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  "body": "\n## Definition\n\nLet $M$ be a [[real analytic manifold|Z7H98E1NHP]], for example an open subset of $\\R^n$.\nA subset $X \\subseteq M$ is *semianalytic* if every point $p \\in M$ has an open [[neighbourhood|0ZCY0Z80B6]] $U$ with the following property.\nThe set $X \\cap U$ is a finite union of sets of the form\n$$\\set{x \\in U : f(x) = 0,\\ g_1(x) \u003e 0, \\dots, g_r(x) \u003e 0},$$\nwhere $r \\ge 0$ and $f, g_1, \\dots, g_r$ are [[real analytic functions|081WKEDK6C]] on $U$.\n\n## Remarks\n\nThe condition is required at every point of $M$, not only at points of $X$.\nEquivalently, $X \\cap U$ belongs to the smallest family of subsets of $U$ that contains the sets $\\set{g \u003e 0}$ for real analytic $g$ on $U$.\nHere the family is required to be closed under finite unions, finite intersections and complements.\nSeveral equations $f_1 = \\cdots = f_s = 0$ can be merged into the single equation $f_1^2 + \\cdots + f_s^2 = 0$, and $f = 0$ gives no equation at all.\nExample: $X = \\set{(x, y) \\in \\R^2 : y \u003e 0,\\ x^2 + y^2 \u003c 1}$ is semianalytic, with $U = \\R^2$, $f = 0$, $g_1 = y$ and $g_2 = 1 - x^2 - y^2$.\nNon-example: $X = \\set{0} \\cup \\set{1/n : n \\ge 1}$ is not semianalytic in $\\R$.\nLet $U$ be any open interval about $0$, and $f, g_i$ real analytic on $U$.\nIf $f$ vanishes at infinitely many of the points $1/n$, then $f = 0$ on $U$ by the identity theorem; otherwise $\\set{f = 0}$ is finite near $0$.\nSo each set $\\set{f = 0, g_i \u003e 0}$ is, near $0$, either finite or a union of finitely many open intervals, and a finite union of such sets cannot equal $X$ near $0$.\nEvery semialgebraic subset of $\\R^n$ and every real analytic subset of $M$ is semianalytic.\n",
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