Let $C(X)$ be the continous function ring and $C*(X)$ be the bounded continous function ring.$Max C(X)$ consisting of all maximal ideals in $C(X)$. Question:why $Max C(X)$ and $Max C*(X)$ are compact but not hausdorff spaces?
closed as off topic by Yemon Choi, Qiaochu Yuan, Andy Putman, Kevin Walker, Andres Caicedo Oct 19 '12 at 22:05
Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.
By $MaxC(X)$ do you mean the space of maximal ideals of some algebra $C(X)$? Does $C(X)$ have a unit? It had better or else the spaces will only be locally compact! In any case the Gel'fandNaimark theorem says that there is an equivalence between the category of commutative $C^*$algebras and *homomorphisms and the category of compact hausdorff spaces and contintinuous maps. If $C(X)$ is commutative then the space of maximal ideals of $C(X)$ can also be thought of as the space of equivalence classes of irreducible representations of $C(X)$. Because $C(X)$ is commutative every representation $\psi$ is one dimensional and can be written as $\psi:C(X)\rightarrow \mathbb{C}$, with $\psi(xy) = \psi(x)\psi(y)$. Now the the space of equivalence classes of irreducible representations, which ill just call $MaxC(X)$ can be given what is called the Gel'fand topology (look it up). $MaxC(X)$ and X can be identified topologically. Each $x\in X$ gives a homomorphism $\psi_x\in MaxC(X)$ $\psi_x:C(X)\rightarrow\mathbb{C}$, $\hspace{1cm}\psi_x(f)=f(x)$ This homomorphsm is a map of $X$ onto $MaxC(X)$ I am sorry this answer is really really rushed as I am running out the door. I dont have time to check back over what I have written thoughtfully  apologies for any small mistakes. Hope it helps in any case. 

