[proofplan] We prove both directions of the sequential characterisation. The forward direction is a direct chaining of the $\epsilon$-$\delta$ definition with the convergence of the sequence. The reverse direction is a contrapositive argument: if $f$ is discontinuous at $a$, we construct a sequence $x_n \to a$ for which $f(x_n)$ fails to converge to $f(a)$, contradicting the sequential condition. [/proofplan] [step:Derive sequential convergence from the $\epsilon$-$\delta$ definition of continuity] Assume $f$ is continuous at $a$. Let $\epsilon > 0$ be given. By continuity, there exists $\delta > 0$ such that for all $x \in E$ with $|x - a| < \delta$, we have $|f(x) - f(a)| < \epsilon$. Let $(x_n)_{n=1}^{\infty}$ be any sequence in $E$ with $x_n \to a$. There exists $N \in \mathbb{N}$ such that $|x_n - a| < \delta$ for all $n \geq N$. For all $n \geq N$: \begin{align*} |f(x_n) - f(a)| < \epsilon. \end{align*} Since $\epsilon > 0$ was arbitrary, $f(x_n) \to f(a)$. [/step] [step:Prove the converse by contrapositive: construct a sequence witnessing discontinuity] We prove the contrapositive: if $f$ is not continuous at $a$, then there exists a sequence $(x_n)$ in $E$ with $x_n \to a$ but $f(x_n) \not\to f(a)$. Since $f$ is discontinuous at $a$, there exists $\epsilon_0 > 0$ such that for every $\delta > 0$, there exists $x \in E$ with $|x - a| < \delta$ but $|f(x) - f(a)| \geq \epsilon_0$. Applying this with $\delta = 1/n$ for each $n \in \mathbb{N}$, we obtain $x_n \in E$ such that \begin{align*} |x_n - a| < \frac{1}{n} \quad \text{and} \quad |f(x_n) - f(a)| \geq \epsilon_0. \end{align*} The first condition gives $x_n \to a$. The second shows $f(x_n) \not\to f(a)$, since $|f(x_n) - f(a)|$ is bounded below by $\epsilon_0 > 0$ for all $n$. This contradicts the hypothesis that every sequence converging to $a$ has $f(x_n) \to f(a)$. Therefore $f$ is continuous at $a$. [guided] Why does the contrapositive approach work? The direct proof of the converse -- starting from "for every $(x_n)$ with $x_n \to a$, $f(x_n) \to f(a)$" and deducing the $\epsilon$-$\delta$ condition -- is awkward because we would need to handle all possible $\delta$ simultaneously. Instead, we negate the conclusion and derive a contradiction. The negation of "$f$ is continuous at $a$" is: there exists $\epsilon_0 > 0$ such that for every $\delta > 0$, some $x \in E$ satisfies $|x - a| < \delta$ and $|f(x) - f(a)| \geq \epsilon_0$. This is an existential statement for each $\delta$, so we instantiate it with $\delta = 1/n$: \begin{align*} |x_n - a| < \frac{1}{n} \quad \text{and} \quad |f(x_n) - f(a)| \geq \epsilon_0. \end{align*} Since $1/n \to 0$, the [Squeeze Theorem](/theorems/627) gives $x_n \to a$. But $|f(x_n) - f(a)| \geq \epsilon_0 > 0$ for all $n$, so $f(x_n)$ cannot converge to $f(a)$. This contradicts the sequential hypothesis, completing the proof. [/guided] [/step]