Maximal ideals of $C\big((0,1)\big)$ We know that all the maximal ideals of $C([0,1])$ have the following form
$$M_a=\big\{f\in C[0,1]\ :\ f(a)=0\ \big\}\ \text{ for some } a\in [0,1]\,.$$
But suppose that if we will replace our domain from a compact to an open set. Then this is not true. I give an example but I'm unable to prove it.
Consider the set $$ I=\big\{f\in C\big((0,1)\big)\ :\ f\big(\tfrac{1}{n}\big)=0\ \ \text{for all  but finitely many }n \big\} $$
Please help me to prove that it is a maximal ideal of $C\big((0,1)\big)$.
 A: Clearly $\;I\;$ is proper, and for example $\;\sin\frac\pi x\in I\;$ Now, fix some $\;m\in\Bbb N\;$ ,  and define
$$g_m(x):=\begin{cases}\sin\frac\pi x,\,&x\le\frac1m\\{}\\x-\frac1m,\,&x>\frac1m\end{cases}$$
Observe that $\;g_m(x)\;$ is continuous everywhere ( in particular, at $\;x=\frac1m\;$), and observe that
$$g_m(x)=0\iff x=\frac1n\le\frac1m\iff n\ge m\;,\;\;$$
$$\text{ but}\,\;\;g_m(x)=x-\frac1m\neq0\,\;\text{for}\;\,x>\frac1m$$
so that $\;g_m(x)\in I\;$
Now, given any maximal ideal of the form (your notation) $\;M_a\;$, if we choose $\;m\;$ big enough -- say, $\;m\in\Bbb N\;$ such that $\;m>\frac1a\iff\frac1m<a\;$ -- then as we saw above $\;g_m(a)\neq0\;$ ,and thus $\;g_m(x)\notin M_a\;$ .
Finally, we use Zorn's Lemma to remember that $\;I\le M\;$ , for some maximal ideal $\;M\le C((0,1))\;$, and by the above $\;M\neq M_a\;$ for any $\;a\in (0,1)\;$ .
I can't remember right now whether I saw at anytime that $\;I\;$ as above is maximal, though. The above shows, nevertheless, that without compactness we cannot have as nice a form for maximal ideals as with compactness.
