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

Question

Problem:

Consider $X$ a compact Hausdorff space, with $C(X)$ the set of continuous functions on $X$. If $A$ is closed subset of $X$, define $I_{A}=\{f \in C(X)| f|_{A} = 0 \}$, and $Z(I):=\{x \in X|f(x)=0\ $ for all $f \in I \}$.

$1$ )Show that $Z(I_{A})=A$.

First if $x \in A$, then for every $f \in I_{A}, f(x)=0$, so $x \in Z(I_{A})$. The other inclusion is confusing me (it could be false). Is it not possible for an ideal of functions on $C(X)$ to be zero on $A$ but also on other sets? Hint appreciated. I know that $I_{A}$ is closed with respect to the norm topology (the sup norm).

Answer 1

For any point $x\notin A$, the sets $\{x\}$ and $A$ are closed and can be separated by a continuous function

Answer 2

If $x \notin A$, then there is by Urysohn's lemma a continuous function $f:X \to [0,1]$ such that $f(x) = 1$ and $f[A] = \{0\}$. Then $f \in I_A$ (as $f|A \equiv 0$) and $x \notin Z(I_A)$ is witnessed by this $f$.

This shows that $Z(I_A) \subseteq A$ by contraposition.

$A \subseteq Z(I_A)$ is true basically by definition: $x \in A$ and for all $f\in I_A$ we have $f(x)=0$ (by definition of $I_A$), so that $x \in Z(I_A)$, by that set's definition. This you already saw.