[proofplan] After unitisation we may assume $A$ is unital. Compactness has two parts: $\sigma_A(x)$ is closed because it is the preimage of the closed set $A \setminus G(A)$ under the continuous map $\lambda \mapsto \lambda 1 - x$, and $\sigma_A(x)$ is bounded by $\|x\|$ because [Invertibility Near the Identity](/theorems/2667) gives invertibility of $\lambda 1 - x$ whenever $|\lambda| > \|x\|$. Non-emptiness is the deep ingredient: if $\sigma_A(x)$ were empty, the resolvent $f(\lambda) = (\lambda 1 - x)^{-1}$ would be an entire $A$-valued function, bounded on $\mathbb{C}$ (by continuity on compact sets and decay at infinity), hence constant by the [Banach Space-Valued Liouville Theorem](/theorems/2635). But this is impossible because $f(\lambda)(\lambda 1 - x) = 1$ forces $f$ to depend on $\lambda$. [/proofplan] [step:Reduce to the unital case via unitisation] If $A$ is non-unital, embed it as a maximal ideal in its unitisation $A^+ := A \oplus \mathbb{C}$ with multiplication $(a, \alpha)(b, \beta) := (ab + \alpha b + \beta a, \alpha \beta)$, unit $(0, 1)$, and norm $\|(a, \alpha)\|_{A^+} := \|a\|_A + |\alpha|$, which makes $A^+$ a unital Banach algebra in which $A$ embeds isometrically as $a \mapsto (a, 0)$. By the standard convention, $\sigma_A(x) := \sigma_{A^+}((x, 0))$ for $x \in A$. Replacing $A$ by $A^+$ if necessary, we may assume $A$ is unital and $x \in A$, with the original norm extended. [/step] [step:Show $\sigma_A(x)$ is closed via continuity of $\lambda \mapsto \lambda 1 - x$] Consider the map \begin{align*} R : \mathbb{C} &\to A \\ \lambda &\mapsto \lambda 1 - x. \end{align*} This is continuous because $\|R(\lambda) - R(\mu)\| = \|(\lambda - \mu) 1\| = |\lambda - \mu| \cdot \|1\|$, so $R$ is Lipschitz. By definition, $\sigma_A(x) = \{\lambda \in \mathbb{C} : R(\lambda) \notin G(A)\} = R^{-1}(A \setminus G(A))$. By [Properties of Invertible Elements](/theorems/2668)(i), $G(A)$ is open in $A$, so $A \setminus G(A)$ is closed. The preimage of a closed set under a continuous map is closed, so $\sigma_A(x)$ is closed in $\mathbb{C}$. [/step] [step:Show $\sigma_A(x) \subseteq \{|\lambda| \leq \|x\|\}$ by direct invertibility for large $\lambda$] Let $\lambda \in \mathbb{C}$ with $|\lambda| > \|x\|$. Then $\|x/\lambda\| = \|x\|/|\lambda| < 1$, so \begin{align*} \|1 - (1 - x/\lambda)\| = \|x/\lambda\| < 1. \end{align*} By [Invertibility Near the Identity](/theorems/2667), the element $1 - x/\lambda$ is invertible in $A$. Multiplying by the non-zero scalar $\lambda$ preserves invertibility (its inverse is $\lambda^{-1}(1 - x/\lambda)^{-1}$): \begin{align*} \lambda 1 - x = \lambda(1 - x/\lambda) \in G(A). \end{align*} Hence $\lambda \notin \sigma_A(x)$. Contrapositively, $\sigma_A(x) \subseteq \{\lambda \in \mathbb{C} : |\lambda| \leq \|x\|\}$. Combined with closedness from Step 2, $\sigma_A(x)$ is a closed subset of the compact disc $\overline{B}(0, \|x\|) \subset \mathbb{C}$, hence compact. [/step] [step:Define the resolvent function and assume $\sigma_A(x) = \varnothing$ for contradiction] Suppose for contradiction that $\sigma_A(x) = \varnothing$. Then $\lambda 1 - x \in G(A)$ for every $\lambda \in \mathbb{C}$, and we may define the **resolvent function** \begin{align*} f : \mathbb{C} &\to A \\ \lambda &\mapsto (\lambda 1 - x)^{-1}. \end{align*} This is well-defined on all of $\mathbb{C}$. [/step] [step:Show $f$ is holomorphic on $\mathbb{C}$ via the resolvent identity] For $\lambda \neq \mu$ in $\mathbb{C}$, multiplying $f(\lambda)$ on the left by $(\mu 1 - x)$ and $f(\mu)$ on the right by $(\lambda 1 - x)$ yields the resolvent identity \begin{align*} f(\lambda) - f(\mu) = f(\lambda)\big[(\mu 1 - x) - (\lambda 1 - x)\big]f(\mu) = (\mu - \lambda) f(\lambda) f(\mu), \end{align*} which we verify by left-multiplying by $(\lambda 1 - x)$ and right-multiplying by $(\mu 1 - x)$: \begin{align*} (\lambda 1 - x)(f(\lambda) - f(\mu))(\mu 1 - x) = (\mu 1 - x) - (\lambda 1 - x) = (\mu - \lambda) 1. \end{align*} Both sides equal $(\mu - \lambda) 1$, confirming the identity. Dividing by $\lambda - \mu$, \begin{align*} \frac{f(\lambda) - f(\mu)}{\lambda - \mu} = -f(\lambda) f(\mu). \end{align*} By continuity of inversion ([Properties of Invertible Elements](/theorems/2668)(ii)) and continuity of the assignment $\mu \mapsto \mu 1 - x$, the map $\mu \mapsto f(\mu)$ is continuous. Hence as $\mu \to \lambda$, \begin{align*} \frac{f(\lambda) - f(\mu)}{\lambda - \mu} \to -f(\lambda)^2. \end{align*} This is convergence in the norm of $A$, so $f$ is complex-differentiable at every $\lambda \in \mathbb{C}$ as an $A$-valued function with $f'(\lambda) = -f(\lambda)^2$. Therefore $f$ is an entire $A$-valued function. [/step] [step:Show $f$ is bounded on $\mathbb{C}$ by combining decay at infinity with compactness on $\overline{B}(0, \|x\|)$] For $|\lambda| > \|x\|$, the proof of Step 3 gives $\lambda 1 - x = \lambda(1 - x/\lambda)$, and the [Invertibility Near the Identity](/theorems/2667) bound applied to $1 - x/\lambda$ (whose distance from $1$ is $\|x/\lambda\| < 1$) yields \begin{align*} \|(1 - x/\lambda)^{-1}\| \leq \frac{1}{1 - \|x/\lambda\|} = \frac{|\lambda|}{|\lambda| - \|x\|}. \end{align*} Hence \begin{align*} \|f(\lambda)\| = |\lambda|^{-1} \|(1 - x/\lambda)^{-1}\| \leq \frac{1}{|\lambda| - \|x\|} \xrightarrow{|\lambda| \to \infty} 0. \end{align*} In particular there exists $R > \|x\|$ such that $\|f(\lambda)\| \leq 1$ for all $|\lambda| \geq R$. On the closed disc $\overline{B}(0, R)$, the function $f$ is continuous (by Step 5) and the disc is compact, so $f$ is bounded on $\overline{B}(0, R)$ by some constant $M_R < \infty$. Combining the two bounds, $\|f(\lambda)\| \leq \max(M_R, 1)$ for all $\lambda \in \mathbb{C}$. So $f$ is bounded. [/step] [step:Apply Liouville to derive a contradiction] By Steps 5 and 6, $f : \mathbb{C} \to A$ is a bounded entire $A$-valued function. The [Banach Space-Valued Liouville Theorem](/theorems/2635) (which states that every bounded entire function from $\mathbb{C}$ to a Banach space is constant) applies: $f$ is constant on $\mathbb{C}$. But the limit in Step 6 gives $\|f(\lambda)\| \to 0$ as $|\lambda| \to \infty$, so the constant value of $f$ is $0$. This means $(\lambda 1 - x)^{-1} = 0$ for all $\lambda \in \mathbb{C}$, in particular for $\lambda = 0$: $(- x)^{-1} = 0$. But then $1 = (-x) \cdot (-x)^{-1} = (-x) \cdot 0 = 0$ in $A$, contradicting that $A$ is unital with $1 \neq 0$ (the unital convention excludes the zero algebra; if $A^+$ is the unitisation we are using, then $1_{A^+} = (0, 1) \neq 0_{A^+}$). This contradiction shows $\sigma_A(x) \neq \varnothing$, completing the proof. [guided] The non-emptiness argument is the heart of the theorem and the part that genuinely uses complex-analytic structure. Let us walk through the contradiction strategy. *Why introduce the resolvent function?* If $\sigma_A(x) = \varnothing$, then $\lambda 1 - x$ is invertible for every $\lambda \in \mathbb{C}$. Defining $f(\lambda) := (\lambda 1 - x)^{-1}$, we obtain an $A$-valued function defined on all of $\mathbb{C}$. The strategy is to show this function is too well-behaved to exist non-trivially: we will show it is bounded and entire, hence constant by Liouville, and derive a contradiction. *Why is $f$ holomorphic?* The resolvent identity \begin{align*} f(\lambda) - f(\mu) = (\mu - \lambda) f(\lambda) f(\mu) \end{align*} is the algebraic source. Verifying it: multiplying $f(\lambda) - f(\mu)$ on the left by $(\lambda 1 - x)$ and on the right by $(\mu 1 - x)$ gives $(\mu 1 - x) - (\lambda 1 - x) = (\mu - \lambda) 1$, which equals what we get by doing the same to the right-hand side $(\mu - \lambda) f(\lambda) f(\mu)$. Dividing by $\lambda - \mu$ and using continuity of inversion ([Properties of Invertible Elements](/theorems/2668)(ii)) gives $f'(\lambda) = -f(\lambda)^2$. *Why is $f$ bounded?* Two regimes. For large $\lambda$, the geometric-series bound from [Invertibility Near the Identity](/theorems/2667) gives explicit decay $\|f(\lambda)\| \leq 1/(|\lambda| - \|x\|) \to 0$. For small $\lambda$, the function is continuous on a compact disc, so bounded by extreme-value considerations. *Why does Liouville give a contradiction?* The [Banach Space-Valued Liouville Theorem](/theorems/2635) says a bounded entire $A$-valued function is constant. Since $\|f(\lambda)\| \to 0$, the constant value is $0$. But $f(\lambda) (\lambda 1 - x) = 1 \neq 0$ in any unital algebra, so $f$ cannot be identically $0$. Contradiction. The conclusion is that the resolvent set is not all of $\mathbb{C}$, i.e.\ $\sigma_A(x) \neq \varnothing$. [/guided] [/step]