Proposition 3

Let $f(x) = f(x_1, x_2, \ldots, x_n)$ be a polynomial with real coefficients, not identically zero, defined on $\mathbf{R}^n$. Then the complement of the set $\overline{f}(0)$ is dense in $\mathbf{R}^n$.

Let $x$ be any point of $\mathbf{R}^n$ and let $y \in \mathbf{R}^n$ be such that $f(y) \neq 0$; $\varphi(t) = f(x + t(y - x))$ is a polynomial in the real variable $t$, not identically zero; hence there exist arbitrarily small values of $t$ such that $\varphi(t) \neq 0$. This shows that $x$ lies in the closure of the complement of $\overline{f}(0)$.