[proofplan] Both assertions follow by unpacking the universal quantifiers in the definitions. For the first inclusion, a point of $V(J)$ is a point at which every polynomial in $J$ vanishes, and therefore every polynomial in the smaller ideal $I$ also vanishes. For the second inclusion, a polynomial in $I(Y)$ vanishes on all of $Y$, and therefore also vanishes on the smaller subset $X$. [/proofplan] [step:Use $I\subset J$ to reverse the inclusion of vanishing sets] Let $a\in V(J)$. By definition of $V(J)$, for every polynomial $f\in J$ one has \begin{align*} f(a)=0. \end{align*} Let $g\in I$. Since $I\subset J$, we have $g\in J$, so the preceding vanishing condition gives \begin{align*} g(a)=0. \end{align*} Thus every polynomial in $I$ vanishes at $a$, which means $a\in V(I)$. Since $a\in V(J)$ was arbitrary, $V(J)\subset V(I)$. [guided] We prove the inclusion $V(J)\subset V(I)$ by proving membership point by point. Let $a\in V(J)$ be an arbitrary point of the [affine space](/page/Affine%20Space) $\mathbb A_k^n$. The definition of $V(J)$ says precisely that every polynomial belonging to $J$ vanishes at $a$; in symbols, for every $f\in J$, \begin{align*} f(a)=0. \end{align*} To show that $a\in V(I)$, we must check the corresponding vanishing condition for every polynomial in $I$. So let $g\in I$ be arbitrary. The hypothesis $I\subset J$ means that membership in $I$ implies membership in $J$, hence $g\in J$. Since every element of $J$ vanishes at $a$, we get \begin{align*} g(a)=0. \end{align*} Because this holds for every $g\in I$, the definition of $V(I)$ gives $a\in V(I)$. Since the original point $a\in V(J)$ was arbitrary, every point of $V(J)$ lies in $V(I)$, so \begin{align*} V(J)\subset V(I). \end{align*} [/guided] [/step] [step:Use $X\subset Y$ to reverse the inclusion of vanishing ideals] Let $f\in I(Y)$. By definition of $I(Y)$, for every point $b\in Y$ one has \begin{align*} f(b)=0. \end{align*} Let $a\in X$. Since $X\subset Y$, we have $a\in Y$, and therefore \begin{align*} f(a)=0. \end{align*} Thus $f$ vanishes at every point of $X$, so $f\in I(X)$. Since $f\in I(Y)$ was arbitrary, $I(Y)\subset I(X)$. [/step]