q-Inv(A) is an Open Subset of A in a Banach Algebra

Theorem 1: Let $\mathfrak{A}$ be a Banach algebra. Then $\mathrm{q-Inv}(\mathfrak{A})$ is an open subset of $\mathfrak{A}$.

• 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 $\mathfrak{A} + \mathbf{F}$. Therefore:

(1)

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

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

(2)

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

• Observe that

(3)

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

• To see this, let $x \in \mathrm{q-Inv}(\mathfrak{A})$ and let $y$ be its quasi-inverse. Then since $(0, 1) - (x, 0)$ is invertible with inverse $(0, 1) - (y, 0)$ in $\mathfrak{A} + \mathbf{F}$. In particular, $f(x) \in \mathrm{Inv}(\mathfrak{A} + \mathbf{F})$ and so clearly:

(4)

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

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

(5)

\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:

(6)

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

• These equations are equivalent to:

(7)

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

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

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

• From the theorem on Inv(A) is an Open Subset of A in a Banach Algebra with Unit page, since $\mathfrak{A} + \mathbf{F}$ is a Banach algebra with unit $(0, 1)$, we have that $\mathrm{Inv}(\mathfrak{A} + \mathbf{F})$ is open in $\mathfrak{A} + \mathbf{F}$. By the continuity of $f$, $f^{-1}(\mathrm{Inv}(\mathfrak{A} + \mathbf{F}))$ is open in $\mathfrak{A}$.

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

Corollary 2: Let $\mathfrak{A}$ be a Banach algebra. Then $\mathrm{q-Sing}(\mathfrak{A})$ is a closed subset of $\mathfrak{A}$.

• Proof: $\mathrm{q-Sing}(\mathfrak{A}) = \mathfrak{A} \setminus \mathrm{q-Inv}(\mathfrak{A})$. $\blacksquare$