Semialgebraic set

Remarks

Equivalently, the semialgebraic subsets of $\R^n$ form the smallest family of subsets that is closed under finite unions, finite intersections and complements. This family must contain all sets $\set{x : g(x) > 0}$ with $g$ a polynomial. The same definition works over any real closed field in place of $\R$.

Example: the closed unit disc $D = \set{(x, y) : x^2 + y^2 \le 1}$ is semialgebraic. It is the union of $\set{1 - x^2 - y^2 = 0}$ (with $k = 0$) and $\set{0 = 0,\ 1 - x^2 - y^2 > 0}$.

Non-example: $\Z \subseteq \R$ is not semialgebraic. A semialgebraic subset of $\R$ is a finite union of points and open intervals. This holds because a nonzero polynomial in one variable has finitely many roots and constant sign between them. But $\Z$ is infinite and discrete.