# A simple question about boundary of spectrum

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.

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 $$\beta e-x\in G$$.