[proofplan] The inclusion $\sigma_B(x) \supseteq \sigma_A(x)$ is the easy half: invertibility in the smaller algebra $B$ implies invertibility in the larger $A$, so non-invertibility in $A$ forces non-invertibility in $B$. The boundary inclusion $\partial \sigma_B(x) \subseteq \partial \sigma_A(x)$ is the substance: a boundary point $\lambda \in \partial \sigma_B(x)$ is the limit of invertible elements in $B$ that are also invertible in $A$, but the limit element $\lambda 1 - x$ is non-invertible in $B$, putting it on the boundary $\partial G(B)$. The **boundary-of-invertibles** statement (part (iv) of the Properties of Invertible Elements theorem) then forces $\lambda 1 - x$ to be non-invertible in *any* unital BA containing $B$ — in particular in $A$ — so $\lambda \in \sigma_A(x)$. The corollary about components of $\mathbb{C} \setminus \sigma_A(x)$ follows from a topological argument: each component of $\mathbb{C} \setminus \sigma_A(x)$ is either entirely contained in $\sigma_B(x)$ or entirely outside it, since $\sigma_B(x) \cap (\mathbb{C} \setminus \sigma_A(x))$ is both open and closed in $\mathbb{C} \setminus \sigma_A(x)$. [/proofplan] [step:Establish the inclusion $\sigma_B(x) \supseteq \sigma_A(x)$ via $G(B) \subseteq G(A)$] Let $\lambda \in \mathbb{C} \setminus \sigma_B(x)$, so $\lambda 1 - x \in G(B)$. Then there exists $y \in B$ with $y(\lambda 1 - x) = (\lambda 1 - x) y = 1$. Since $B \subseteq A$ as a unital subalgebra (so $1_B = 1_A$ and the operations agree), the same identity $y(\lambda 1 - x) = (\lambda 1 - x) y = 1$ holds in $A$, so $\lambda 1 - x \in G(A)$, i.e.\ $\lambda \in \mathbb{C} \setminus \sigma_A(x)$. Contrapositively, $\sigma_A(x) \subseteq \sigma_B(x)$. In particular, $G(B) \subseteq G(A) \cap B$. [/step] [step:Take $\lambda \in \partial \sigma_B(x)$ and approximate by points outside $\sigma_B(x)$] Let $\lambda \in \partial \sigma_B(x)$. Since $\sigma_B(x)$ is closed in $\mathbb{C}$ (by [Spectrum is Non-Empty and Compact](/theorems/2669) applied in $B$), boundary points are contained in $\sigma_B(x)$, so $\lambda \in \sigma_B(x)$, i.e.\ $\lambda 1 - x \notin G(B)$. By the definition of boundary, $\lambda$ is also a limit point of the open set $\mathbb{C} \setminus \sigma_B(x)$, so there exists a sequence $(\lambda_n) \subset \mathbb{C} \setminus \sigma_B(x)$ with $\lambda_n \to \lambda$. Then $\lambda_n 1 - x \in G(B)$ for every $n$, while \begin{align*} \lambda_n 1 - x \to \lambda 1 - x \quad \text{in } B \end{align*} since the map $\mu \mapsto \mu 1 - x$ is continuous (linear with bounded image of the constant $1$). [/step] [step:Place $\lambda 1 - x$ on the boundary $\partial G(B)$] From Step 2: $\lambda_n 1 - x \in G(B)$ and $\lambda_n 1 - x \to \lambda 1 - x \notin G(B)$. By the definition of topological boundary, \begin{align*} \lambda 1 - x \in \overline{G(B)} \setminus G(B) = \partial G(B), \end{align*} where the closure is taken in $B$. [/step] [step:Use the boundary-of-invertibles property to deduce $\lambda 1 - x \notin G(A)$] We apply part (iv) of [Properties of Invertible Elements](/theorems/2668) in the Banach algebra $B$ (verifying: $B$ is a unital Banach algebra since it is a closed unital subalgebra of $A$, hence complete). The hypothesis of (iv) is that $\lambda 1 - x \in \partial G(B) = \overline{G(B)} \setminus G(B)$, verified in Step 3. The conclusion of (iv) is that **$\lambda 1 - x$ has no left or right inverse in any unital Banach algebra $C$ containing $B$ as a unital subalgebra**. Taking $C := A$ (which contains $B$ as a unital subalgebra by hypothesis), we conclude $\lambda 1 - x \notin G(A)$, i.e. \begin{align*} \lambda \in \sigma_A(x). \end{align*} [/step] [step:Upgrade to $\lambda \in \partial \sigma_A(x)$] We now show $\lambda$ lies on the boundary of $\sigma_A(x)$, not in its interior. Since $\sigma_A(x) \subseteq \sigma_B(x)$ (Step 1), we have \begin{align*} \mathbb{C} \setminus \sigma_B(x) \subseteq \mathbb{C} \setminus \sigma_A(x). \end{align*} The sequence $(\lambda_n) \subset \mathbb{C} \setminus \sigma_B(x)$ from Step 2 therefore lies entirely in $\mathbb{C} \setminus \sigma_A(x)$, with $\lambda_n \to \lambda$. So $\lambda$ is a limit point of $\mathbb{C} \setminus \sigma_A(x)$. Combined with $\lambda \in \sigma_A(x)$ from Step 4, $\lambda$ is in $\sigma_A(x)$ and is a limit point of its complement. Since $\sigma_A(x)$ is closed (theorem 2669 in $A$), this means \begin{align*} \lambda \in \overline{\sigma_A(x)} \cap \overline{\mathbb{C} \setminus \sigma_A(x)} = \partial \sigma_A(x). \end{align*} Since $\lambda \in \partial \sigma_B(x)$ was arbitrary, $\partial \sigma_B(x) \subseteq \partial \sigma_A(x)$. [guided] The logic is a chain: (boundary of $\sigma_B$) → (boundary of $G(B)$) → (boundary-of-invertibles theorem) → (non-invertibility in any larger algebra) → (in $\sigma_A$). The crucial insight is that *boundary points of the spectrum* correspond, after the affine shift $\lambda \mapsto \lambda 1 - x$, to *boundary points of the invertibles*. Why? The map $f(\lambda) := \lambda 1 - x$ is continuous and a homeomorphism onto its image $\{1 - x + \mu 1 : \mu \in \mathbb{C}\}$ (an affine line). So $f$ pulls back open sets to open sets and closed sets to closed sets, preserving boundaries: $f(\partial \sigma_B(x)) \subseteq \partial f(\sigma_B(x)) = \partial \{f(\lambda) : \lambda \in \sigma_B(x)\}$. And by definition $\sigma_B(x) = f^{-1}(B \setminus G(B))$, so $f(\sigma_B(x)) = (B \setminus G(B)) \cap \mathrm{Image}(f)$, putting $f(\partial \sigma_B(x))$ on the boundary of $B \setminus G(B)$, equivalently on $\partial G(B)$. Once we have $\lambda 1 - x \in \partial G(B)$, theorem 2668(iv) does the heavy lifting: in *any* Banach algebra $C \supseteq B$, an element on the boundary of invertibles in $B$ remains non-invertible. The proof of 2668(iv) goes via constructing approximate annihilators $z_n$ of unit norm in $B$ with $z_n(\lambda 1 - x), (\lambda 1 - x) z_n \to 0$; if $(\lambda 1 - x)$ had a left inverse $w$ in $C$, then $z_n = w (\lambda 1 - x) z_n \to 0$, contradicting $\|z_n\| = 1$. So we get non-invertibility in $A$ for free. The final step is showing $\lambda$ is on the boundary of $\sigma_A(x)$, not in its interior. The sequence $\lambda_n$ approximating $\lambda$ from outside $\sigma_B(x)$ also approximates from outside $\sigma_A(x)$ (by the inclusion $\sigma_A \subseteq \sigma_B$). So $\lambda$ has the property of being in $\sigma_A(x)$ but with arbitrarily nearby points outside it — which is exactly the topological definition of $\partial \sigma_A(x)$. [/guided] [/step] [step:Deduce that $\sigma_B(x)$ is $\sigma_A(x)$ together with bounded components of $\mathbb{C} \setminus \sigma_A(x)$] Set $V := \mathbb{C} \setminus \sigma_A(x)$, an open subset of $\mathbb{C}$. We claim that for each connected component $W$ of $V$: \begin{align*} \text{either } W \subseteq \sigma_B(x), \quad \text{or} \quad W \cap \sigma_B(x) = \varnothing. \end{align*} Fix a component $W$. The set $\sigma_B(x) \cap W$ is closed in $W$ (since $\sigma_B(x)$ is closed in $\mathbb{C}$ and intersection with $W$ inherits the relative topology). We show $\sigma_B(x) \cap W$ is also open in $W$. Suppose $\lambda \in \sigma_B(x) \cap W$, i.e.\ $\lambda \notin \sigma_A(x)$ but $\lambda \in \sigma_B(x)$. Since $\lambda \notin \sigma_A(x)$, $\lambda$ cannot lie in $\partial \sigma_B(x)$ (which by the boundary inclusion proven above is contained in $\partial \sigma_A(x) \subseteq \sigma_A(x)$). Hence $\lambda$ is in the **interior** of $\sigma_B(x)$, so a small open ball $B(\lambda, r)$ is contained in $\sigma_B(x)$; intersecting with $W$ shows $\sigma_B(x) \cap W$ is open in $W$. Since $W$ is connected and $\sigma_B(x) \cap W$ is both open and closed in $W$, either $\sigma_B(x) \cap W = W$ or $\sigma_B(x) \cap W = \varnothing$. Finally, the **unbounded** component of $V$ (the component containing $\{|\mu| > \|x\|\}$, which is non-empty because $\sigma_A(x)$ is bounded by theorem 2669) cannot be contained in $\sigma_B(x)$, since $\sigma_B(x)$ is itself bounded by theorem 2669 applied in $B$. Therefore the unbounded component of $V$ is disjoint from $\sigma_B(x)$, and only some bounded components may be contained in $\sigma_B(x)$. Thus \begin{align*} \sigma_B(x) = \sigma_A(x) \cup \bigcup_{W \in \mathcal{W}} W, \end{align*} where $\mathcal{W}$ is the (possibly empty) collection of bounded components of $\mathbb{C} \setminus \sigma_A(x)$ contained in $\sigma_B(x)$. This completes the proof. [/step]