Added the spectral mapping theorem.

src/op/banach/spectrum.tex

@@ -118,3 +118,12 @@
     (1): Let $\lambda \in \complex \setminus \bracs{0}$, then by \autoref{proposition:swap-invertible}, $\lambda - xy \in G(A)$ if and only if $\lambda - yx \in G(A)$. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:spectrum-continuous}
+    Let $A$ be a unital Banach algebra and $U \subset \complex$, then $\bracs{x \in A| \sigma_A(x) \subset U}$ is open. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition I.2.9]{Takesaki1}}}. ]
+    Let $x \in A$ with $\sigma_A(x) \subset U$, and $\lambda \in U^c$. By \autoref{proposition:banach-algebra-inverse}, for any $y \in A$ with $\norm{y}_A \le \normn{(\lambda - x)^{-1}}_A^{-1}$, $\lambda - x - y \in G(A)$ as well. Since the mapping $\lambda \mapsto \normn{(\lambda - x)^{-1}}_A$ vanishes at infinity and $U^c$ is closed, $\delta = \inf_{\lambda \in U^c}\normn{(\lambda - x)^{-1}}_A^{-1} > 0$. Therefore for every $z \in B_A(x,\delta)$, $\sigma_A(x) \subset U$. 
+\end{proof}
+
+