(x) \lt y$ for $x \in (c - \delta, c + \delta)$. But then $c + \delta/2 \in S$ Contradicting that $c = \sup(S)$.

If $f(c) > y$ Then by continuity, there exists $\delta > 0$ such that $f(x) > y$ for $x \in (c - \delta, c + \delta)$. But then $c - \delta/2$ is an upper bound for $S$ Contradicting that $c = \sup(S)$.

Therefore $f(c) = y$. $\blacksquare$

Alternative …/1-number-and-algebra/3_proof-and-logic (bisection). Set $a_0 = a$, $b_0 = b$. Given $[a_n, b_n]$ with $f(a_n) \lt y \lt f(b_n)$ Let $m_n = (a_n + b_n)/2$. If $f(m_n) \geq y$ Set $a_{n+1} = a_n$, $b_{n+1} = m_n$. If $f(m_n) \lt y$ Set $a_{n+1} = m_n$, $b_{n+1} = b_n$. Either way, $f(a_n) \lt y \leq f(b_n)$ and $b_n - a_n = (b-a)/2^n \to 0$. By the nested interval property, $a_n \to c$ and $b_n \to c$. By continuity, $f(c) = \lim f(a_n) \leq y$ And $f(c) = \lim f(b_n) \geq y$ So $f(c) = y$. $\blacksquare$

Theorem 4.4 (EVT). If $f : [a,b] \to \mathbb{R}$ is continuous, then $f$ attains its maximum and Minimum on $[a,b]$: there exist $c_1, c_2 \in [a,b]$ such that $f(c_1) \leq f(x) \leq f(c_2)$ for all $x \in [a,b]$.

Proof. We first show $f$ is bounded. Suppose not; then for each $n \in \mathbb{N}$ There exists $x_n \in [a,b]$ with $|f(x_n)| > n$. By Bolzano-Weierstrass, $(x_n)$ has a convergent subsequence $x_{n_k} \to c \in [a,b]$. By continuity, $f(x_{n_k}) \to f(c)$ So $(f(x_{n_k}))$ is bounded. But $|f(x_{n_k})| > n_k \to \infty$ A contradiction.

Now we show $f$ attains its supremum. Let $M = \sup\{f(x) : x \in [a,b]\}$. For each $n$ Choose $x_n \in [a,b]$ with $f(x_n) > M - 1/n$. By Bolzano-Weierstrass, $(x_n)$ has a subsequence $x_{n_k} \to c \in [a,b]$. By continuity, $f(c) = \lim f(x_{n_k})$. Since $M - 1/n_k \lt f(x_{n_k}) \leq M$ for all $k$ The squeeze theorem gives $f(c) = M$. The argument for the infimum is similar (consider $-f$). $\blacksquare$

Definition. $f$ is uniformly continuous on $D$ if for every $\varepsilon > 0$ There exists $\delta > 0$ such that for all $x, y \in D$:

The key distinction: for ordinary continuity, $\delta$ may depend on both $\varepsilon$ and the point $a$; for uniform continuity, $\delta$ depends only on $\varepsilon$.

Theorem 4.5 (Heine-Cantor). If $f : [a,b] \to \mathbb{R}$ is continuous on the closed, bounded Interval $[a,b]$ Then $f$ is uniformly continuous on $[a,b]$.

Proof. Suppose $f$ is continuous on $[a,b]$ but not uniformly continuous. Then there exists $\varepsilon > 0$ such that for every $n \in \mathbb{N}$ There exist $x_n, y_n \in [a,b]$ with $|x_n - y_n| \lt 1/n$ but $|f(x_n) - f(y_n)| \geq \varepsilon$.

By the Bolzano-Weierstrass theorem, $(x_n)$ has a convergent subsequence $x_{n_k} \to c \in [a,b]$. Since $|x_{n_k} - y_{n_k}| \lt 1/n_k \to 0$ We have $y_{n_k} \to c$ as well.

By continuity of $f$ at $c$: there exists $\delta > 0$ such that $|x - c| \lt \delta$ implies $|f(x) - f(c)| \lt \varepsilon/2$. For $k$ sufficiently large, $|x_{n_k} - c| \lt \delta$ and $|y_{n_k} - c| \lt \delta$ So:

Contradicting $|f(x_{n_k}) - f(y_{n_k})| \geq \varepsilon$. $\blacksquare$

Problem. Prove that $f(x) = \sqrt{x}$ is uniformly continuous on $[0, \infty)$.

Solution. For $x, y \geq 0$: $|\sqrt{x} - \sqrt{y}| = \frac{|x - y|}{\sqrt{x} + \sqrt{y}} \leq |x - y|^{1/2}$.

