I know that in $\mathbb R^n$, the collection of open sets $\{(Z(\{f\}))^C: f\in k[x_1,x_2,\dots, x_n]\}$ is a basis for the Zariski topology in $\mathbb R^n$. When $n=1$, I could find a subbasis for the Zariski topology as follows: Since each polynomial $f:\mathbb R\to \mathbb R$ as at most finitely many roots in $\mathbb R$, then let $x_1,x_2,\dots, x_n$ be the real roots of $f$. It is easy to see that $(Z(\{f\}))^c=(Z(\{f_1\}))^c\cap (Z(\{f_2\}))^c\cap \dots (Z(\{f_n\}))^c $, where each of $f_1,f_2,\dots,f_n$ is a linear polynomial with root $x_1,x_2,\dots, x_n$ respectively. Hence $\{(Z(\{f\}))^c: f \text{ is a linear polynomial}\}$ is a subbasis for the Zariski topology on $\mathbb R$.
My question is that if $n\ge 2$, it is possible that a polynomial $f:\mathbb R^n\to \mathbb R$ may have infinitely many roots, so my construction for a subbasis in $\mathbb R$ cannot be applied. How should we construct a subbasis in this case? Do we just take $\{(Z(\{f\}))^C: f\in k[x_1,x_2,\dots, x_n]\}$ as a subbasis?