Let $X=(0, 1)$ be the open unit interval and $C(X, \mathbb R)$ be the ring of continuous functions from $X$ to $\mathbb R$. For any $x\in(0, 1)$, let $I(x)=\{f\in C(X, \mathbb R)|f(x)=0\}$. Then which of the following is true?
- $I(x)$ is prime.
- $I(x)$ is maximal.
- Every maximal ideal of $C(X, \mathbb R)$ is equal to $I(x)$ for some $x\in X$.
- $C(X, \mathbb R)$ is an integral domain
Option 4 isn't true, its well known. Even option 1, 2 also well known. I mean they are true. What troubles me is the option 3. Really I've no idea how to disprove option 3. I've seen option 3 true for $[0, 1]$, but why here it is not true.
Please help in answer section rather comments. It would be great achievement if I got option 3.Thanks in advance.
Claim 3. is false, as can be seen below.
Define $$ Z(g)\,:=\,\{x\in(0;1) : g(x)=0\} $$ Let
$$ \forall_{n=1}^\infty\quad f_n(x)\,:=\, \sin\left(\frac\pi{2^n\cdot x}\right) $$ so that $$ Z(f_n) \,=\, \left\{\frac1{k\cdot 2^n} :\, k\in\Bbb N\right\} $$
are non-empty and decreasing, and
$$ \bigcap_{n=1}^\infty Z(f_n)\,=\,\emptyset $$
Consider the ideal $\,J\,$ generated by the sequence of all functions $\,f_n.\ $ It is proper and not contained in any $\, I(x)\,$ while it is contained in a maximal ideal (by a very general theorem).
That's it. Great!