Given $\varepsilon > 0$ Choose $\delta = \varepsilon^2$. Then $|x - y| \lt \delta$ implies $|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x-y|} \lt \sqrt{\delta} = \varepsilon$. Since $\delta$ depends only on $\varepsilon$ The continuity is uniform. $\blacksquare$

The epsilon-delta definition of a limit is best understood as a challenge game between two players. Player A (the challenger) picks any tolerance epsilon — no matter how small. Player B (the responder) must then find a delta such that all points within delta of the target produce function values within epsilon of the claimed limit. If Player B can always win regardless of how small epsilon is, the limit exists. This adversarial framing captures the logical structure of the universal quantifiers: “for every epsilon, there exists a delta.”

The squeeze theorem extends this intuition: if a function is trapped between two other functions that both converge to the same limit, it must converge there too. A physical analogy is a grape squeezed between two approaching walls — the grape has nowhere to go but forward. Uniform continuity removes the dependence of delta on the point of evaluation: the same delta works everywhere on the domain simultaneously. This is a much stronger condition, and it fails on open intervals like (0,1) because functions like 1/x can change arbitrarily fast near the boundary. The Heine-Cantor theorem guarantees that continuity on a closed bounded interval automatically yields uniform continuity — compactness eliminates the boundary trouble.

The following table summarizes the key counterexamples in the theory of continuity:

The pattern: A continuous function on a bounded interval fails to be uniformly continuous if and only if it can oscillate arbitrarily fast near some point (in standard practice a boundary point where the function is not defined, or a point where the derivative is unbounded). On closed, bounded intervals, this cannot happen because the function is bounded and the interval is compact.

Three counterexamples that break the intuition.

Thomae’s function deserves a closer look, because the proof is instructive. Given $\varepsilon > 0$ and an irrational $x_0$, only finitely many rationals $p/q$ have $1/q \ge \varepsilon$ — namely those with $q \le 1/\varepsilon$. Choose $\delta$ avoiding that finite set. Then every $x$ within $\delta$ of $x_0$ has either $t(x) = 0$ (irrational) or $t(x) = 1/q < \varepsilon$.

That argument uses compactness twice, and it is the reason the function works on a bounded interval. It is also the reason Thomae’s function is Riemann integrable with integral zero — a fact that becomes the Lebesgue criterion in measure theory.

Weierstrass’ function is why “continuous” does not license “has a gradient” in any downstream argument. Every theorem about differentiable functions needs differentiability as a hypothesis, and this is the example that forces it.

Mix these with material from Sequences and Limits, Measure Theory and Probability Spaces.

1. (Sequences.) Show $D(x)$ is discontinuous everywhere using the sequential criterion, taking $x_n \to x$ rational and $y_n \to x$ irrational.
2. (Sequences.) Show $t(x)$ is discontinuous at every rational $p/q$, by taking a sequence of irrationals converging to $p/q$.
3. (Measure theory.) Show Thomae’s function is Riemann integrable on $[0,1]$ with integral zero, using the fact that its discontinuities have measure zero.
4. (Probability.) Let $X$ be uniform on $[0,1]$. Show $P(\lim_{n}\mathbf{1}_{X_n \in \mathbb{Q}} = D(X)) = 0$ is false for any sequence — connecting Dirichlet’s function to almost-sure convergence.
5. (Differentiability.) Show that if $f$ is differentiable at $x_0$ then $f$ is continuous at $x_0$, and explain why Weierstrass’ function does not contradict this.

• The Heine-Cantor theorem requires a closed, bounded interval. The function $f(x) = 1/x$ on $(0, 1)$ is continuous but not uniformly continuous. The theorem requires a closed and bounded interval. Also, a function can be uniformly continuous on an unbounded domain (e.g., $f(x) = \sqrt{x}$ on $[0, \infty)$) --- boundedness of the domain is sufficient but not necessary.
• Do not confuse continuity with uniform continuity. Every uniformly continuous function is continuous, but the converse is false on unbounded or open domains.
• The sequential criterion is the most common tool for disproving continuity or limits. Find two sequences converging to the same point whose images converge to different limits.
• The IVT requires the function to be continuous on a closed interval. The function $f(x) = 1/x$ on $(0, 1)$ satisfies $f(0.1) = 10 > 0$ and $f(0.9) = 1.11 > 0$, but there is no $c \in (0.1, 0.9)$ with $f(c) = 0$. The IVT does not apply because $f$ is not defined on $[0.1, 0.9]$ with $f$ taking both positive and negative values.
• A function can be continuous at every point but not uniformly continuous. The classic example is $f(x) = x^2$ on $\mathbb{R}$. It is continuous everywhere, but on any interval $[-M, M]$, the maximum slope is $2M$, which grows without bound as $M \to \infty$.

• Differentiability: Differentiability implies continuity, but the converse is false; the relationship between these concepts is explored via the mean value theorem.
• Riemann Integration: Continuous functions on closed intervals are Riemann integrable, and the Heine-Cantor theorem guarantees uniform continuity on compact domains.
• Sigma-Algebras and Measurable Spaces: Continuity is defined in terms of open sets, which generate the Borel sigma-algebra used in measure theory.
• Classical Mechanics
• Electromagnetism

• Differentiability
• Problem Set
• Riemann Integration
• Sequences and Limits