I'm studying Conway's Functional Analysis.
In Chaper VII.7 (The Spectral Theory of a Compact Operator) first paragraph,
$\mathcal{B}_0(\mathcal{X})$ is the algebra of all compact operators. If $A \in \mathcal{B}_0(\mathcal{X})$ , then $\sigma(A)$ refers to the spectrum of $A$ as an element of $ \mathcal{B}(\mathcal{X})$. If $\mathcal{A} =\mathcal{B}_0(\mathcal{X}) + \mathbb{C} $, then $\mathcal{A}$ is Banach algebra with identity, and we consider $\sigma_{\mathcal{A}}(A)$. Since $\sigma(A) \subset \sigma_{\mathcal{A}}(A)$ and $\partial \sigma_{\mathcal{A}}(A) \subset \partial\sigma(A)$. Using fact that $\sigma(A)$ is countable, $\sigma(A) = \partial \sigma(A) = \sigma(A)\hat{}$. Thus $\sigma(A) = \sigma_{\mathcal{A}}(A)$
I have question about, why $\sigma(A) = \partial \sigma(A) = \sigma(A)\hat{}$ holds indicate $\sigma(A) = \sigma_{\mathcal{A}}(A)$. Thanks. I think this logic would be easy and intuitively right, but how to prove it..?