= Solution
A <character of an algebra> is a nonzero complex-linear multiplicative map $\varphi:A\to\mathbb C$. The <character space of an algebra> $\Phi_A$ carries the <Gelfand topology>, namely pointwise convergence on $A$, equivalently the subspace weak-star topology in $A^*$.
For a unital complex <Banach algebra>, $\varphi(1)=1$: multiplicativity and nonzeroness force this. If $a-\varphi(a)1$ were invertible, applying $\varphi$ to its product with its inverse would give $0=1$. Thus $\varphi(a)\in\sigma_A(a)$. The <Neumann series> shows $\sigma_A(a)\subset\{|z|\le\|a\|\}$, so $|\varphi(a)|\le\|a\|$ and the <character of an algebra> is automatically bounded. The same conclusion for a nonunital algebra follows by extending $\varphi$ to its <unitization of an algebra> via $\widetilde\varphi(a,t)=\varphi(a)+t$ and using the standard <norm> $\|(a,t)\|=\|a\|+|t|$.
In the algebra $R(K)$ let $u(z)=z$ and fix $\varphi\in\Phi_{R(K)}$. Write $\lambda=\varphi(u)$. If $\lambda\notin K$, the <rational function> $(u-\lambda)^{-1}$ belongs to $R(K)$, which is impossible because a <character of an algebra> cannot vanish on an invertible element. Thus $\lambda\in K$. For a <rational function> $r=p/q$ without poles in $K$, choose a reduced denominator having no zeros in $K$. Multiplicativity gives $\varphi(r)=p(\lambda)/q(\lambda)=r(\lambda)$. Density of these <rational functions> and boundedness of $\varphi$ then yield $\varphi(h)=h(\lambda)$ for every $h\in R(K)$.
Conversely, evaluation at each $\lambda\in K$ is a <character of an algebra>. The map $\lambda\mapsto\delta_\lambda$ is continuous in the Gelfand topology because every $h\in R(K)$ is continuous; its inverse is $\varphi\mapsto\varphi(u)$, also continuous. Hence the <character space of R(K)> is
$$
\boxed{\Phi_{R(K)}\cong K,\qquad\varphi=\delta_\lambda.}
$$
The identification includes the topology, not merely a bijection of sets.
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