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  "body": "\n## Definition\n\nLet $X$ be a [[topological space|0CQHFPGJ45]].\nFor $y \\in X$, write $\\overline{\\set{y}}$ for the [[closure|0QEAJB91GZ]] of $\\set{y}$.\nThe *specialization preorder* on $X$ is the relation $\\le$ defined by\n$$x \\le y \\iff x \\in \\overline{\\set{y}}, \\qquad x, y \\in X.$$\n\n## Remarks\n\nThe relation is a [[preorder|0VABYWJWH9]]: $x \\in \\overline{\\set{x}}$, and if $x \\in \\overline{\\set{y}}$ and $y \\in \\overline{\\set{z}}$ then $\\overline{\\set{y}} \\subseteq \\overline{\\set{z}}$, so $x \\in \\overline{\\set{z}}$.\nEquivalently, $x \\le y$ if and only if every open set that contains $x$ also contains $y$.\n\nThe preorder is antisymmetric, hence a partial order, if and only if $X$ is a [[T0 space|07SG12TQH3]].\nIt is the equality relation if and only if every point of $X$ is closed, that is, $X$ is T1.\nEvery continuous map is monotone for the specialization preorders.\n\nExample: in the spectrum of a commutative ring with the Zariski topology, the closure of a prime $\\mathfrak q$ is the set of primes that contain $\\mathfrak q$.\nSo $\\mathfrak p \\le \\mathfrak q$ holds exactly when $\\mathfrak q \\subseteq \\mathfrak p$.\n\nNon-example: on $X = \\set{a, b}$ with the indiscrete topology, $a \\le b$ and $b \\le a$ but $a \\ne b$.\nSo the specialization preorder is not a partial order, and indeed $X$ is not T0.\n\nSome authors use the opposite convention, with $x \\le y$ when $y \\in \\overline{\\set{x}}$.\n",
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