Consider the polynomial $p(x) = x^3 - 2$. Since $p(1) = -1 < 0$ and $p(2) = 6 > 0$, the function changes sign on $[1, 2]$, so there must be a root somewhere in between — namely $\sqrt[3]{2}$. This reasoning is the Intermediate Value Theorem, and it requires continuity. If we replace $p$ with the discontinuous function

\begin{align*} f(x) = \begin{cases} -1 & \text{if } x < \sqrt[3]{2}, \\ 1 & \text{if } x \geq \sqrt[3]{2}, \end{cases} \end{align*}

then $f(1) = -1 < 0$ and $f(2) = 1 > 0$, yet $f$ never equals zero — the jump at $\sqrt[3]{2}$ allows the function to "skip" the value $0$ entirely. The discontinuity creates a gap in the range that continuity would have prevented.

This simple observation — that continuous [functions](/page/Function) cannot skip values — is the entry point to a remarkably deep theory. Continuity is the condition under which [limits](/page/Limit) commute with function evaluation, and it is the minimal regularity needed for the major theorems of real analysis: the Intermediate Value Theorem, the Extreme Value Theorem, and the [Fundamental Theorem of Calculus](/theorems/632) all require it. Yet continuity is also surprisingly permissive: continuous functions can oscillate wildly at every scale (the Weierstrass function), be discontinuous on a dense set (the Thomae function), and defy every geometric intuition one might bring from drawing smooth curves.

This page develops continuity for functions $f: E \to \mathbb{R}$ where $E \subseteq \mathbb{R}$, the concrete setting of Cambridge IA Analysis. The general metric-space and [topological](/page/Topology) definitions are developed on the parent page [Continuity](/page/Continuity).

## The $\varepsilon$-$\delta$ Definition

The informal idea of continuity — "small changes in input produce small changes in output" — must be made precise. The question is: how small is "small"? The $\varepsilon$-$\delta$ definition answers this by demanding that *for any tolerance $\varepsilon$ on the output*, there exists a corresponding tolerance $\delta$ on the input that guarantees the output stays within $\varepsilon$ of $f(a)$.

[definition:Continuity At A Point Real Line] Let $E \subseteq \mathbb{R}$ and let $f: E \to \mathbb{R}$. The function $f$ is **continuous at** $a \in E$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that \begin{align*} |x - a| < \delta \quad \text{and} \quad x \in E \quad \implies \quad |f(x) - f(a)| < \varepsilon. \end{align*} The function $f$ is **continuous on $E$** if it is continuous at every point $a \in E$. [/definition]

The quantifier structure $\forall \varepsilon > 0 \; \exists \delta > 0$ is the engine of the definition. Earlier mathematicians, including Newton and Leibniz, spoke of "infinitesimal changes" — quantities smaller than any positive number yet nonzero. Cauchy's insight was to eliminate infinitesimals entirely: instead of asserting that the difference $|f(x) - f(a)|$ *becomes* infinitesimal, he demanded a verifiable test for *every* positive $\varepsilon$, however small. This quantifier machinery works in abstract spaces where geometric intuition fails, and it became the foundation of modern analysis.

Notice that $\delta$ is allowed to depend on both $\varepsilon$ *and* the point $a$. This dependence is essential: for $f(x) = x^2$ near $a = 100$, a much smaller $\delta$ is needed than near $a = 1$ to achieve the same $\varepsilon$. The question of when $\delta$ can be chosen independently of $a$ — *uniform* continuity — is one of the central themes of this page.

[example:Epsilon Delta Estimation For A Quadratic] We prove that $f(x) = x^2$ is continuous at $a = 3$. The key technique is to *factor* $|f(x) - f(a)|$ and control each factor separately. **Step 1: Factor the distance.** $|x^2 - 9| = |x - 3| \cdot |x + 3|$. The first factor $|x - 3|$ is directly controlled by $\delta$. The second factor $|x + 3|$ must be bounded, which requires restricting $x$ to a neighbourhood of $3$. **Step 2: Bound the uncontrolled factor.** If $|x - 3| < 1$, then $|x| < 4$, so $|x + 3| \leq |x| + 3 < 7$. This gives $|x^2 - 9| < 7|x - 3|$. **Step 3: Choose $\delta$.** Given $\varepsilon > 0$, set $\delta = \min(1, \varepsilon/7)$. If $|x - 3| < \delta$, then both restrictions are active: $|x - 3| < 1$ (so the bound from Step 2 applies) and $|x - 3| < \varepsilon/7$. Therefore $|x^2 - 9| < 7 \cdot \varepsilon/7 = \varepsilon$. This "factor and bound" technique is the standard method for $\varepsilon$-$\delta$ proofs. The trick $\delta = \min(1, \varepsilon/7)$ — imposing a preliminary bound to control the "bad" factor — recurs throughout analysis. [/example]

## The Sequential Characterisation

The $\varepsilon$-$\delta$ definition is precise but unwieldy for many arguments. In $\mathbb{R}$, there is a cleaner equivalent: continuity means preserving convergent [sequences](/page/Sequence).

[quotetheorem:179]

The sequential characterisation is the standard tool for proving discontinuity: to show $f$ is discontinuous at $a$, exhibit a single [sequence](/page/Sequence) $x_n \to a$ with $f(x_n) \not\to f(a)$. This is far easier than showing that *no* $\delta$ works for some $\varepsilon$ in the $\varepsilon$-$\delta$ definition. The forward direction (continuity implies sequence preservation) is immediate; the reverse constructs a "bad sequence" by taking $\delta = 1/n$ and finding points $x_n$ violating the $\varepsilon$-$\delta$ condition, yielding $x_n \to a$ but $f(x_n) \not\to f(a)$.

## Pathological Examples

The power of the $\varepsilon$-$\delta$ definition is that it handles functions far wilder than anything one might draw by hand. The following three examples, each a milestone in the history of analysis, reveal just how strange continuous and discontinuous functions can be.

### The Dirichlet Function: Discontinuous Everywhere

[example:The Dirichlet Function] Define $\mathbf{1}_\mathbb{Q}: \mathbb{R} \to \mathbb{R}$ by \begin{align*} \mathbf{1}_\mathbb{Q}(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q}, \\ 0 & \text{if } x \notin \mathbb{Q}. \end{cases} \end{align*} This function is discontinuous at every point. At any $a \in \mathbb{R}$, the density of both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ in $\mathbb{R}$ provides two sequences converging to $a$ with different image limits. **Proof of discontinuity.** Fix any $a \in \mathbb{R}$. By the density of the rationals, there exists a sequence of rationals $r_n \to a$; by the density of the irrationals, there exists a sequence of irrationals $s_n \to a$. Then $\mathbf{1}_\mathbb{Q}(r_n) = 1 \to 1$ and $\mathbf{1}_\mathbb{Q}(s_n) = 0 \to 0$. Since $1 \neq 0$, $\mathbf{1}_\mathbb{Q}$ cannot be continuous at $a$: if it were, both sequences would have to converge to $\mathbf{1}_\mathbb{Q}(a)$, but they converge to different values. The Dirichlet function is the standard example showing that a function can be discontinuous at *every* point. It is also not [Riemann integrable](/page/Riemann%20Integral): on every subinterval $[x_{j-1}, x_j]$, the supremum of $\mathbf{1}_\mathbb{Q}$ is $1$ and the infimum is $0$, so the upper and lower sums never agree. [/example]

### The Thomae Function: Continuous on a Dense Set