Let $X$ a locally compact Hausdorff space, $Y\subseteq X$ a closed subset and $I_Y:=\{f\in C_0(X): f_{|Y}=0\}$.
a)Let $U=X\setminus Y$. Prove that $I_Y\cong C_0(U)$.
b)For every closed ideal $I$ in $C_0(X)$ exists exactly one closed subset $Y\subset X$ such that $I=I_Y$.
for a) I want to prove that the map $\phi:I_Y\to C_0(U),\; f\mapsto f_{|U}$ is an $\ast$-isomorphism of algebras (well-defined, linear, multiplicative, bijective, preserves involution). Here I'm stuck to prove that $\phi$ is well-defined. Everything else was ok.
My try: Let $f\in I_Y$, this means:f continuous on $X$, for all $\epsilon >0$ exists $K\subseteq X$ compact such that $|f(x)|<\epsilon$ for all $x\in X\setminus K$ and $f_{|Y}=0$.
For given $\epsilon >0$, I have to define a compact setz $L\subset U$ such that $|f(x)|<\epsilon$ for all $x\in U\setminus L$. How to define $L$?
I know that $Y\subseteq f^{-1}(\{0\})\subseteq f^{-1}(B_\epsilon(0))$ and therefore $f^{-1}(\{\lambda\in\mathbb{C}: |\lambda |\ge \epsilon\})=:M\subseteq U$. I think I need $K$ and $M$ to define $L$, but I don't know, how to choose $L$.
for b) Let $I\subseteq C_0(X)$ be a closed ideal. Define $Y=\{x\in X:f(x)=0\;\text{for all}\;f\in I\}=\bigcap\limits_{f\in I}f^{-1}(\{0\})$. $Y$ is closed in $X$. Consider $I_Y=\{f\in C_0(X): f_{|Y}=0\}$, it is $I\subseteq I_Y$. But how to prove $I_Y\subseteq I$? I think it's similar to the unital case, isn't it?
regards
For a), note that the intersection $M\cap K$ is compact and contained in $U$.