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Suppose $\Omega_1$ and $\Omega_2$ are two disjoint unbounded domains in $\mathbb{R}^{2n}$, $n \in \mathbb{N}$. Can there be conditions on $\Omega_1$ and $\Omega_2$ so that these two domains can be separated by the zero set of a polynomial $p(x)$ such that $p|_{\Omega_1}>0$ and $p|_{\Omega_2}<0$ ? Is there any reference to any such result ?

Is there any such result for complex setting i.e. can such domains in $\mathbb{C}^{n}$ be separated by a complex algebraic variety ?

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    $\begingroup$ Your question about complex varieties has a trivial answer: it is an easy exercise to show that the complement of a complex affine variety is connected. $\endgroup$ Commented Jun 5 at 23:26

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One sufficient condition is when $\Omega_1$ and $\Omega_2$ are bounded: Suppose $\Omega_1,\Omega_2\subset\Bbb{R}^m$ are such that $\overline{\Omega_1}$ and $\overline{\Omega_2}$ are disjoint, bounded subsets of $\Bbb{R}^m$. By the Urysohn lemma, there exists a continuous function $f:\Bbb{R}^m\rightarrow\Bbb{R}$ satisfying $f\restriction_{\overline{\Omega_1}}\,\equiv 1$ and $f\restriction_{\overline{\Omega_2}}\,\equiv -1$. Applying the Stone-Weierstrass theorem, there exists a polynomial $p$ such that $|p(x)-f(x)|<\frac{1}{2}$ for every $x\in \overline{\Omega_1}\cup\overline{\Omega_2}$. Thus $p(x)>\frac{1}{2}$ for $x\in\Omega_1$ while $p(x)<-\frac{1}{2}$ for $x\in\Omega_2$.

An immediate corollary: $\Omega_1,\Omega_2\subset\Bbb{R}^m$ can be separated by a polynomial if there exists a polynomial map $F:\Bbb{R}^m\rightarrow\Bbb{R}^k$ such that $F(\Omega_1),F(\Omega_2)$ are bounded subsets of $\Bbb{R}^k$ with disjoint closures.

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  • $\begingroup$ This is a great answer. Unfortunately, our domains are unbounded. I have changed that in the main post. $\endgroup$ Commented Jun 6 at 23:55
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I know two very basic facts about the growth of polynomial functions, which put mild restrictions on such polynomial-separable pairs of domains.

One reference I know is Asymptotic properties of polynomials and algebraic functions of several variables by E. A. Gorin https://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=rm&paperid=6566; beware, English translation has several misprints.

Theorem (folklore). For any real polynomial $f$ the function $\mu_f(r) := min_{|x| = r}f(x)$ is a piecewise algebraic* function, which has a form $Cr^{\frac m n}(1 + o(1))$ for some $C \in \Bbb R_+, m, n \in \Bbb Z$.

It follows quite easily from Tarski–Seidenberg quantifier elimination.

Theorem (Hörmander). Let $M$ be the zero set of a polynomial $f$, and denote by $d_N(x)$ the distance from $x$ to $M$; if polynomial has no zeroes, let $d_{\emptyset}(x) = 1$. Then $$|f(x)| \geq C(1 + |x|^2)^{\alpha}d_N(x)^{\beta}$$ for some constants $\alpha, \beta$.


There are likely far better results which can be obtained from various versions of qualitative Hartogs' theorem describing properties of domains $\Omega \subset \Bbb C^n$ such that $\mathcal O(\Bbb C^n) \to \mathcal O(\Omega)$ is an isomorphism.

* A function $\Bbb R^m \to \Bbb R^n$ is piecewise algebraic if its graph is a finite union of intersections of algebraic submanifolds of $\Bbb R^{m+n}$ with boxes — products of intervals and rays.

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  • $\begingroup$ I am not sure. How can this result of Hormander be used to separate two unbounded domains by a real algebraic variety ? $\endgroup$ Commented Jun 12 at 19:38
  • $\begingroup$ @SoumyaGanguly Of course, it can only be used to show that something can NOT be separated. $\endgroup$
    – Denis T
    Commented Jun 12 at 19:57
  • $\begingroup$ I am not sure I follow your argument. Can you please explain a bit more if possible ? $\endgroup$ Commented Jun 12 at 20:00

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