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  "body": "\n## Counterexample\n\nLet $\\Gamma = \\set{(x, y) \\in \\R^2 : y \\ge 0}$ be the closed upper half-plane.\nGive a point $(x, y)$ with $y \u003e 0$ the usual Euclidean open discs around it, of radius less than $y$, as basic [[neighbourhoods|0ZCY0Z80B6]].\nGive a point $(x, 0)$ the basic neighbourhoods $\\set{(x, 0)} \\cup D$, where $D$ is an open disc of radius $r \u003e 0$ centred at $(x, r)$.\nSuch a disc $D$ is tangent to the $x$-axis at $(x, 0)$.\nThe space $\\Gamma$ with this topology is the Moore plane, also called the Niemytzki plane.\nIt is a [[Tychonoff space|C5N2J2K6Z3]], hence [[Hausdorff|0S1FJ8VEX9]] and regular, but it is not [[normal|0NSR9V8TED]].\n\n## Remarks\n\nThe $x$-axis $A = \\set{(x, 0) : x \\in \\R}$ is [[closed|0ZCK8DQYN1]], since every point with $y \u003e 0$ has a disc neighbourhood missing $A$.\nThe subspace $A$ is discrete, because a tangent disc neighbourhood of $(x, 0)$ meets $A$ only in $(x, 0)$.\nSo every subset of $A$ is closed in $\\Gamma$.\nThe points with rational coordinates and $y \u003e 0$ form a countable dense set, so $\\Gamma$ is [[separable|0Z2P41SVDP]].\nBy Jones' lemma, a separable normal space has no closed discrete subset $A$ with $2^{\\abs{A}} \u003e 2^{\\aleph_0}$.\nHere $\\abs{A} = 2^{\\aleph_0}$, so $\\Gamma$ is not normal.\nFor example, the points of $A$ with rational $x$ and the points of $A$ with irrational $x$ are disjoint closed sets that cannot be separated by disjoint open sets.\nSince $\\Gamma$ is Tychonoff, it is a subspace of a compact Hausdorff space, which is normal; so a subspace of a normal space need not be normal.\n\n## Explanations\n\n### L1 human\n\nAway from the bottom edge, the Moore plane is the ordinary half-plane.\nOn the bottom edge, a point is only near the points of small discs that sit on top of it and touch the edge there.\nSo the points of the edge cannot see each other, and the edge breaks up into isolated points.\nThis makes it impossible to pull apart some disjoint closed pieces of the edge with open sets.\n",
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