Is a square a semi-algebraic set?

Question

Consider the square $[-1,1]^2$ in $\mathbb R^2$. Is this set semi-algebraic?

Answer 1

Yes. For instance, it may be written as $$\{-x_1+1>0\land x_1+1>0\land -x_2+1>0\land x_2+1>0\}\cup \\ \cup\{x_1+1=0\land x_2+1>0\land -x_2+1>0\}\cup\\\cup \{x_1-1=0\land x_2+1>0\land -x_2+1>0\}\cup\\\cup\{x_2+1=0\land x_1+1>0\land -x_1+1>0\}\cup\\ \cup\{x_2-1=0\land x_1+1>0\land -x_1+1>0\}\cup\\\cup \{x_1+1=0\land x_2+1=0\}\cup\\ \cup \{x_1-1=0\land x_2+1=0\}\cup\\\cup \{x_1+1=0\land x_2-1=0\}\cup\\\cup \{x_1-1=0\land x_2-1=0\}$$

Answer 2

While Gae. S.'s answer is complete and correct, I think you might be able to learn a little more about semialgebraic sets with this question.

Here are a list of basic properties of semialgebraic sets:

Any algebraic set (zero of any collection of polynomials) is semialgebraic.

The semialgebraic subsets of $\Bbb R^n$ form a Boolean algebra of the powerset of $\Bbb R^n$ (they're closed under complements, finite unions, and finite intersections).

The product of semialgebraic sets $A\subset \Bbb R^n$ and $B\subset \Bbb R^m$ is a semialgebraic subset $A\times B\subset \Bbb R^{n+m}$.

For the projection $\Bbb R^n\times \Bbb R\to\Bbb R^n$, the image of any semialgebraic subset is again semialgebraic.

Let's use these to build some intermediate results and some intuition that make this a little easier to understand.

Lemma 1: Any subset $\{x\in\Bbb R^n \mid f(x) > 0\}$ where $f$ is a polynomial is semialgebraic.

Proof: If $f(x)> 0$, then there is a $t\in\Bbb R$ so that $f(x)t^2=1$. So $\{x\in\Bbb R^n \mid f(x) > 0\}$ is the projection of $\{(x,t)\in\Bbb R^n\times \Bbb R \mid f(x)t^2 =1\}$.

Corollary: For any polynomial $f$, any subset $\{x\in\Bbb R^n\mid f(x)\ ?\ c\}$ where $?$ is any one of $>,\geq,\leq, <$ is semialgebraic.

Proof: To interchange $\leq$ or $<$ and $\geq$ or $>$, swap $f$ for $-f$. To get to $\geq$ from $>$, take the union of $\{x\in\Bbb R^n \mid f(x) > 0\}$ with $\{x\in\Bbb R^n \mid f(x) = 0\}$.

Corollary: $\{x\in\Bbb R^n \mid a \leq f(x) \leq b\}$ is semialgebraic for any real numbers $a,b$.

Proof: Apply the previous corollary, adjusting $f$ as necessary.

These results imply that your square is semialgebraic in several different ways.

Proof 1: $[-1,1]^2\subset \Bbb R^2 = \{(x,y)\in\Bbb R^2\mid -1\leq x\leq 1\}\cap \{(x,y)\in\Bbb R^2\mid -1\leq y\leq 1\}$.

Proof 2: $[-1,1]^2\subset \Bbb R^2 = \{x \in\Bbb R\mid -1\leq x\leq 1\}\times\{x \in\Bbb R\mid -1\leq x\leq 1\}$

etc...

I hope this helps you develop better intuition for when you can write things as semialgebraic sets. For futher reading, I like Michel Coste's notes and this paper of Denkowska and Denkowski.