[proofplan] The four claims all reduce to the geometric-series lemma [Invertibility Near the Identity](/theorems/2667). Openness (i) follows by translating the lemma along $x \in G(A)$: a small perturbation $y$ satisfies $\|1 - x^{-1}y\| < 1$, so $x^{-1}y$ is invertible. Continuity (ii) uses the resolvent identity $y^{-1} - x^{-1} = y^{-1}(x - y)x^{-1}$ together with a uniform bound on $\|y^{-1}\|$ from the lemma. Claim (iii) is the contrapositive of (i). For (iv), the elements $z_n := x_n^{-1}/\|x_n^{-1}\|$ are unit vectors with $z_n x_n = 1/\|x_n^{-1}\| \to 0$ since (iii) forces $\|x_n^{-1}\| \to \infty$; the boundary obstruction transfers to any larger algebra because a left/right inverse of $x$ would force $z_n$ to vanish. [/proofplan] [step:Establish openness of $G(A)$ by perturbing in a ball of radius $1/\|x^{-1}\|$] Fix $x \in G(A)$ and let $y \in A$ with \begin{align*} \|y - x\| < \frac{1}{\|x^{-1}\|}. \end{align*} Multiplying $x - y$ on the left by $x^{-1}$ and using submultiplicativity, \begin{align*} \|1 - x^{-1} y\| = \|x^{-1}(x - y)\| \leq \|x^{-1}\| \cdot \|x - y\| < \|x^{-1}\| \cdot \frac{1}{\|x^{-1}\|} = 1. \end{align*} By [Invertibility Near the Identity](/theorems/2667), the element $x^{-1} y$ is invertible in $A$, so $x^{-1} y \in G(A)$. Since $G(A)$ is closed under products and contains $x$, we have $y = x \cdot (x^{-1} y) \in G(A)$. This shows $B(x, 1/\|x^{-1}\|) \subseteq G(A)$, so every point of $G(A)$ has an open neighbourhood contained in $G(A)$, i.e.\ $G(A)$ is open. [/step] [step:Bound $\|y^{-1}\|$ uniformly for $y$ in a ball around $x$] For $y \in A$ with $\|y - x\| < \frac{1}{2\|x^{-1}\|}$, the calculation in Step 1 gives $\|1 - x^{-1} y\| \leq \|x^{-1}\| \|y - x\| < 1/2$, so $x^{-1} y \in G(A)$ and the norm bound from [Invertibility Near the Identity](/theorems/2667) applies: \begin{align*} \|(x^{-1} y)^{-1}\| \leq \frac{1}{1 - \|1 - x^{-1} y\|} \leq \frac{1}{1 - \|x^{-1}\|\|y - x\|}. \end{align*} Since $y^{-1} = (x^{-1} y)^{-1} x^{-1}$, by submultiplicativity \begin{align*} \|y^{-1}\| \leq \|(x^{-1} y)^{-1}\| \cdot \|x^{-1}\| \leq \frac{\|x^{-1}\|}{1 - \|x^{-1}\|\|y - x\|}. \end{align*} [/step] [step:Prove continuity of inversion via the resolvent identity] For $x \in G(A)$ and $y \in B(x, 1/(2\|x^{-1}\|))$, the resolvent identity \begin{align*} y^{-1} - x^{-1} = y^{-1} (x - y) x^{-1} \end{align*} holds: multiplying out, $y^{-1}(x - y)x^{-1} = y^{-1} x x^{-1} - y^{-1} y x^{-1} = y^{-1} - x^{-1}$. Taking norms and using submultiplicativity together with the bound from Step 2, \begin{align*} \|y^{-1} - x^{-1}\| \leq \|y^{-1}\| \cdot \|x - y\| \cdot \|x^{-1}\| \leq \frac{\|x^{-1}\|^2 \|x - y\|}{1 - \|x^{-1}\|\|x - y\|}. \end{align*} As $y \to x$ in $A$, the right-hand side tends to $0$. So $y^{-1} \to x^{-1}$ as $y \to x$, proving continuity of inversion at $x$. Since $x \in G(A)$ was arbitrary, the inversion map is continuous on $G(A)$. [guided] We want to show inversion is continuous. The natural approach is to write the difference $y^{-1} - x^{-1}$ in a form that makes its smallness manifest when $y$ is close to $x$. The resolvent identity does exactly this. Let us verify it: starting from $y^{-1} - x^{-1}$, we factor through $y^{-1} \cdot 1 \cdot x^{-1}$ by inserting $1 = x x^{-1} = y^{-1} y$: \begin{align*} y^{-1} - x^{-1} = y^{-1}(x x^{-1}) - (y^{-1} y)x^{-1} = y^{-1}(x - y)x^{-1}. \end{align*} Now we estimate. By submultiplicativity, $\|y^{-1} - x^{-1}\| \leq \|y^{-1}\| \cdot \|x - y\| \cdot \|x^{-1}\|$. The factors $\|x^{-1}\|$ and $\|x - y\|$ are well-controlled — the first is fixed, the second tends to $0$. The remaining issue is to control $\|y^{-1}\|$ uniformly: a priori, $\|y^{-1}\|$ could blow up as $y \to x$. This is where Step 2 enters. From [Invertibility Near the Identity](/theorems/2667) applied to $x^{-1} y$, we get the explicit bound \begin{align*} \|y^{-1}\| \leq \frac{\|x^{-1}\|}{1 - \|x^{-1}\|\|y - x\|}, \end{align*} which is finite and bounded (say, by $2\|x^{-1}\|$) on the smaller ball $\|y - x\| < 1/(2\|x^{-1}\|)$. Substituting, \begin{align*} \|y^{-1} - x^{-1}\| \leq \frac{\|x^{-1}\|^2 \|x - y\|}{1 - \|x^{-1}\|\|x - y\|} \to 0 \end{align*} as $y \to x$. So inversion is continuous at $x$. [/guided] [/step] [step:Deduce blow-up of $\|x_n^{-1}\|$ at boundary points by contradiction] Let $(x_n) \subset G(A)$ with $x_n \to x \notin G(A)$. Suppose for contradiction that $\|x_n^{-1}\|$ is not unbounded, so there is a subsequence (still denoted $(x_n)$ for notational simplicity) and $M > 0$ with $\|x_n^{-1}\| \leq M$ for all $n$. Since $x_n \to x$, for $n$ sufficiently large \begin{align*} \|x - x_n\| < \frac{1}{M} \leq \frac{1}{\|x_n^{-1}\|}. \end{align*} Applying the openness argument of Step 1 with $x_n \in G(A)$ in the role of $x$: the bound $\|x - x_n\| < 1/\|x_n^{-1}\|$ implies $\|1 - x_n^{-1} x\| < 1$, so $x_n^{-1} x \in G(A)$ by [Invertibility Near the Identity](/theorems/2667), and consequently $x = x_n (x_n^{-1} x) \in G(A)$. This contradicts $x \notin G(A)$. Hence $\|x_n^{-1}\| \to \infty$. [/step] [step:Construct unit vectors $z_n$ with $z_n x, x z_n \to 0$ at boundary points] Let $x \in \partial G(A) = \overline{G(A)} \setminus G(A)$. Choose $(x_n) \subset G(A)$ with $x_n \to x$; this is possible since $x \in \overline{G(A)}$. Since $x \notin G(A)$, by Step 4 we have $\|x_n^{-1}\| \to \infty$. In particular $x_n^{-1} \neq 0$ for all $n$ (and $x_n^{-1}$ is well-defined since $x_n \in G(A)$). Define \begin{align*} z_n := \frac{x_n^{-1}}{\|x_n^{-1}\|}, \end{align*} so $\|z_n\| = 1$. We compute \begin{align*} z_n x &= \frac{x_n^{-1}}{\|x_n^{-1}\|} \cdot x = \frac{x_n^{-1} x_n + x_n^{-1}(x - x_n)}{\|x_n^{-1}\|} = \frac{1}{\|x_n^{-1}\|} + z_n (x - x_n). \end{align*} Both terms tend to $0$: $1/\|x_n^{-1}\| \to 0$ since $\|x_n^{-1}\| \to \infty$; and $\|z_n(x - x_n)\| \leq \|z_n\| \|x - x_n\| = \|x - x_n\| \to 0$. Hence $z_n x \to 0$. The symmetric calculation $x z_n = 1/\|x_n^{-1}\| + (x - x_n) z_n \to 0$ gives the right-sided convergence. [/step] [step:Show $x$ has no one-sided inverse in any larger unital Banach algebra] Suppose $A$ embeds isometrically as a closed subalgebra of a unital Banach algebra $B$ with $1_B = 1_A$, and that $x$ has a right inverse $w \in B$, i.e.\ $xw = 1_B$. Multiplying the convergence $z_n x \to 0$ on the right by $w$ and using continuity of multiplication in $B$, \begin{align*} 0 = \lim_{n \to \infty} (z_n x) w = \lim_{n \to \infty} z_n (x w) = \lim_{n \to \infty} z_n \cdot 1_B = \lim_{n \to \infty} z_n, \end{align*} which contradicts $\|z_n\|_B = \|z_n\|_A = 1$ for all $n$ (the embedding is isometric, so $z_n$ does not converge to $0$ in $B$). Hence $x$ has no right inverse in $B$. The symmetric argument with $z_n$ on the right and a hypothetical left inverse $w$ of $x$ uses $x z_n \to 0$ to derive $w (x z_n) = (wx) z_n = z_n \to 0$, again contradicting $\|z_n\| = 1$. Hence $x$ has no left inverse in $B$ either. This completes the proof of (i)–(iv). [/step]