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  "body": "\n## Proof\n\nWe argue by induction, from constants and the identity, using that [[sums and products of continuous functions are continuous|0VGMHCF4SC]].\n\n1. Let $c \\in \\R$ and let $k_c \\colon \\R \\to \\R$ be the constant function $k_c(x) = c$. Let $a \\in \\R$ and let $V$ be an open interval containing $k_c(a) = c$.\n2. Take $U = (a - 1, a + 1)$. For all $x \\in U$, $k_c(x) = c \\in V$. So $k_c$ is [[continuous at|0EEE2CZB2A]] $a$. Hence $k_c$ is [[continuous|0P27W9RW0G]].\n3. Let $\\id \\colon \\R \\to \\R$ be $\\id(x) = x$. Let $a \\in \\R$ and let $V$ be an open interval containing $a$. Take $U = V$. Then $\\id(x) = x \\in V$ for all $x \\in U$. So $\\id$ is continuous.\n4. For each natural number $i$, let $m_i(x) = x^i$. We show by induction on $i$ that $m_i$ is continuous.\n5. Base case: $m_0 = k_1$, which is continuous by step 2.\n6. Induction step: $m_{i+1}(x) = m_i(x) \\cdot x$, so $m_{i+1}$ is the product of $m_i$ and $\\id$. It is continuous by the induction hypothesis, step 3 and the product rule.\n7. For each $i$, the function $x \\mapsto a_i x^i$ is the product of $k_{a_i}$ and $m_i$. So it is continuous, by steps 2 and 6 and the product rule.\n8. For $j \\le n$, let $p_j(x) = \\sum_{i=0}^{j} a_i x^i$. By step 7, $p_0$ is continuous.\n9. If $p_j$ is continuous and $j \u003c n$, then $p_{j+1} = p_j + a_{j+1} x^{j+1}$ is a sum of two continuous functions, by step 7. So it is continuous.\n10. By induction on $j$, $p_n = p$ is continuous.\n\n## Idea\n\nA polynomial is built from constants and the identity $x \\mapsto x$ by finitely many sums and products. Both building blocks are continuous, and sums and products keep continuity. Induction on the degree carries this through.\n",
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