Let A and B be a Banach algebra with $1\in A\subset B$ (1 is the identity). Let $a\in A $ then show that $Spectrum B\subset SpectrumA$ Let A and B be a Banach algebra with $1\in A\subset B$ (1 is the identity). Let $a\in A $ then show that
$$1).Spectrum_a\ B\subset Spectrum_aA$$
$$2). Boundary(Spectrum_aA)\subset Boundary(Spectrum_aB)$$
First, let's take $x\in Spectrum_a\ B$, I'm going to show that is in $Spectrum_aA$.
since $x\in Spectrum_a\ B$
$a-x*1$ is not invertible. Since $a\in A$ , $_\ $ contains x.
This implies $Spectrum_a\ B\subset Spectrum_aA$.
Is this correct?
And how we show the second part?
 A: Lemma.  Assume that $A$ is a unital Banach algebra and that $\{a_n\}_n\subseteq A$ is a sequence of invertible elements,  converging to
a non-invertible element $a$.  Then $a$ is a topological zero-divisor, meaning that there exists a sequence $\{x_n\}_n$ in
$A$, with $\|x_n\|=1$, and $\displaystyle\lim_{n\to \infty }ax_n = 0$.
Proof.  We begin by claiming that there is no positive constant $c$, such that
$$
  \|ax\|\geq  c\|x\|,\quad\forall x\in  A.
  $$
In order to prove the claim we assume otherwise and hence
$$
  c\|a_n^{-1}\|\leq
  \|aa_n^{-1}\| =
  \|(a-a_n+a_n)a_n^{-1}\| \leq  $$$$ \leq
  \|a-a_n\|\|a_n^{-1}\| +  1,
  $$
from where it follows that
$$
  \|a_n^{-1}\| \leq  \frac 1{c-\|a-a_n\|},
  $$
as long as the denominator is positive, which it sure is for large enough $n$.   This  also shows that   $\|a_n^{-1}\|$ is bounded.  Given any $n$
and $m$ we then have that
$$
  \Vert a_n^{-1}-a_m^{-1}\Vert  =
  \Vert a_n^{-1}(a_m-a_n)a_m^{-1}\Vert  \leq  $$$$ \leq
  \Vert a_n^{-1}\|\|a_m-a_n\|\|a_m^{-1}\Vert  \leq
  \|a_m-a_n\|\big(\sup_k\|a_k^{-1}\Vert \big)^2.
  $$
This says that $\{a_n^{-1}\}_n$ is a Cauchy sequence.   Setting $\displaystyle b= \lim_{n\to \infty }a_n^{-1}$, we then have that
$$
  ab = \big (\lim_{n\to \infty }a_n\big )\big (\lim_{n\to \infty }a_n^{-1}\big ) = \lim_{n\to \infty }a_na_n^{-1} = 1,
  $$
and likewise $ba=1$, so $a$ is invertible, a contradiction,  hence proving the claim.
Consequently, for every $n>0$, the inequality  "$\|ax\|\geq  (1/n)\|x\|$" does not always hold, so there exists some $x_n$
such that
$$
  \|ax_n\|< (1/n)\|x_n\|.
  $$
The conclusion then  follows by replacing the $x_n$ with  $x_n/\|x_n\|$. QED

Back to the question,   suppose that $\lambda $ is in the boundary of $\sigma _A(a)$.  Then $a-\lambda $ is not invertible, but it is
clearly a limit of invertible elements of the form $a-\rho _n$, where $\rho _n$ lies in complement of $\sigma (A)$.  By the Lemma,  we
deduce that $a-\lambda $ is a topological zero-divisor, and  this obviously prevents $a-\lambda $ from being invertible in $B$, so we see that
$\lambda \in \sigma _B(a)$.
This shows that
$$
  \partial \sigma _A(a)\subseteq \sigma _B(a),
  $$
and from this one easily shows that in fact
$$
  \partial \sigma _A(a)\subseteq \partial \sigma _B(a).
  $$
