Let $X$ be a compact set. Suppose $A$ is a proper, closed subalgebra of $C(X)$ that separates points.
- Then $A \subseteq I_{x_0} = \{f \in C(X) : f(x_0) = 0\}$ for some unique $x_0 \in X$.
I understand this part / the proof. But the next two parts of the question confuse me.
- Show that $B = \{ f + \lambda : f \in A, \lambda \in \mathbb{R}\}$ is equal to $C(X)$.
I think I understand his if we had $\lambda \neq 0$ but the question didn’t. Was this a typo or is this true?
Edit to clarify: if we had that $g = f + \lambda$ for a non-zero lambda then I think I get it. We could (I think) show $B$ is a closed algebra that separates points and vanishes nowhere. So, by the Stone-Weierstrass theorem, $B=C(X)$.
- Use part 2 to show that $A = I_{x_0}$.
I don’t get this but it probably doesn’t help that I’m confused about 2.
As to 2, the set $B$ is all functions of the form $f+\lambda$ where we can pick any $f \in A$ and any $\lambda \in \mathbb{R}$. It's clear $B$ vanishes nowhere, so if you can show $B$ is a closed subalgebra you're done (as $A \subseteq B$, $B$ also separates points, as $A$ does).
As to 3. pick any $g$ in $I_{x_0}$. Then by 2. $g = f+\lambda$ for some $f \in A$ and some $\lambda \in \mathbb{R}$. But $0 = g(x_0) =f(x_0) + \lambda$. By 1. $f \in I_{x_0}$ as well, so $f(x_0) = 0$ and so $\lambda = 0$ and hence $g =f \in A$ and we have $I_{x_0} \subseteq A$ and 1. already gives us the other inclusion.