Analysis of the ideals of $C[0,1]$

Question

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 $C[0,1]$ be equipped with the sup metric and $[0,1]$ be equipped with the usual euclidean metric . I am trying to show the following , please help .

(i) $I(A)$ is closed in $C[0,1]$ for every $A \subseteq [0,1]$

(ii) $Z(I)=Z(\bar I)$

(iii) $I(Z(I))=\bar I$

where $\bar I$ denotes the closure i.e. the set of all adherent points of $I$ . Can someone please provide some link for some material where I can get these things . Thanks in advance

Answer 1

Hints:

(i) If $\forall x\in A: f_n(x)=0$, what happens with $f=\lim f_n$?

(ii) The nontrivial inclusion is $Z(I)\subset Z(\bar I)$. Read again (i).

(iii) Any $f\in\bar I$ is an uniform limit of $f_n\in I$.