The Set of Quasi-Invertible Elements q-Inv(X) is an Open Subset of X

# The Set of Quasi-Invertible Elements q-Inv(X) is an Open Subset of X

Theorem 1: Let $X$ be a Banach algebra. Then the set of quasi-invertible elements in $X$, $\mathrm{q-Inv}(X)$ is an open subset of $X$. |

**Proof:**Recall that $x$ is quasi-invertible with quasi-inverse $y$ if and only if $(0, 1) - (x, 0)$ is invertible with inverse $(0, 1) - (y, 0)$ in $X + \mathbf{F}$. Therefore:

\begin{align} \quad \mathrm{q-Inv}(X) = \left \{ x : (0, 1) - (x, 0) \in \mathrm{Inv}(X + \mathbf{F}) \right \} \end{align}

- Let $f : X \to X + \mathbf{F}$ be the function defined for all $x \in X$ by:

\begin{align} \quad f(x) = (0, 1) - (x, 0) \end{align}

- Observe that

\begin{align} \quad \mathrm{q-Inv}(X) \overset{(*)} = f^{-1}(\mathrm{Inv}(X + \mathbf{F}) \end{align}

- To see this, let $x \in \mathrm{q-Inv}(X)$ and let $y$ be its quasi-inverse. Then by one of the theorems on the Basic Theorems Regarding Quasi-Invertible Elements in an Algebra, since $(0, 1) - (x, 0)$ is invertible with inverse $(0, 1) - (y, 0)$ in $X + \mathbf{F}$. In particular, $f(x) \in \mathrm{Inv}(X + \mathbf{F})$ and so clearly:

\begin{align} \quad x \in f^{-1}((0, 1) - (x, 0)) \in f^{-1}(\mathrm{Inv}(X + \mathbf{F}) \end{align}

- Now let $x \in f^{-1}(\mathrm{Inv}(X + \mathbf{F})$. Then $f(x) \in \mathrm{Inv}(X + \mathbf{F})$ and so $f(x) = (0, 1) - (x, 0)$ is invertible. Let $(z, \alpha)$ be it's inverse. Then by definition of inverse:

\begin{align} \quad (0, 1) &= [(0, 1) - (x, 0)](z ,\alpha) = (-x, 1)(z, \alpha) = (-xz + z - \alpha x, \alpha) \\ \quad (0, 1) &= (z, \alpha)[(0, 1) - (x, 0)] = (z, \alpha)(-x, 1) = (-zx - \alpha x + z, \alpha) \end{align}

- From the above two equations we immediately see that $\alpha = 1$, and so we obtain the following two equations:

\begin{align} \quad xz - z + x &= 0 \\ \quad zx + x - z &= 0 \end{align}

- These equations are equivalent to:

\begin{align} \quad x \circ (-z) &= 0 \\ \quad (-z) \circ x &= 0 \end{align}

- So $x$ is quasi-invertible (with quasi-inverse $-z$), so $x \in \mathrm{q-Inv}(X)$. So indeed $(*)$ holds.

- The function $f$ is continuous on $X$. To see this, observe that the function $g : X \to X + \mathbf{F}$ defined by $g(x) = (x, 0)$ is an isomorphism of $X$ onto the subspace $\{ (x, 0) : x \in X \}$ of $X + \mathbf{F}$ (as we have mentioned earlier), and thus $g$ is continuous on $X$, so clearly $f = (0, 1) + g$ is continuous on $X$.

- From the theorem on The Set of Invertible Elements Inv(X) is an Open Subset of X, since $X + \mathbf{F}$ is a Banach algebra with unit $(0, 1)$, we have that $\mathrm{Inv}(X + \mathbf{F})$ is open in $X + \mathbf{F}$. By the continuity of $f$, $f^{-1}(\mathrm{Inv}(X + \mathbf{F}))$ is open in $X$.

- So from $(*)$, $\mathrm{q-Inv}(X)$ is an open subset of $X$. $\blacksquare$