[proofplan] The forward direction follows by applying the $\varepsilon$-$\delta$ definition of [continuity](/page/Continuity) to each term of the convergent [sequence](/page/Sequence). The reverse direction is proved by contrapositive: if $f$ fails the $\varepsilon$-$\delta$ condition at $a$, negating the definition produces a sequence $x_n \to a$ for which $f(x_n)$ stays bounded away from $f(a)$, contradicting the sequential hypothesis. [/proofplan] [step:Show that $\varepsilon$-$\delta$ continuity implies sequential continuity] Suppose $f$ is [continuous](/page/Continuity) at $a$ and $x_n \to a$ in $(X, d)$. Fix $\varepsilon > 0$. By continuity of $f$ at $a$, there exists $\delta > 0$ such that $d(x, a) < \delta$ implies $d'(f(x), f(a)) < \varepsilon$. Since $x_n \to a$, there exists $N \in \mathbb{N}$ with $d(x_n, a) < \delta$ for all $n \geq N$. For all $n \geq N$, $d'(f(x_n), f(a)) < \varepsilon$. Since $\varepsilon > 0$ was arbitrary, $f(x_n) \to f(a)$. [/step] [step:Show that sequential continuity implies $\varepsilon$-$\delta$ continuity by contrapositive] Suppose for contradiction that $f$ is not [continuous](/page/Continuity) 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 $n \in \mathbb{N}$, we obtain a [sequence](/page/Sequence) $(x_n)_{n=1}^\infty$ in $X$ with \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. \end{align*} The first condition gives $x_n \to a$. By hypothesis, $f(x_n) \to f(a)$, contradicting $d'(f(x_n), f(a)) \geq \varepsilon_0$ for all $n$. [guided] We prove the contrapositive: if $f$ is not continuous at $a$, then there exists a sequence $x_n \to a$ with $f(x_n) \not\to f(a)$. What does it mean for $f$ to fail continuity at $a$? The negation of the $\varepsilon$-$\delta$ definition says: there exists $\varepsilon_0 > 0$ such that for every $\delta > 0$, some point $x$ satisfies $d(x, a) < \delta$ yet $d'(f(x), f(a)) \geq \varepsilon_0$. We exploit this by choosing $\delta = 1/n$ for $n = 1, 2, 3, \ldots$, producing 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 first condition forces $x_n \to a$ (since $d(x_n, a) < 1/n \to 0$). The second condition forces $f(x_n) \not\to f(a)$ (the image sequence stays at least $\varepsilon_0$ away from $f(a)$). This contradicts the assumption that $f$ preserves convergent sequences at $a$, completing the contrapositive argument. [/guided] [/step]