Zariski Topology: Prove that $V(I(X))=\bar{X}$ I am studying Zariski Topology. Here is a problem I am trying to work on:
Let $X\subset A^n$ be an arbitrary subset. Prove that $V(I(X))=\bar{X}$. 
This is my work so far:
(1) Since $\bar{X}$ is the intersection of all varieties containing $X$, it is a variety. So $V(I(\bar{X}))=\bar{X}$. 
(2) Now $X\subset\bar{X}$, so $V(I(X))\subset V(I(\bar{X}))$. 
(3) It remains to show that $\bar{X} \subset V(I(X))$. 
(4) For any $x\in \bar{X}$, it is in all varieties that contains $X$. Specifically, $V(I(X))$ is a variety that contains $X$, so we are done. 
Is the proof correct? I am especially not sure about step (b). Please help. Thanks!

If $X$ is contained in $V(J)$ then any function in $J$ vanishes on $X$ so it is in $I(X)$.  In fact we have the equivalence:
$$ X \subset V(J) \iff  J \subset I(X)$$
Therefore amongst the $J$ so that $V(J) \supset X$ there exists a largest one, $I(X)$.  This will give the smallest closed subset $V(I(X))$  containing $X$
