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  "body": "\n## Definition\n\nLet $n \\ge 1$ and let $\\R[x_1, \\dots, x_n]$ be the ring of [[real polynomials|07GNX8GCYP]] in $n$ variables.\nA subset $S \\subseteq \\R^n$ is *semialgebraic* if it is a finite union of sets of the form\n$$\\set{x \\in \\R^n : f(x) = 0,\\ g_1(x) \u003e 0, \\dots, g_k(x) \u003e 0},$$\nwith $k \\ge 0$ and $f, g_1, \\dots, g_k \\in \\R[x_1, \\dots, x_n]$.\n\n## Remarks\n\nEquivalently, the semialgebraic subsets of $\\R^n$ form the smallest family of subsets that is closed under finite unions, finite intersections and complements.\nThis family must contain all sets $\\set{x : g(x) \u003e 0}$ with $g$ a polynomial.\nThe same definition works over any [[real closed field|DDY3Q3PJ6W]] in place of $\\R$.\n\nExample: the closed unit disc $D = \\set{(x, y) : x^2 + y^2 \\le 1}$ is semialgebraic.\nIt is the union of $\\set{1 - x^2 - y^2 = 0}$ (with $k = 0$) and $\\set{0 = 0,\\ 1 - x^2 - y^2 \u003e 0}$.\n\nNon-example: $\\Z \\subseteq \\R$ is not semialgebraic.\nA semialgebraic subset of $\\R$ is a finite union of points and open intervals.\nThis holds because a nonzero polynomial in one variable has finitely many roots and constant sign between them.\nBut $\\Z$ is infinite and discrete.\n",
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