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  "body": "\n## Definition\n\nLet $D \\subseteq \\C^n$ be an [[open set|02BH3MCBQB]].\nThen $D$ is a *Runge domain* if the following holds for every [[holomorphic function|507C34W0ZT]] $f$ on $D$.\nFor every compact set $K \\subseteq D$ and every $\\varepsilon \u003e 0$ there is a [[polynomial|07GNX8GCYP]] $p$ in $z_1, \\dots, z_n$ with $\\abs{f(z) - p(z)} \u003c \\varepsilon$ for all $z \\in K$.\n\nEquivalently, the polynomials are dense in the space of holomorphic functions on $D$ for the topology of [[uniform convergence|08VZ57PT26]] on compact subsets.\n\n## Remarks\n\nExample: the unit polydisc $\\Delta^n$.\nEvery holomorphic $f$ on $\\Delta^n$ is the sum of its Taylor series at $0$, and this series converges uniformly on each compact subset of $\\Delta^n$; its partial sums are polynomials.\nMore generally every convex open set is a Runge domain, by the [[Oka–Weil theorem|7AWZJNQGT2]] applied to the convex hull of $K$, which is compact and polynomially convex.\n\nNon-example: $D = \\set{z \\in \\C^n : 1 \u003c \\abs{z_1} \u003c 2}$ and $f(z) = 1/z_1$.\nOn the compact circle $K = \\set{\\abs{z_1} = 3/2, \\ z_2 = \\cdots = z_n = 0}$, every polynomial $p$ satisfies $\\int_{\\abs{\\zeta} = 3/2} p(\\zeta, 0, \\dots, 0) \\, d\\zeta = 0$, while the same integral of $f$ is $2\\pi i$.\nSo $f$ is not a uniform limit of polynomials on $K$.\n\nFor $n = 1$, [[Runge's theorem|6DG4CTGXJD]] gives a complete answer.\nAn open set $D \\subseteq \\C$ is a Runge domain exactly when every connected component of $D$ is simply connected.\n",
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