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  "body": "\n## Proof\n\nWe compare open intervals with symmetric intervals $(c - r, c + r)$.\n\n1. Let $D \\subseteq \\R$, let $f \\colon D \\to \\R$, and let $a \\in D$.\n2. Claim: if $V = (p, q)$ is an [[open interval|0TK8Y3J5RW]] containing a point $c$, then $r = \\min(c - p, q - c)$ is positive and $(c - r, c + r) \\subseteq V$.\n3. Proof of the claim: $p \u003c c \u003c q$ gives $r \u003e 0$. If $\\abs{y - c} \u003c r$, then $y \u003e c - r \\ge p$ and $y \u003c c + r \\le q$, so $y \\in V$.\n4. Assume first that $f$ is [[continuous at|0EEE2CZB2A]] $a$. Let $\\varepsilon \u003e 0$.\n5. The set $V = (f(a) - \\varepsilon, f(a) + \\varepsilon)$ is an open interval containing $f(a)$. The definition gives an open interval $U$ containing $a$ with $f(x) \\in V$ for all $x \\in U \\cap D$.\n6. By the claim, there is $\\delta \u003e 0$ with $(a - \\delta, a + \\delta) \\subseteq U$.\n7. Let $x \\in D$ with $\\abs{x - a} \u003c \\delta$. Then $x \\in U \\cap D$, so $f(x) \\in V$, that is, $\\abs{f(x) - f(a)} \u003c \\varepsilon$.\n8. So the epsilon-delta condition holds.\n9. Conversely, assume the epsilon-delta condition. Let $V$ be an open interval containing $f(a)$.\n10. By the claim, there is $\\varepsilon \u003e 0$ with $(f(a) - \\varepsilon, f(a) + \\varepsilon) \\subseteq V$. The condition gives $\\delta \u003e 0$ for this $\\varepsilon$.\n11. Let $U = (a - \\delta, a + \\delta)$, an open interval containing $a$.\n12. Let $x \\in U \\cap D$. Then $\\abs{x - a} \u003c \\delta$, so $\\abs{f(x) - f(a)} \u003c \\varepsilon$. Hence $f(x) \\in V$.\n13. So $f$ is continuous at $a$ in the sense of the definition. This proves the equivalence.\n\n## Idea\n\nThe definition speaks of arbitrary open intervals, the condition of symmetric ones of radius $\\varepsilon$ or $\\delta$. Every symmetric interval is an open interval, and every open interval around a point contains a symmetric one. So the two conditions are the same.\n",
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