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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?

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closed as off topic by Yemon Choi, Qiaochu Yuan, Andy Putman, Kevin Walker, Andres Caicedo Oct 19 '12 at 22:05

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1 Answer 1

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'fand-Naimark 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.

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thanks for your answer.it really helps me . –  sh gh Oct 20 '12 at 7:00

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