 Solution

A uniform algebra on a compact Hausdorff space $K$ is a closed complex subalgebra $A\subseteq C(K)$ containing the constants and separating points of $K$, with the supremum norm. It is therefore a commutative unital Banach algebra. Its character space of an algebra ${\Phi }_A$, also called its carrier space, consists of nonzero multiplicative linear functionals, equipped with the Gelfand topology.

For $x\in K$, evaluation gives the character of an algebra ${\delta }_x(f)=f(x)$. The map $\delta :K\to {\Phi }_A$ is continuous: composing it with the coordinate $\phi \mapsto \phi (f)$ gives the continuous function $f$ for every $f\in A$. It is injective because $A$ separates points. The Gelfand topology is Hausdorff because distinct algebra characters differ on some element of $A$. A continuous injection from compact $K$ into a Hausdorff space is a homeomorphism onto its image: closed subsets of $K$ have compact, hence closed, images. Its image is also compact and closed. Thusis the asserted homeomorphism. A natural uniform algebra is one for which this image is all of ${\Phi }_A$.