# property of an affine algebraic set in Zariski topology

The following problem is giving me trouble:

• Suppose $X \subset \mathbb{A}^{n}$ is an affine algebraic set, and $S \subset X$ is a subset. Show that if $\bar{S}$ is the closure of $S$ in the Zariski topology, then $\bar{S}= V(I(S))$.

I have no idea where to start.

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Remember that the closure of a set $S$ is the smallest closed set containing $S$ –  Jonathan May 3 '12 at 19:46
For the inclusion $\overline S \subseteq VI(S)$, note that it suffices to show that $S \subseteq VI(S)$ and that $VI(S)$ is closed. By definition of the Zariski topology, $VI(S)$ is closed, so we only have to show $S \subseteq VI(S)$. Let $P \in S$. If $f$ is a polynomial in $I(S)$ it vanishes at every point of $S$, in particular $f(P) = 0$. This shows the first inclusion.
For the converse, remember that $I$ and $V$ reverse inclusions, i.e. if $S_1 \subseteq S_2$ are subsets of $X$ then $I(S_2) \subseteq I(S_1)$, and if $A_1 \subseteq A_2$ are subsets of $K[X_1,\ldots,X_n]$ then $V(A_2) \subseteq V(A_1)$. Since $\overline S$ is closed, there is some ideal $A \subseteq K[X_1,\ldots,X_n]$ s.t. $\overline S = V(A)$. Applying $I$ to the inclusion $S \subseteq V(A)$ yields $IV(A) \subseteq I(S)$. It is clear that $A \subseteq IV(A)$, hence we have $A \subseteq I(S)$. Applying $V$, we finally get $VI(S) \subseteq V(A) = \overline S$.