Semianalytic set

Remarks

The condition is required at every point of $M$, not only at points of $X$. Equivalently, $X \cap U$ belongs to the smallest family of subsets of $U$ that contains the sets $\set{g > 0}$ for real analytic $g$ on $U$. Here the family is required to be closed under finite unions, finite intersections and complements. Several 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. Example: $X = \set{(x, y) \in \R^2 : y > 0,\ x^2 + y^2 < 1}$ is semianalytic, with $U = \R^2$, $f = 0$, $g_1 = y$ and $g_2 = 1 - x^2 - y^2$. Non-example: $X = \set{0} \cup \set{1/n : n \ge 1}$ is not semianalytic in $\R$. Let $U$ be any open interval about $0$, and $f, g_i$ real analytic on $U$. If $f$ 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 $0$. So each set $\set{f = 0, g_i > 0}$ is, near $0$, either finite or a union of finitely many open intervals, and a finite union of such sets cannot equal $X$ near $0$. Every semialgebraic subset of $\R^n$ and every real analytic subset of $M$ is semianalytic.