[proofplan] Both assertions are immediate order-reversal consequences of the definitions. For vanishing sets, a point satisfying all equations from the larger ideal $J$ also satisfies all equations from the smaller ideal $I \subset J$. For vanishing ideals, a polynomial vanishing on the larger set $Y$ also vanishes on the smaller set $X \subset Y$. [/proofplan] [step:Pass from the larger ideal to the smaller system of equations] Assume $I \subset J$. Let $p \in V(J)$. By definition of $V(J)$, every polynomial $f \in J$ satisfies $f(p)=0$. Since $I \subset J$, every polynomial $f \in I$ also belongs to $J$, and hence satisfies $f(p)=0$. Therefore $p \in V(I)$ by definition of $V(I)$. Since $p \in V(J)$ was arbitrary, this proves \begin{align*} V(J) \subset V(I). \end{align*} [guided] Assume $I \subset J$. To prove the set inclusion $V(J) \subset V(I)$, we start with an arbitrary point $p \in V(J)$ and prove that it also lies in $V(I)$. The definition of $V(J)$ says that $p$ is a common zero of every polynomial in $J$. Thus, for every polynomial $f \in J$, one has \begin{align*} f(p)=0. \end{align*} Now take an arbitrary polynomial $f \in I$. The hypothesis $I \subset J$ implies that this same polynomial also lies in $J$. Since all polynomials in $J$ vanish at $p$, it follows that \begin{align*} f(p)=0. \end{align*} Because this holds for every $f \in I$, the point $p$ is a common zero of all polynomials in $I$. By definition, this means $p \in V(I)$. Since the original point $p \in V(J)$ was arbitrary, we conclude \begin{align*} V(J) \subset V(I). \end{align*} [/guided] [/step] [step:Restrict vanishing from the larger set to the smaller set] Assume $X \subset Y$. Let $f \in I(Y)$. By definition of $I(Y)$, the polynomial $f \in k[x_1,\ldots,x_n]$ satisfies $f(p)=0$ for every $p \in Y$. Since $X \subset Y$, every point $p \in X$ also belongs to $Y$, so $f(p)=0$ for every $p \in X$. Therefore $f \in I(X)$ by definition of $I(X)$. Since $f \in I(Y)$ was arbitrary, this proves \begin{align*} I(Y) \subset I(X). \end{align*} [/step]