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  "body": "\n## Proof\n\nWe use [[sequential continuity|0PAQM3H57X]] and the limit rules for sequences.\n\n1. Let $D \\subseteq \\R$ and let $f, g \\colon D \\to \\R$ be [[continuous|0P27W9RW0G]].\n2. By the equivalence of continuity and sequential continuity, $f$ and $g$ are [[sequentially continuous|0MVDQEAB60]].\n3. Let $(x_n)$ be a sequence in $D$ that converges to a point $a \\in D$. By step 2, $f(x_n) \\to f(a)$ and $g(x_n) \\to g(a)$.\n4. By the [[sum rule for limits|0RAM4E07NR]], $(f + g)(x_n) = f(x_n) + g(x_n) \\to f(a) + g(a) = (f + g)(a)$.\n5. By the [[product rule for limits|0734RSVF2V]], $(f g)(x_n) = f(x_n) g(x_n) \\to f(a) g(a) = (f g)(a)$.\n6. Since $(x_n)$ and $a$ were arbitrary, $f + g$ and $f g$ are sequentially continuous.\n7. By the equivalence again, $f + g$ and $f g$ are continuous.\n\n## Idea\n\nSequential continuity turns the question into one about limits of sequences. The sum and product rules for limits of sequences then give the result at once.\n",
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