[proofplan] We prove both implications separately. The forward direction is a direct unfolding of definitions: the $\varepsilon$-$\delta$ condition provides a ball around $a$ in which [function](/page/Function) values are controlled, and convergence $x_n \to a$ eventually places all terms inside this ball. The reverse direction is a contrapositive: if $\varepsilon$-$\delta$ [continuity](/page/Continuity) fails, the negation with $\delta = 1/n$ constructs a [sequence](/page/Sequence) converging to $a$ whose image does not converge to $f(a)$. [/proofplan] [step:Prove $\varepsilon$-$\delta$ continuity implies sequential continuity] Assume $f$ is continuous at $a$ in the $\varepsilon$-$\delta$ sense. Let $(x_n)$ be a sequence in $X$ with $x_n \to a$, and fix $\varepsilon > 0$. By continuity, there exists $\delta > 0$ such that $d(x, a) < \delta$ implies $d'(f(x), f(a)) < \varepsilon$. Since $x_n \to a$, there exists $N \in \mathbb{N}$ such that $d(x_n, a) < \delta$ for all $n \geq N$. For all such $n$, the implication gives $d'(f(x_n), f(a)) < \varepsilon$. Since $\varepsilon > 0$ was arbitrary, $f(x_n) \to f(a)$. [/step] [step:Prove sequential continuity implies $\varepsilon$-$\delta$ continuity by contrapositive] We prove the contrapositive: if $f$ is not $\varepsilon$-$\delta$ continuous at $a$, then there exists a sequence violating the sequential condition. Suppose $f$ is not continuous at $a$. Then there exists $\varepsilon_0 > 0$ such that for every $\delta > 0$, there exists $x \in X$ with $d(x, a) < \delta$ and $d'(f(x), f(a)) \geq \varepsilon_0$. Applying this with $\delta = 1/n$ for each $n \in \mathbb{N}$ produces a sequence $(x_n)$ satisfying \begin{align*} d(x_n, a) < \frac{1}{n} \quad \text{and} \quad d'(f(x_n), f(a)) \geq \varepsilon_0 \quad \text{for all } n \in \mathbb{N}. \end{align*} The first condition gives $x_n \to a$ as $n \to \infty$. The second condition shows $f(x_n) \not\to f(a)$, since $d'(f(x_n), f(a)) \geq \varepsilon_0 > 0$ for all $n$. This violates the sequential condition, completing the contrapositive. [guided] Why use the contrapositive rather than a direct argument? The direct direction ($\varepsilon$-$\delta$ implies sequential) works by "plugging in" the convergent sequence to the $\delta$-ball. The reverse direction is harder to prove directly because we would need to produce a $\delta$ from the sequential condition, and there is no natural way to do so. The contrapositive approach is cleaner: we assume $\varepsilon$-$\delta$ continuity fails and construct a specific sequence that violates the sequential condition. The negation of $\varepsilon$-$\delta$ continuity says: there exists $\varepsilon_0 > 0$ such that for every $\delta > 0$, some point $x$ satisfies $d(x,a) < \delta$ yet $d'(f(x), f(a)) \geq \varepsilon_0$. Setting $\delta = 1/n$ gives a sequence $(x_n)$ with \begin{align*} d(x_n, a) < \frac{1}{n} \quad \text{and} \quad d'(f(x_n), f(a)) \geq \varepsilon_0. \end{align*} The condition $d(x_n, a) < 1/n$ forces $x_n \to a$ (squeeze: $0 \leq d(x_n, a) < 1/n \to 0$). But $d'(f(x_n), f(a)) \geq \varepsilon_0$ for all $n$ means the distances $d'(f(x_n), f(a))$ are bounded away from zero, so $f(x_n) \not\to f(a)$. This is a concrete witness to the failure of the sequential condition. [/guided] [/step]