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Let $C[0,1]$ be the ring of continuous real-valued functions on $[0,1]$, with addition and multiplication defined pointwise. For any subset $S$ of $C[0,1]$ let $Z(S)=\{f\in C[0,1]: f(x)=0 \text{ for all }x\in S\}$. Then which of the following statements are true?
(a) If $Z(S)$ is an ideal in $C[0,1]$ then $S$ is closed in $[0,1]$.
(b) If $Z(S)$ is a maximal ideal then $S$ has only one point.
(c) If $S$ has only one point then $Z(S)$ is a maximal ideal.

(a) Not necessary: if I take $S=(1/2,1/3)$ still $Z(S)$ is an ideal.

(b) I know that maximal ideals in $C[0,1]$ come in this way (I don't know the proof rigorously), i.e. $C_a=\{f\in C[0,1]:f(a)=0\}$ so $S$ may be finite or countable set? So I guess $(b)$ is a true statement and for the same reason $(c)$ is also true. But I will be happy if someone can explain me a bit about (b) and (c). Thank you.

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    it is not from a book just a question paper of past year exam paper2012-12-18

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To show that $C_a$ is maximal define $\phi:C[0,1]\longrightarrow\mathbb R$ with $\phi(f)=f(a)$ and then use the first isomorphism theorem.
For the converse define $\phi:C[0,1]\longrightarrow C(S), \ f \mapsto f|_S$ to show that $C[0,1]\ /Z(S)$ is isomorphic to $C(S)$.

Then use that $m \leq R$ is maximal $\iff R/m$ is a field.

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    @Kuttus: $Z(S)$ is maximal iff $S$ is an one point set.2012-12-18
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Another approach to proving maximality: If $S, T$ are two nonempty subsets of $[0,1]$ with $T \subset S$, then $Z(S) \subset Z(T)$. So $Z(S)$ is maximal if and only if it does not contain any proper subsets -- i.e., it must be a one-point set.

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    I suppose you want to say $T\subsetneq S\implies Z(S)\subsetneq Z(T)$. Well, this deserves a proof, don't you think?2016-12-29