I'm studying Banach algebras and I'm confused by the following statement:
Let $\mathcal{A,B}$ be Banach algebras with common identity and suppose $\mathcal{B}\subset\mathcal{A}$
What does that mean? As far as I understand, it is the same as "let $\mathcal{B}$ be a subalgebra of $\mathcal{A}$ that contains the identity element". Does the statement also imply that $\mathcal{B}$ is a closed (topologically speaking) subalgebra of $\mathcal{A}$?
Also, in the exercises it says "Let $\mathcal{A,B}$ be Banach algebras with common identity and suppose $\mathcal{B}\subset\mathcal{A}$ and let $B\in\mathcal{B}$. Prove that $\sigma_\mathcal{A}(B)\subset\sigma_\mathcal{B}(B)$ and $\partial\sigma_{\mathcal{B}}(B)\subset\partial\sigma_\mathcal{A}(B)$." Even though the first inclusion is easy and true without $\mathcal{B}$ being necessarily a closed subalgebra, I can't prove the second one without the implication that $\mathcal{B}$ is closed and the only thing I found here is this (that also uses a closed subalgebra).
I would appreciate any help.
Any subspace of a Banach space which is complete is closed. A Banach algebra is complete so $\mathcal B$ is closed.