Let $R$ be the ring of all continuous real valued functions on the unit interval $[0,1]$ (with pointwise operations), and let $I$ be a proper ideal of $R$. Show that there exists $λ\in [0,1]$ such that $$I\subseteq M_{λ}= \left\{f \in R\mid f(λ)=0\right\}.$$
Can't really get anywhere with this one. Appreciate the help.
One way to approach this is to show directly that the intersection of all zero sets of functions in $I$ is nonempty.
Denote the zero set of $f$ as $z(f)$. Of course $z(f)$ is always closed and nonempty for $f\in I$ (If it's empty, the function is a unit.)
The next observation is that the intersection of two zero sets for functions $I$ is nonempty. If to the contrary $z(f)\cap z(g)=\emptyset$, then $z(f^2+g^2)=\emptyset$, but this isn't possible since $f^2+g^2\in I$. From this proof you can see that for every $f$ and $g$ in $I$, $f^2+g^2$ is another element of $I$ whose zero set is just $z(f)\cap z(g)$. By induction, the intersection of finitely many zero sets is again a nonempty zero set of a function in $I$.
Now I claim that in a compact space, $\cap\{z(f)\mid f\in I\}$ is nonempty. If it were empty, then the complements of the $z(f)$ form an open covering of the space. Extracting a finite subcover, we have $\cap_{i=1}^n z(f_i)=\emptyset$. But we've already said this isn't possible, by the observations in the last paragraph.
Therefore $\cap\{z(f)\mid f\in I\}\neq\emptyset$, and any $\lambda$ in this set will do the trick for your question.
Since $[0,1]$ is compact, all maximal ideals look like $M_\lambda$ (see, for example, these notes). Now your question follows from the fact that any ideal is contained in a maximal ideal.
If possible, let $I\not\subset M_\lambda$ for all $\lambda$. Then for each $\lambda$ there will be a function $f_\lambda\in I$ such that $f_\lambda(\lambda)\neq 0$. By continuity there is a neighborhood $V_\lambda$ of $\lambda$ where $f_\lambda$ is nonzero. These neighbourhoods cover your compact set $[0,1]$. So there are $V_1,..,V_n$ covering $[0,1]$ and the function $g=f_1^2+...+f_n^2$ (tired of typing 'lambda' over and over again) therefore does not vanish anywhere in the interval. And it is an element of $I$. Since it doesn't vanish, $\frac{1}{g}$ is an element of your ring. It follows that $1\in I$ which is a contradiction.
Here is a proof using nets: By Lemma of Zorn we can assume that $I$ is a maximal ideal and hence prime. Thus, if $f \in I$, then so is $|f|^2 = \bar{f}f$ and hence also $|f| \in I$.
Define a direction $\preceq$ on $I$ by $f \preceq g$ iff $|f| \leq |g|$ pointwise. Then $(I,\preceq)$ is a directed set, since if $f,g \in I$, then $|f|+|g| \in I$ and $f,g \preceq |f|+|g|$. Now, for each $f \in I$, there is some $x_f \in [0,1]$ such that $f(x_f) = 0$, because otherwise $I$, containing an invertible element, could not be maximal. Then $(x_f)_{f \in I}$ is a net and since $[0,1]$ is compact, it has a cluster point $x \in [0,1]$, i.e. given any open set $U$ with $x \in U \subseteq [0,1]$ and any fixed $g \in I$, we can find some $f \in I$ with $g \preceq f$ and $x_f \in U$. We now claim: $I = M_x$. Indeed, if there is some $g \in I$ with $g(x) \neq 0$, then there is some open set $U$ with $x \in U$ and $g \neq 0$ on $U$. Now take $f$ as previously stated, then $$ 0 = |f(x_f)| \geq |g(x_f)|\,, $$ contradicting the fact that $x_f \in U$.
