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I'm working on Part D of the following question:

I am trying to use the following variation on Stone-Weierstrass:

Let $U$ be a non-compact locally compact hausdorff space. If $mathcal{A}$ is a closed subalgebra of $C_0(U,mathbb{R})$ (the continuous functions from $U$ to $mathbb{R}$ which vanish at infinity) which separates points. Then $mathcal{A}=C_0(U,mathbb{R})$ or $mathcal{A}={f in C_0(U,mathbb{R}):f(x_0)=0}$ for some $x_oin U$

I'm having trouble seeing how to use the theorem and why $U$ should not be compact. Thoughts?

A solution is given here: https://pdfs.semanticscholar.org/b602/17d061e2f6ca4c592edf5efb7bcb19320b04.pdf

I am having trouble following it though.

The first part a) is easy: We have $h(J) = bigcap_{f in J} f^{-1}({0})$. Since every $f in J$ is continuous, this set is closed as the intersection of the closed sets $f^{-1}({0})$. In b) obviously $k(E)$ is also an ideal. Here we denote $pi_x colon C(X) rightarrow mathbb{R}$ the projection on the point $x$, i.e. $pi_x(f) = f(x)$. Then $pi_x$ is continuous and $k(E) = bigcap_{x in E} pi_x^{-1}({0})$, i.e. $k(E)$ is closed. (One can also argue by using sequences in order to prove that $k(E)$ is sequentially closed and thus closed, since $C(X)$ is a metric space.)

c) can be shown as follows: First, let $x in E$, then for any $f in k(E)$ we have $f(x)=0$. Thus $x in h(k(E))$ by definition. SInce $h(k(E))$ is closed, we have $overline{E} subset h(k(E))$. On the other hand, for any $x notin overline{E}$, there exists by Urysohn's lemma a function $f colon X rightarrow [0,1]$ with $f(x) =1$ and $f(y) =0$ for all $y in overline{E}$. Thus $f in k(E)$ by definition and, because $f(x) =1$, we get $x notin h(k(E))$.

Let us prove now d): Let $f in J$, then for any $x in h(J)$ we have $f(x) =0$. Thus $f in k(h(J))$. Since this set is closed, we also get $overline{J} subset k(h(J))$. Now, we use the hint: Let $U = X subset h(J)$. This is a open set and is not necessary compact.

Counterexample: If $X=[0,1]$ we can take $x_0 =0$ and the maximal ideal $J = {f in C([a,b]): f(x_0) =0}$, then $h(J) = {0}$. Thus $X setminus h(J) = (0,1]$ is not compact.

For any $varepsilon >0$ we can cover $h(J)$ with finite many open sets $V_1,ldots,V_n$ such that $|f(x)| < varepsilon$ for fixed $f in k(h(J))$, because $h(J)$ is compact. Set $V = V_1 cup ldots cup V_n$. Then $K^c = V^c = V_1^c cap ldots cap V_n^c$ is compact in $X$ and hence also in $U$. We verified that $k(h(J)) subset C_0(U)$. Note that this set is a closed subalgebra of $C_0(U)$.

Thus $mathcal{A}:= overline{J}$ is a closed subalgebra of $C_0(U)$. By definition for any $x in U$ there exists at least one $f_x in J$ with $f_x(x) ne 0$. Let $x,y in U$ are different points, we can use Urysohn's lemma to find functions $h_x$ and $h_y$ with $h_x(x)=1$ and $h_x(z) = 0 $ for $z in h(J) cup {y}$, resp. $h_y(y)=1$ and $h_y(y) = 0$ for $z in h(J) cup {x}$. Now define
$$h(z) := h_x(z) frac{f_x(z)}{f_x(x)} - h_y(z) frac{f_y(z)}{f_y(y)}.$$
Then $h in J$ with $h(x) =1 neq -1 =h(y)$. Hence $mathcal{A}$ separates points and also vanishes nowhere. Stone-Weierstraß already shows that $mathcal{A} = C_0(U)$ and thus $mathcal{A} = k(h(J))$.

The last part is now a consequence of the previous ones: If $E$ is a closed set in $X$, then $k(E)$ is a closed ideal in $C(X)$ with $h(k(E)) =E$. On the other hand, any closed ideal $J$ in $C(X)$ gives a closed set $h(J)$ with $k(h(J)) = J$.