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  "body": "\n## Counterexample\n\nLet $\\omega$ be the first infinite [[ordinal|0YSCSX5YZC]] and $\\omega_1$ the first uncountable ordinal.\nGive $[0, \\omega_1]$ and $[0, \\omega]$ the order topology, and let $T = [0, \\omega_1] \\times [0, \\omega]$ with the [[product topology|0GBG8B49AA]].\nThis is the Tychonoff plank.\nThe deleted Tychonoff plank is the [[subspace|0WS9P12FQR]] $T_\\infty = T \\setminus \\set{(\\omega_1, \\omega)}$.\nThe plank $T$ is [[compact|0TVJZYB6SH]] and [[Hausdorff|0S1FJ8VEX9]], hence [[normal|0NSR9V8TED]], but the open subspace $T_\\infty$ is not normal.\n\n## Remarks\n\nThe sets $A = \\set{\\omega_1} \\times [0, \\omega)$ and $B = [0, \\omega_1) \\times \\set{\\omega}$ are disjoint and closed in $T_\\infty$.\nLet $U$ be an open set of $T_\\infty$ containing $A$.\nFor each $n \u003c \\omega$ there is $\\alpha_n \u003c \\omega_1$ with $(\\alpha_n, \\omega_1] \\times \\set{n} \\subseteq U$.\nThe supremum $\\alpha$ of the countably many $\\alpha_n$ is still countable, so $\\alpha + 1 \u003c \\omega_1$ and $(\\alpha + 1, n) \\in U$ for every $n$.\nAny open set $V$ containing $(\\alpha + 1, \\omega) \\in B$ contains $(\\alpha + 1, n)$ for all large $n$, so $V$ meets $U$.\nHence $A$ and $B$ cannot be separated, and $T_\\infty$ is not normal.\nSince $T_\\infty$ is a subspace of the compact Hausdorff space $T$, it is a [[Tychonoff space|C5N2J2K6Z3]]; so an open subspace of a normal space need not be normal.\n\n## Explanations\n\n### L1 human\n\nThe plank is a product of two well-ordered \"lines\", one countably long and one uncountably long, each with a top point.\nIt is compact, so it behaves well.\nRemoving the single top corner point leaves two edges that run towards the missing corner.\nAny open set around one edge must swallow a whole column near the corner, and so it always touches the other edge.\n",
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