Inv(A) is an Open Subset of A in a Banach Algebra with Unit

# Inv(A) is an Open Subset of A in a Banach Algebra with Unit

Recall from the Invertibility of 1 - a When r(a) < 1 in a Banach Algebra with Unit page that if $\mathfrak{A}$ is a Banach algebra with unit and if $a \in \mathfrak{A}$ is such that $r(a) < 1$ then $1 - a$ is invertible and:

(1)\begin{align} \quad (1 - a)^{-1} = 1 + \sum_{n=1}^{\infty} a^n \end{align}

We also proved as a corollary that if $\mathfrak{A}$ is a Banach algebra with unit and if $\| 1 - a \| < 1$ then $a$ is invertible.

We will use these results to prove that $\mathrm{Inv}(\mathfrak{A})$ is always an open set when $\mathfrak{A}$ is a Banach algebra with unit.

Theorem 1: Let $\mathfrak{A}$ be a Banach algebra with unit. Then the set of invertible elements $\mathrm{Inv}(\mathfrak{A})$ is open in $\mathfrak{A}$. |

**Proof:**Let $a \in \mathrm{Inv}(\mathfrak{A})$. Let $y \in B \left ( a, \frac{1}{\| a^{-1} \|} \right )$. Then:

\begin{align} \quad \| a - y \| < \frac{1}{\| a^{-1} \|} \quad (*) \end{align}

- Observe that:

\begin{align} \quad a - y = a - aa^{-1}y = a(1 - a^{-1}y) \end{align}

- Therefore the inequality at $(*)$ can be rewritten as:

\begin{align} \quad \| a(1 - a^{-1}y) \| &< \frac{1}{\| a^{-1} \|} \\ \quad \| a^{-1} \| \|a (1 - a^{-1}y) \| &< 1 \\ \quad \| a^{-1}a(1 - a^{-1}y) \| & < 1 \\ \quad \| 1 - a^{-1}y \| & <1 \end{align}

- Since $\mathfrak{A}$ is a Banach algebra and since $r(1 - a^{-1}y) \leq \| 1 - a^{-1}y \| < 1$, by the Theorem on the Invertibility of 1 - a When r(a) < 1 in a Banach Algebra with Unit page we have that $1 - (1 - a^{-1}y) = a^{-1}y$ is invertible in $\mathfrak{A}$.

- Since $a$ and $a^{-1}y$ are invertible elements of $\mathfrak{A}$ and since $\mathrm{Inv}(\mathfrak{A})$ is closed under multiplication (since it is a group), we have that the product, $aa^{-1}y = y \in \mathrm{Inv}(\mathfrak{A})$.

- Thus we see that:

\begin{align} \quad B \left (a, \frac{1}{\| a^{-1}\|} \right ) \subseteq \mathrm{Inv}(\mathfrak{A}) \end{align}

- So for every $a \in \mathrm{Inv}(\mathfrak{A})$ there exists an open ball centered at $a$ which is fully contained in $\mathrm{Inv}(\mathfrak{A})$ and hence $\mathrm{Inv}(\mathfrak{A})$ is open in $\mathfrak{A}$. $\blacksquare$

Corollary 2: Let $a$ be a Banach algebra with unit. Then the set of singular elements $\mathrm{Sing}(\mathfrak{A})$ is closed in $\mathfrak{A}$. |

**Proof:**By Theorem 1, $\mathrm{Inv}(\mathfrak{A})$ is open. And so $\mathrm{Sing}(\mathfrak{A}) = \mathfrak{A} \setminus \mathrm{Inv}(\mathfrak{A})$ is closed. $\blacksquare$