I'm studying Murphy's book: C*-Algebras and Operator Theory, and got stuck on exercise 8 from chapter 1:
"Show that if $B$ is a maximal abelian subalgebra of a unital Banach algebra $A$, then $B$ is closed and contains the unit. Show that $\sigma_A(b)=\sigma_B(b)$ for all $b\in B$." (where $\sigma_A(b)=\left\{\lambda\in\mathbb{C}:\lambda 1-b\text{ is not invertible in }A\right\}$, and $\sigma_B(b)$ is defined analogously)
I tried to argue by contradiction: suppose that $B$ is not closed. Then $\overline{B}$ is a closed commutative subalgebra of $A$ that contains $B$ strictly. By maximality, we have $\overline{B}=A$, hence $B$ is a dense maximal subalgebra of a unital commutative Banach algebra. But I don't see how this leads to a contradiction, although it smells like Gelfand Transform...
EDIT: Actually, I tried to show that $B$ must contain the unit, and maybe the problem is wrong: Let $B$ be a commutative non-unital Banach algebra, for example, $B=C_0(\Omega)$, where $\Omega$ is a non-compact, locally compact, Hausdorff topological space, say $\mathbb{R}$. Let $A=B\oplus\mathbb{C}$ be its unitization, that is $A=B\times\mathbb{C}$ as a set and the operations on $A$ are defined as $$(b,\alpha)+\gamma(c,\beta)=(b+\gamma c,\alpha+\gamma\beta),\quad (b,\alpha)(c,\beta)=(bc+\alpha c+\beta b,\alpha\beta)$$ and $A$ has the norm $\Vert(b,\alpha)\Vert=\Vert b\Vert+|\alpha|$. It is easily checked that $A$ is a commutative Banach algebra with unit $(0,1)$.
If we identify $B$ with the subalgebra $B\oplus\left\{ 0\right\}$ of $A$, then it's clear that $B$ is a maximal commutative subalgebra of $A$ which does not contain the unit.
If my arguments were correct, then it is not true in general that $B$ contains the unit. But maybe we can still assure that $B$ is closed in $A$ and that, if $B$ contains the unit, then $\sigma_A(b)=\sigma_B(b)$ for every $b\in B$.