My question is on the topological closure of any subset $X$ of the affine space $\mathbb{A}_k^n$. I tried to prove that $\overline{X} = Z(I(X))$. The attempt goes as follows:
Notice that we have the inclusions $$ X \subset \overline{X} \implies I(\overline{X}) \subset I(X) \implies Z(I(X)) \subset Z(I(\overline{X})), $$ and, also $$ X \subset Z(I(X)) \subset Z(I(\overline{X})) = \overline{X}, $$ since $\overline{X}$ is closed. So we just need to show that $Z(I(X)) \supset Z(I(\overline{X}))$. Take $(a_1,\dots,a_n) \in Z(I(\overline{X}))$, then $f(a_1,\dots,a_n) = 0$ for all $f \in I(\overline{X})$, id est, $f(a_1,\dots,a_n) = 0$ for all $(a_1,\dots,a_n) \in \overline{X}$. Since $X \subset \overline{X}$, we have $f(a_1,\dots,a_n) = 0$ for all $(a_1,\dots,a_n) \in X$, then $f(a_1,\dots,a_n) = 0$ for all $f \in I(X)$, ergo $(a_1,\dots,a_n) \in Z(I(X))$. $\blacksquare$
What I am suspicious about is that i did not use the fact that $\overline{X}$ is the smallest closed set containing $X$, because of that I am almost certain it is not correct (I'm specially suspicious about the $f(a_1,\dots,a_n) = 0$ for all $f \in I(X)$ part).
Can you show me where I got wrong and how I could prove the equality $Z(I(X))=Z(I(\overline{X})$? Also, this is probably the hard way of solving it, if there is an easier way, could you explain briefly?
Thank you very much!
I just figured the easier way: Since $Z(I(X))$ is a closed set containing $X$, then $\overline{X} \subset Z(I(X))$. $\blacksquare$
But, still, is it possible to show that $Z(I(\overline{X})) = Z(I(X))$ directly?