I am trying to prove that the product of semialgebraic sets is semialgebraic.
If $X \in \mathbb{R}^{n}$ and $Y \subset \mathbb{R}^{m}$ are semialgebraic, I can't see the polynomial conditions involved on the representation of $X \times Y$. My first idea was taking the product of the polynomials $P_{i,j} s_{i,j} 0$ of X and $Q_{i,j} s'_{i,j} 0$ of Y, but it doesn't seem to make sense. I have no clue of how it goes, and I have only the definition and the characterization of semialgebraic sets as the minimal collection of sets that contains $x \in \mathbb{R}^{n} : P(x) > 0 \}$ where $P$ is a polynomial and is closed under finite intersections, unions and complements.
A semi-algebraic set is not necessary described by polynomials, but it's boolean combination of sets described by polynomials ( equalities and inequalities ) ( see http://gcomte.perso.math.cnrs.fr/M2/CosteIntroToSemialGeo.pdf ).
Take $X\subset \mathbb{R}^n$, $Y\subset \mathbb{R}^m$ semi-algebraic, you want to prove that $X\times Y$ is semi-algebraic. You can go stup by stup:
Stup1: it's enough to assume that $X=\{x\in \mathbb{R}^n \; : \; P(x)=0 \; and \; P_1(x)>0 \; and \; ...\; and \; P_l(x)>0 \}$ and $Y=\{y\in \mathbb{R}^m \; : \; Q(y)=0 \; and \; Q_1(y)>0 \; and \; ...\; and \; Q_p(y)>0 \}$, where $P,P_1 ,..., P_l \in \mathbb{R}[x]$ and $Q,Q_1 ,..., Q_p \in \mathbb{R}[y]$. ( you can using properties of the Cartesian product).
Stup2: Note that if $f:\mathbb{R}^k \longrightarrow \mathbb{R}$ is a semi-algebraic map ( means that the graph of $f$ is a semi-algebraic set ), then the sets $\{z \; : \; f(z)=0 \}$, $\{z \; : \; f(z)>0 \}$, and $\{z \; : \; f(z)<0 \}$ are semi-algebraic.
Stup3: the map $\mid \; \mid :\mathbb{R}\longrightarrow \mathbb{R}$, $x\mapsto \mid x \mid$ is semi-algebraic ( easy to show ), and therefore any set described by composition of semi-algebraic maps with $\mid \; \mid$ is semi-algebraic.
Stup4: the $Min$ map is semi-algebraic, indeed:
$Min(z,w)=\frac{z+w- \mid z-w \mid}{2}$.
Stup5: $X\times Y$ can be described by semi-algebraic functions as the following:
$X\times Y= \{(x,y)\in \mathbb{R}^{n+m} \; : \; P^2 (x)+Q^2 (y)=0 \; and \; Min(P_1 (x),...,P_l (x),Q_1 (y),...,Q_p (y))>0 \}$.
Hence, $X\times Y$ is a semi-algebraic set.