maximal ideals of $C((0,1))$ For any compact set $\mathbb{K}$, set of all continuous functions from $\mathbb{K} \rightarrow \mathbb{R}$, we have all the maximal ideals of form $M_{\gamma}=\{ f: f(\gamma)=0\}$ for some $\gamma \in K$.
The problem is to show for $C((0,1))$, there exists infinitely many maximal ideals which are not of the form $M_{\gamma}$, for some $\gamma \in (0,1)$.
 Also, I want to ask, does it have any connection with compactification of (0,1)?
Please help me with this.
 A: The set $M$ you describe is an ideal, but need not be maximal. However, it must be contained in some maximal ideal $\tilde M$. We want to achieve that $\tilde M$ is not of the form $M_\gamma$. Unfortunately, we have $f(\frac12)=0$ for all $f\in M$ (because $f$ is continuous), hence certainly $M\subset M_{\frac12}$ and so possibly $\tilde M= M_{\frac12}$. However, if you adjust your definition and replace $\frac12$ with $0$ (or $1$), then $\tilde M$ cannot be equal to any $M_\gamma$ because we readily find $f\in M$ with $f(\gamma)\ne 0$.
But how do we find infinitely many different maximal ideals? By picking different sequences $x_n\to 0$, of course!
Note that any $f\in C((0,1))$ with $f(x_n)\ne 0$ for all $x_n$ is $\notin \tilde M$: 
Let $g(x)=\min\{\,|x-x_n|:n\in \Bbb N\,\}$. Then $g\in M$ and $f^2+g>0$, so that from  $f\cdot\frac f{f^2+g}=1-g$, we conclude $f\notin \tilde M$.
Hence by using pairwise disjoint sequences converging to $0$, we obtain pairwise distinct maximal ideals because we can easily find continuous functions (similar to $g$ above) that are zero on the one sequence but nowhere zero on the other sequence and vice versa.
Concretely, we find continuum-many such sequences by letting $x_n=\frac1{n+c}$ with a parameter $0<c\le 1$.
