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Let $A$ be a Banach algebra with identity and $G$ is set of all invertible elements of $A$.

$\sigma(x)=\{z\in \Bbb C : ze-x\ \textrm{is not invertible} \}$ is spectrum of $x\in A$ where $e$ is identity.

If $\lambda$ is an element of boundary of $\sigma(x)$ then $\lambda e-x$ is element of $G$'s boundary.

Let $\lambda \in \partial \sigma(x)=\overline{\sigma(x)}\setminus \sigma(x)^{\circ}=\sigma(x)\setminus \sigma(x)^{\circ}$ (last equality holds since spectrum is closed) so $\lambda e-x$ is not invertible.

How can I show that $\lambda e-x \in \partial G=\overline G \setminus G^{\circ}=G=\overline G \setminus G$ (last equality holds since $G$ is open)

I appreciate any help.

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1 Answer 1

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By assumption, $\lambda$ can be approximated arbitrarily closely by points $\alpha\in\sigma(x)$ and also by points $\beta\notin\sigma(x)$. Then $\lambda e-x$ is approximated closely by $\alpha e-x\notin G$ and by $\beta e-x\in G$.

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