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  "body": "\n## Proof\n\nWe use the [[epsilon-delta form of continuity|0W8YHHGCJG]] and the [[epsilon-N form|09G6KHS6GY]] for sequences. The converse is proved by contraposition.\n\n1. Let $D \\subseteq \\R$ and $f \\colon D \\to \\R$. By definition, $f$ is [[continuous|0P27W9RW0G]] when it is continuous at each point of $D$.\n2. Assume that $f$ is continuous. Let $(x_n)$ be a sequence in $D$ that [[converges|0XKWSSY8P2]] to $a \\in D$. Let $\\varepsilon \u003e 0$.\n3. Continuity at $a$ in epsilon-delta form gives $\\delta \u003e 0$ with $\\abs{f(x) - f(a)} \u003c \\varepsilon$ for all $x \\in D$ with $\\abs{x - a} \u003c \\delta$.\n4. Since $x_n \\to a$, there is $N$ with $\\abs{x_n - a} \u003c \\delta$ for all $n \\ge N$. Then $\\abs{f(x_n) - f(a)} \u003c \\varepsilon$ for all $n \\ge N$.\n5. So $f(x_n) \\to f(a)$, and $f$ is [[sequentially continuous|0MVDQEAB60]].\n6. Conversely, assume that $f$ is not continuous. Then $f$ is not continuous at some point $a \\in D$.\n7. By the epsilon-delta form, there is $\\varepsilon \u003e 0$ with this property: for each $\\delta \u003e 0$ there is $x \\in D$ with $\\abs{x - a} \u003c \\delta$ and $\\abs{f(x) - f(a)} \\ge \\varepsilon$.\n8. For each natural number $n$, apply step 7 with $\\delta = 1/(n+1)$. This gives $x_n \\in D$ with $\\abs{x_n - a} \u003c 1/(n+1)$ and $\\abs{f(x_n) - f(a)} \\ge \\varepsilon$.\n9. Let $\\eta \u003e 0$. There is $m \\ge 1$ with [[$1/m \u003c \\eta$|0WH916TJNY]]. For $n \\ge m$, $\\abs{x_n - a} \u003c 1/m \u003c \\eta$. So $x_n \\to a$.\n10. But $\\abs{f(x_n) - f(a)} \\ge \\varepsilon$ for all $n$, so $f(x_n) \\not\\to f(a)$. Hence $f$ is not sequentially continuous.\n11. By contraposition, sequential continuity implies continuity. This proves the equivalence.\n\n## Idea\n\nContinuity at $a$ says that points near $a$ have values near $f(a)$, and a convergent sequence eventually stays near $a$. If continuity fails at $a$, choosing one bad point within $1/(n+1)$ of $a$ for each $n$ gives a sequence that converges to $a$ while its values stay away from $f(a)$.\n",
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