A set is not semialgebraic

Question

A subset $A$ of $\mathbb R^n$ is called semi-algebraic if it can be represented as a finite union of sets of the form \begin{equation*} \{x\in \mathbb R^n\; |\; p_i(x)=0, q_i(x)<0\; \mbox{for all }i=1, \ldots, m\}, \end{equation*} where $p_i$ and $q_i$ for $i=1,\ldots, m$ are polynomial functions on $\mathbb R^n$.

I know that the projection of a semialgebraic set in $\mathbb R^{n+1}$ to $\mathbb R^n$ is also a semialgebraic set (Tarski - Seidenberg Theorem). Examples of semialgebraic sets seem to be easy to find, but the opposite seems to be harder. I try to prove that the following set $$A=\{(x,y)\in \mathbb R^2: y=\sin(x)\}$$ is not semialgebraic. But the difficulty is that I can not show that $A$ can not be of the above form or can not to be the projection of any semialgebraic set in $\mathbb R^{3}$.

How can I continue?

Answer 1

Lemma. Let $p\in\mathbb R[x,y]$ be a polynomial and let $T\subset \mathbb R$ be an infinite set with an accumulation point $a$. Assume $p(t,\sin t)=0$ for all $t\in T$. Then $p$ is the zero polynomial.

Proof. Because $t\mapsto p(t,\sin t)$ is analytic, it follows that $p(t,\sin t)=0$ for all $t\in\mathbb R$. Write $$p(x,y)=\sum_{k=1}^n x^kr_k(y)$$ with $r_k\in\mathbb R[y]$. For fixed $t\in\mathbb R$ the polynomial $$\sum_{k=1}^n r_k(\sin t)\cdot x^k\in\mathbb R[x]$$ has infinitely zeroes, namely at $x=2n\pi+t$ with $n\in\mathbb Z$, hence must be the zero polynomial, i.e. all $r_k(\sin t)=0$. Then $r_k$ is identically zero on $[-1,1]$ and hence is the zero polynomial. So ultimately $p$ is also the zero polynomial. $_\square$

Now assume $A=\{\,(x,\sin x)\mid x\in\mathbb R\,\}$ is semi-algebraic. Each point $a\in A$ must be an accumulation point for one of the finitely many sets $\{\,x\in\mathbb R^2\mid p_i(x)=0, q_i(x)<0,1\le i\le m\,\}$ we take the union of. By the lemma, all $p_i$ must be zero and can be ignored, i.e., $a$ is an accumulation point of $\{\,x\in\mathbb R^2\mid q_i(x)<0,1\le i\le m\,\}\subseteq A$. In other words, $A$ contains a nonempty open set, contradiction.

Answer 2

Here is a different approach to showing $A$ is not semialgebraic.

When you intersect $A$ with the semialgebraic set $\{(x,y)\in\mathbb{R}^2:y =1\}$, you get an infinite collection of points. Projecting onto the $x$-axis then also gives an infinite collection of points in the real line. If we assume that $A$ is semialgebraic, that implies that the infinite collection of points on the real line is semialgebraic, since semialgebraic sets are closed under intersections and projections. But it is not too hard to show that any semialgebraic set in $\mathbb{R}$ can have at most finitely many connected components, so we have reached a contradiction. Thus $A$ is not semialgebraic.