[proofplan] Both containments are direct consequences of the definitions. To prove $\mathfrak a\subset I(V(\mathfrak a))$, we fix a polynomial in the ideal and check that it vanishes on every point where all elements of the ideal vanish. To prove $X\subset V(I(X))$, we fix a point of $X$ and check that every polynomial vanishing on $X$ also vanishes at that point. [/proofplan] [step:Unpack the two vanishing constructions] Let $R=k[x_1,\dots,x_n]$. For an ideal $\mathfrak a\trianglelefteq R$, the affine vanishing set of $\mathfrak a$ is \begin{align*} V(\mathfrak a)=\{a\in \mathbb A_k^n: f(a)=0 \text{ for every } f\in \mathfrak a\}. \end{align*} For a subset $Y\subset \mathbb A_k^n$, the vanishing ideal of $Y$ is \begin{align*} I(Y)=\{f\in R: f(a)=0 \text{ for every } a\in Y\}. \end{align*} In particular, $I(V(\mathfrak a))$ means the vanishing ideal of the subset $V(\mathfrak a)$, while $\mathfrak a$ denotes the fixed ideal. [/step] [step:Show that every polynomial in the ideal vanishes on its zero set] Fix $f\in \mathfrak a$. Let $a\in V(\mathfrak a)$. By the definition of $V(\mathfrak a)$, every polynomial in $\mathfrak a$ vanishes at $a$. Since $f\in \mathfrak a$, this gives \begin{align*} f(a)=0. \end{align*} Thus $f$ vanishes at every point of $V(\mathfrak a)$, so by the definition of the vanishing ideal, \begin{align*} f\in I(V(\mathfrak a)). \end{align*} Since $f\in \mathfrak a$ was arbitrary, it follows that \begin{align*} \mathfrak a\subset I(V(\mathfrak a)). \end{align*} [guided] We want to prove an inclusion of sets of polynomials: \begin{align*} \mathfrak a\subset I(V(\mathfrak a)). \end{align*} By the definition of subset inclusion, it is enough to start with an arbitrary element $f\in \mathfrak a$ and prove that $f\in I(V(\mathfrak a))$. Now $f\in I(V(\mathfrak a))$ means exactly that $f$ vanishes on every point of $V(\mathfrak a)$. So let $a\in V(\mathfrak a)$. The definition of $V(\mathfrak a)$ says that $a$ is a common zero of all polynomials belonging to $\mathfrak a$. Since the polynomial $f$ was chosen from $\mathfrak a$, the defining condition gives \begin{align*} f(a)=0. \end{align*} Because this holds for every $a\in V(\mathfrak a)$, the polynomial $f$ lies in the vanishing ideal of $V(\mathfrak a)$: \begin{align*} f\in I(V(\mathfrak a)). \end{align*} Since the original choice of $f\in \mathfrak a$ was arbitrary, we have proved \begin{align*} \mathfrak a\subset I(V(\mathfrak a)). \end{align*} [/guided] [/step] [step:Show that every point of the subset satisfies its vanishing ideal] Fix $a\in X$. Let $f\in I(X)$. By the definition of $I(X)$, the polynomial $f$ vanishes at every point of $X$. Since $a\in X$, this gives \begin{align*} f(a)=0. \end{align*} Thus every polynomial in $I(X)$ vanishes at $a$, so by the definition of the vanishing set, \begin{align*} a\in V(I(X)). \end{align*} Since $a\in X$ was arbitrary, it follows that \begin{align*} X\subset V(I(X)). \end{align*} Together with the first containment, this proves both assertions. [/step]