Theorem 33.5.13 (”Runge’s Theorem”).label Let $A$ be a unital Banach algebra, $x \in A$, $P \subset \complex \setminus \sigma_{A}(x)$ such that $P$ intersects every bounded component of $\complex \setminus \sigma_{A}(x)$, and $B \subset A$ be a closed algebra containing $1$, $x$, and $\bracsn{(\lambda - x)^{-1}|\lambda \in P}$, then $\sigma_{A}(x) = \sigma_{B}(x)$.
Proof, [Theorem 4.9, Mar21]. By construction, $P \subset \complex \setminus \sigma_{B}(x)$. In addition, for any polynomial $p \in \complex[z]$, $p(x) \in B$. Thus for every rational function $f \in \complex(z) \cap H(\complex_{\infty} \setminus (P \cup \bracs{\infty}); \complex)$, $f(x) \in B$.
By Runge’s Theorem, $H(\complex_{\infty} \setminus (P \cup \bracs{\infty}); \complex)$ is dense in $H(\sigma_{A}(x); \complex)$. The continuity of the holomorphic functional calculus then implies that $f(x) \in B$ for all $f \in H(\sigma_{A}(x); \complex)$. In particular, $(\lambda - x)^{-1}\in B$ for all $\lambda \in \complex \setminus \sigma_{A}(x)$. Therefore $\sigma_{B}(x) \subset \sigma_{A}(x)$, and $\sigma_{B}(x) = \sigma_{A}(x)$ by Proposition 33.5.12.$\square$
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