Let $X$ be a local compact Hausdorff space and let $$(p_n)_{n\in\mathbb{N}}\subset C_0(X)=C^*\left(\sum_{n\in\mathbb{N}}\frac{p_n}{3^n}\right)$$ a sequence of projections such that $C^*(\{p_n:n\in\mathbb{N}\})=C_0(X)$. Prove that $X$ is totally disconnected, i.e. for every subspace $Y\subset X$ with at least two elements there exists a continuous, non-constant function $f:Y\to \{0,1\}$.
My idea was to use Urysohn's Lemma: For $x,y\in Y$, $x\neq y$, there exists a continuous function $f:Y\to [0,1]$ such that $f(x)=1,\; f(y)=0$ (with Urysohn).
Now we know that the $p_n's$ generate $C_0(X)$, therefore for the $f$ above, there exists a linear combination of the $p_n$, $\sum_{n}\lambda_np_n$ such that $\sum_{n}\lambda_np_n\to f$ uniformly. Our question is: can you find a $n\in \mathbb{N}$ such that $p_n(x)=1$ and $p_n(y)=0$? And if not, how to prove this claim?
Regards
If $p_n(x)=p_n(y)$ for all $n$, then $f(x)=f(y)$, which we assumed was false. So $p_n(x)\neq p_n(y)$ for some $n$. Since $p_n$ is a projection, it can only take the values $0$ and $1$. If $p_n(x)=1$ and $p_n(y)=0$, we're done. The only other possibility is $p_n(x)=0$ and $p_n(y)=1$, in which case you can just consider the projection $1-p_n$.