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  "body": "\n## Proof\n\nWe work directly with the [[definition of continuity at a point|0EEE2CZB2A]] by open intervals.\n\n1. Let $D, E \\subseteq \\R$, let $f \\colon D \\to \\R$ with $f(D) \\subseteq E$, and let $g \\colon E \\to \\R$. Assume that $f$ is continuous at $a \\in D$ and $g$ is continuous at $f(a)$.\n2. Let $V$ be an open interval containing $g(f(a))$.\n3. Continuity of $g$ at $f(a)$ gives an open interval $W$ containing $f(a)$ with $g(y) \\in V$ for all $y \\in W \\cap E$.\n4. Continuity of $f$ at $a$, applied to $W$, gives an open interval $U$ containing $a$ with $f(x) \\in W$ for all $x \\in U \\cap D$.\n5. Let $x \\in U \\cap D$. Then $f(x) \\in W$ by step 4, and $f(x) \\in E$ since $f(D) \\subseteq E$. So $f(x) \\in W \\cap E$.\n6. By step 3, $(g \\circ f)(x) = g(f(x)) \\in V$.\n7. So $U$ works for $V$, and $g \\circ f$ is continuous at $a$.\n8. If $f$ and $g$ are [[continuous|0P27W9RW0G]], then step 7 applies at each $a \\in D$, since $g$ is continuous at $f(a) \\in E$. So $g \\circ f$ is continuous.\n\n## Idea\n\nTo land in a target interval $V$ around $g(f(a))$, it suffices that $f(x)$ lands in an interval $W$ that $g$ maps into $V$. Continuity of $f$ provides an interval around $a$ that $f$ maps into $W$. The two steps chain together.\n",
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