For every ideal $I$ of $C[0,1]$ , define
$Z(I):=\{x \in [0,1] :f(x)=0 , \forall f \in I\}$ and for every $A \subseteq [0,1]$ , let
$I(A):=\{f \in C[0,1] : f(x)=0 , \forall x \in A\}$ . Then clearly $I(A)$ is always an ideal . Let $[0,1]$ be equipped with usual euclidean metric . I want to show that $Z(I(A))=\bar A , \forall A \subseteq [0,1]$ . Now one
side I have done ; let $x \in \bar A$ , if $f \in I(A)$ i.e. $f \in C[0,1]$ such that $f(A)=\{0\}$ implying
$A \subseteq f^{-1} \{0\}$ , then since $f$ is continuous and $\{0\}$ is closed , so $f^{-1} \{0\}$ is closed , so $\bar A \subseteq { f^{-1} \{0\}} $ ,
hence $x \in \bar A$ implies $x \in f^{-1} \{0\}$ i.e. $f(x)=0$ , so $f(x)=0 , \forall f \in I(A)$ i.e. $x \in Z(I(A))$ ; hence
$\bar A \subseteq Z(I(A))$ . But I am having trouble in showing the other side . Let $x \in Z(I(A))$ , then
$f(x)=0 , \forall f \in I(A)$ i.e. $f(x)=0 , \forall f\in C[0,1] : f(A)=0$ , now as $f$ is continuous ,
$f(\bar A) \subseteq \bar {f(A)}=\bar{\{0\}}=\{0\}$ , so $f(\bar A)=\{0\}$ thus $f(x)=0 , \forall f\in C[0,1] : f(\bar A)=0$ , but then I
am stuck . Please help .
Suppose $x\notin \overline{A}$, then by Urysohn's Lemma, $\exists f\in C[0,1]$ such that $f \equiv 0$ on $\overline{A}$ and $f(x) = 1$. Hence, $f\in I(A)$ and $f(x) \neq 0$, so $x \notin Z(I(A))$. Thus, $Z(I(A)) \subset \overline{A}$