Show that $I_{Z(I)}=I$

Question

Problem

Let $X$ be a compact Hausdorff space with $C(X)$ the ring of continuous functions on $X$ mapped to $\Bbb R$. If $A$ is closed subset of $X$, define

$$I_{A}=\{f \in C(X)| f|_{A} = 0 \}.$$

If $I\subseteq C(X)$ is an ideal, define

$$Z(I):=\{x \in X|f(x)=0\ \text{for all } f \in I \}.$$

Given a proper, non-zero ideal $I\subseteq C(X)$ that is closed in $C(X)$ (where we give $C(X)$ the sup-norm topology), is it true that $I_{Z(I)} = I$?

Work

If $f \in I$, then $f$ is zero for any $x \in Z(I)$, but the other direction I am not sure. If it were true, it would mean that for any function $g$ which is zero on $Z(I)$ would have to also be part of the ideal $I$. Any tips appreciated.

Answer 1

This is true and follows from a characterization of closed ideals of $C(X)$: every closed ideal is of the form $I_A$ for some closed $A$.

So $I$ in your problem = $I_A$ for some closed $A$. And $Z(I_A) = A$, which finishes it.