[proofplan] We prove equality by mutual inclusion. The inclusion $V((S)) \subset V(S)$ follows from the containment $S \subset (S)$. Conversely, if a point vanishes on every polynomial in $S$, then every polynomial in the generated ideal vanishes at that point because it is a finite $R$-linear combination of elements of $S$. [/proofplan] [step:Use the containment $S \subset (S)$ to get the first inclusion] Set $R := k[x_1, \ldots, x_n]$. Let $a \in V((S))$. By definition of vanishing set, $g(a)=0$ for every $g \in (S)$. Since $S \subset (S)$ by definition of the ideal generated by $S$, every $f \in S$ satisfies $f(a)=0$. Therefore $a \in V(S)$, and hence \begin{align*} V((S)) \subset V(S). \end{align*} [/step] [step:Evaluate finite ideal combinations at a point of $V(S)$] Let $a \in V(S)$. We prove that $a \in V((S))$. Let $g \in (S)$. By definition of the ideal generated by $S$, there exist a finite index set $E$, polynomials $f_i \in S$ for each $i \in E$, and polynomials $h_i \in R$ for each $i \in E$, such that \begin{align*} g = \sum_{i \in E} h_i f_i. \end{align*} Evaluating at $a$ gives \begin{align*} g(a) = \sum_{i \in E} h_i(a) f_i(a). \end{align*} Since $a \in V(S)$, each $f_i(a)=0$. Therefore \begin{align*} g(a) = \sum_{i \in E} h_i(a) \cdot 0 = 0. \end{align*} Thus every $g \in (S)$ vanishes at $a$, so $a \in V((S))$. [guided] Let $a \in V(S)$. The goal is to prove that $a$ also lies in $V((S))$, which means that every polynomial in the ideal generated by $S$ vanishes at $a$. Take an arbitrary polynomial $g \in (S)$. The meaning of $g \in (S)$ is that $g$ can be written as a finite $R$-linear combination of elements of $S$. Thus there exist a finite index set $E$, polynomials $f_i \in S$ for each $i \in E$, and polynomials $h_i \in R$ for each $i \in E$, such that \begin{align*} g = \sum_{i \in E} h_i f_i. \end{align*} Now evaluate this polynomial identity at the point $a \in \mathbb{A}^n_k$. Polynomial evaluation respects addition and multiplication, so \begin{align*} g(a) = \sum_{i \in E} h_i(a) f_i(a). \end{align*} Because $a \in V(S)$, every polynomial in $S$ vanishes at $a$. In particular, for each $i \in E$, the polynomial $f_i \in S$ satisfies $f_i(a)=0$. Substituting this into the displayed expression gives \begin{align*} g(a) = \sum_{i \in E} h_i(a) \cdot 0 = 0. \end{align*} The polynomial $g \in (S)$ was arbitrary, so every element of $(S)$ vanishes at $a$. By the definition of vanishing set, this proves $a \in V((S))$. [/guided] [/step] [step:Conclude equality by mutual inclusion] The previous steps prove both inclusions \begin{align*} V((S)) \subset V(S) \end{align*} and \begin{align*} V(S) \subset V((S)). \end{align*} Therefore \begin{align*} V(S)=V((S)). \end{align*} [/step]