By this question, we know that a non-trivial affine variety in $\mathbb{C}^n$ has empty interior. But the argument uses the (strong) fact that a holomorphic function vanishing in a non empty set $U$ vanishes everywhere. Therefore, my question is:
Is this result valid for varieties in $\mathbb{R}^n$?
Thank you in advance!
Yes.
Let $f\in\mathbb R[X_1,\ldots,X_n]$ be a polynomial that vanishes on an open subset of $\mathbb R^n$, wlog. an open neighbourhood of $0$. Then for any $(a_1,\ldots,a_n)\in\mathbb R^n$ the polynomial $g(T)=f(a_1T,\ldots, a_nT)\in\mathbb R][T]$ vanishes in a neighbourhood of $0$, hence in infinitely many points, hence is the zero polynomial. Now assume $c X_1^{k_1}\cdots X_n^{k_n}$ is a nonzero monomial of $f$ and is lexically maximal (i.e., has maximal degree in $X_1$, among those of degree $k_1$ in $X_1$ has maximal degree in $X_2$, and so on). Then for a suitable choice with $a_1\gg a_2\gg\ldots \gg a_n>0$, this monomial will not cancel against the other monomials of total degree $k_1+\ldots +k_n$ when we form $g(T)$, that is, $g$ will contain a nonzero monomial of degree $k_1+\ldots +k_n$, contradiction.
