I got the following statements to show.
Let $S \neq \emptyset$ equiped with the discrete topology and let $\ell_\infty(S) = \{f: S \to \mathbb C \mid f \text{ bounded}\}$. Not $\ell_\infty(S)$ with the pointwise operations is a C$^*$-algebra and a Banach lattice. Let $G_S =\{\varphi: \ell_\infty(S) \to \mathbb C \mid \varphi \text{ character} \}$ the Gelfand space of $\ell_\infty(S)$. Then $G_S$ with the weak-$*$ topology is a compact space and due to the Gelfand-Naimark theorem $G: \ell_\infty(S) \to C(G_S)$ is a isometric $*$-isomorphism.
Now I want to embed $S$ into $G_S$ by the map $\pi: S \to G_S, s \mapsto \delta_s$ where $\delta_s: \ell_\infty(S) \to \mathbb C, x \mapsto x(s)$ is the dirac functional. I could show that $\pi$ is injective, and surjective onto the Image of $\pi$. It is obviously continous but I need further that $\pi^{-1}$ is continuos too. But I don't see that.
Further I want to show that every $p \in G_S$ is positive, i.e. $p(x) \geq 0$ for all $x \in \ell_\infty(S)$ with $x \geq 0$, where $\geq$ is the order in the lattice. I managed to do that for $p = \delta_s$ for some $s \in S$, but I don't see how that works if $p \in G_S \setminus \pi(S)$.
Thanks in advance!
I could solve the problems:
We show that $\pi$ is open onto its image $\pi(S)$. Let $\{s\} \subseteq S$, then we have $\{\delta_s\} = \pi(s) \cap (\operatorname{ev}_{1_{\{s\}}})^{-1}(\mathbb C \setminus \{0\})$, where $\operatorname{ev}_{1_{\{s\}}}(\varphi) = \varphi(1_{\{s\}})$ for all $\varphi \in G_S$. Hence $\{\delta_s\}$ is open in $\pi(S)$ and $\pi(S)$ is equiped with the discrete topology as relative topology. Hence $\pi(S)$ is open.
Let $p \in G_S$ and $x \geq 0$. Then we have $x = \sqrt x \overline{\sqrt x}$. Since the Gelfand transform is a $*$-isomorphism we get for all $y \in \ell^\infty(S)$ that $\overline{p(y)} = \overline{G[y] (p)} = G[\overline y] (p) = p(\overline y)$. Since $p$ is a character we can conclude: $p(x) = p(\sqrt x) p(\overline{\sqrt x}) = p(x) \overline{p(x)} \geq 0$.
Be free to use!