Let $X$ be a locally compact Hausdorff space. We say that a function $f\colon X \to \mathbb{R}$ vanishes at $\infty$ if for each $\epsilon >0$ there exists a compact $K_\epsilon \subset X$ such that $|f(x)|< \epsilon, \forall x\in X-K_\epsilon$.
We denote by $C_0(X)$ the set of all such functions and we endow it with the usual operations of addition and multiplication of functions, and multiplication of functions by scalars.
(i) Show that one has a well-defined norm
$\left|\left|\cdot\right|\right|:C_0(X) \to \mathbb{R}, f\to ||f||:=\sup\left\{ {|f(x)|:x \in X}\right\}$
and together with this norm, $C_0(X)$ becomes a Banach space.
(ii) Assume that $\mathbb{A}\subset C_0(X)$ has the following properties:
- $\mathbb{A}$ is a non-unital subalgebra of $C_0(X)$.
- $\mathbb{A}$ is strongly point-separating: it is point separating and, $\forall x \in X$, there exists $f\in \mathbb{A}$ such that $f(x)\neq0$.
Show that $\mathbb{A}$ is dense in $C_0(X)$.
(i) was easy, but I'm completely stumped on (ii). Since Stone Weierstrass requires an unital subalgebra in a nontrivial way I don't know how to proceed. Due the way the question is phrased and the conditions given, it is very suggestive of using one-point compactification and then either producing a unit (which doesn't work as far as I can see) or proceeding in an entirely different way. I'd very much appreciate a hint.
Please note that this is not homework.
EDIT: I've found the solution.
Proof. $X$ is a locally compact Hausdorff space and so can be one-point compactified. We do so and obtain $X^*$. We imbed $A$ into $C(X^*)$, using the fact that any element $a$ of $A$ vanishes at the extra point $\infty$. Send that $a$ to the $b\in C(X^*)$ that is is equal to $a$ for all point in $X$ and zero at $\infty$. This defines $A^*$, a subalgebra of $C(X^*)$. Similarly we can imbed $C_0(X)$ into $C_0(X^*)$. This obviously respects the norm. Moreover we define $B^*:=$Span$(A^*)\cup \left\{{1}\right\}$. $B^*$ is a point separating unital subalgebra of $C(X^*)$, because $A$ is strongly point separating. We now apply Stone-Weierstrass and the fact that $X^*$ is compact Hausdorff to find that $B^*$ is dense in $C(X^*)$.
Observe that the closure of $A^*$ lies in $C(X^*)\cap $Cl$(B^*)$, because $C(X^*)\cap $Cl$(B^*)$ is closed and and contains $A^*$. Let $\gamma \in C_0(X^*)=C(X^*)\cap $Cl$(B^*)$. $\gamma$ lies in $C_0(X^*)$, so $\gamma$ is zero at $\infty$, and it is a limit point of $B$. Note however that every sequence to $\gamma$ lies in $A^*$ after a certain index. So the $\gamma$ is in the closure of $A^*$. As noted earlier, bijection between $A$ and $A^*$ respects the norm. It follows that $A$ lies dense in $C_0(X)$.
Let $\tilde{X}$ be the one-point compactification of $X$, embed the linear span of $A$ and constants in $C(\tilde{X})$, and use the ordinary Stone-Weierstrass theorem there.