Let $A$ be a complex Banach algebra and $a \in A.$ Let $\sigma(a)$ denote the spectrum of $a.$ Let $V$ be the union of all bounded connected components of $\sigma (a)^c.$ Let $K = \sigma (a) \cup V.$ Then show that $\partial K = \partial (\sigma (a)).$
We know that $\partial K \subseteq \partial (\sigma (a)) \cup \partial V.$ Also since $\sigma(a)^c \subseteq_{\text {open}} \mathbb C$ it's components are all open and their boundaries are all contained in $\sigma(a).$ Hence $\partial V \subseteq \sigma (a).$ So $\partial K \subseteq \sigma (a),$ since $\partial (\sigma (a)) \subseteq \overline {\sigma (a)} = \sigma (a),$ as $\sigma (a)$ is closed. But I can't conclude anything more than that. Could anyone please help me in this regard?
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