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  "body": "\n## Counterexample\n\nThe *Hartogs triangle* is the domain\n$$H = \\set{(z, w) \\in \\C^2 : \\abs{z} \u003c \\abs{w} \u003c 1}.$$\nIt is a bounded [[pseudoconvex|AHYZMSBBNH]] domain whose closure $\\overline{H}$ has no basis of pseudoconvex neighbourhoods.\n\nCheck that $H$ is pseudoconvex.\nOn $H$ the function $g(z, w) = z/w$ is holomorphic with $\\abs{g} \u003c 1$.\nLet\n$$\\varphi(z, w) = -\\log\\big(1 - \\abs{z/w}^2\\big) - \\log\\big(1 - \\abs{w}^2\\big) - \\log \\abs{w}.$$\nThe first two terms are [[plurisubharmonic|FF2P7ER3QT]], since $-\\log(1 - \\abs{h}^2) = \\sum_{k \\ge 1} \\abs{h^k}^2 / k$ for a holomorphic $h$ with $\\abs{h} \u003c 1$, and the last term is pluriharmonic.\nThe boundary of $H$ consists of the points with $\\abs{z} = \\abs{w} \\le 1$ and the points with $\\abs{w} = 1$, $\\abs{z} \\le 1$.\nAll three terms are nonnegative on $H$.\nNear the first part the first term tends to $+\\infty$, except at the origin, where the last term does.\nNear the second part the second term tends to $+\\infty$.\nSo $\\varphi$ is a continuous [[plurisubharmonic exhaustion function|FGSKR6ZS3Q]], and $H$ is pseudoconvex, hence a [[domain of holomorphy|4487CSC0MB]] by [[Oka's theorem|6BJB149MDK]].\n\nCheck the neighbourhoods.\nLet $U$ be an open set containing $\\overline{H} = \\set{\\abs{z} \\le \\abs{w} \\le 1}$.\nSince $\\overline{H}$ is compact, $U$ contains, for some $\\varepsilon \u003e 0$, the set $\\set{\\abs{z} \u003c \\varepsilon, \\abs{w} \u003c 1 + \\varepsilon}$ and the set $\\set{\\abs{z} \u003c 1 + \\varepsilon, 1 - \\varepsilon \u003c \\abs{w} \u003c 1 + \\varepsilon}$.\nThe union of these two sets is a [[Hartogs figure|GHJC53RFBT]].\nLet $f$ be holomorphic on $U$.\nThe Laurent series of $w \\mapsto f(z, w)$ on the annulus has no negative powers when $\\abs{z} \u003c \\varepsilon$.\nIts coefficients are holomorphic in $z$, so by the identity theorem it has none when $\\abs{z} \u003c 1$.\nSo $f$ extends holomorphically to the bidisc $\\set{\\abs{z} \u003c 1, \\abs{w} \u003c 1 + \\varepsilon}$.\nHence any domain of holomorphy that contains $\\overline{H}$ contains the whole unit bidisc, and the pseudoconvex neighbourhoods of $\\overline{H}$ cannot shrink to $\\overline{H}$.\n\n## Remarks\n\nThe boundary of $H$ is not a Lipschitz graph near the origin, and this singular point is the source of the pathology.\nFor bounded pseudoconvex domains with smooth boundary, the closure can still fail to have a pseudoconvex neighbourhood basis.\nThe examples are much harder, such as the worm domains of Diederich and Fornæss.\n\n## Explanations\n\n### L1 human\n\nThe Hartogs triangle is the set of pairs $(z, w)$ of complex numbers with $\\abs{z} \u003c \\abs{w} \u003c 1$.\nIf one draws it with axes $\\abs{z}$ and $\\abs{w}$, it is a triangle.\nIt is a perfectly good domain for holomorphic functions, a domain of holomorphy.\nBut every domain of holomorphy that contains its closure must contain the whole unit bidisc, because holomorphic functions near the closure always extend that far.\nSo the closure of a domain of holomorphy need not be approximable from outside by domains of holomorphy.\n",
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