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

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

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

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

We also proved as a corollary that if $X$ is a Banach algebra with unit and if $\| 1 - x \| < 1$ then $x$ is invertible.

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

 Theorem 1: Let $X$ be a Banach algebra with unit. Then the set of invertible elements $\mathrm{Inv}(X)$ is open in $X$.
• Proof: Let $x \in \mathrm{Inv}(X)$. Let $y \in B \left ( x, \frac{1}{\| x^{-1} \|} \right )$. Then:
(2)
\begin{align} \quad \| x - y \| < \frac{1}{\| x^{-1} \|} \quad (*) \end{align}
• Observe that:
(3)
\begin{align} \quad x - y = x - xx^{-1}y = x(1 - x^{-1}y) \end{align}
• Therefore the inequality at $(*)$ can be rewritten as:
(4)
\begin{align} \quad \| x(1 - x^{-1}y) \| &< \frac{1}{\| x^{-1} \|} \\ \quad \| x^{-1} \| \|x (1 - x^{-1}y) \| &< 1 \\ \quad \| x^{-1}x(1 - x^{-1}y) \| & < 1 \\ \quad \| 1 - x^{-1}y \| & <1 \end{align}
• Since $x$ and $x^{-1}y$ are invertible elements of $X$ and since $\mathrm{Inv}(X)$ is closed under multiplication (since it is a group), we have that the product, $xx^{-1}y = y \in \mathrm{Inv}(X)$.
• Thus we see that:
(5)
\begin{align} \quad B \left (x, \frac{1}{\| x^{-1}\|} \right ) \subseteq \mathrm{Inv}(X) \end{align}
• So for every $x \in \mathrm{Inv}(X)$ there exists an open ball centered at $x$ which is fully contained in $\mathrm{Inv}(X)$ and hence $\mathrm{Inv}(X)$ is open in $X$. $\blacksquare$
 Corollary 2: Let $X$ be a Banach algebra with unit. Then the set of singular elements $\mathrm{Sing}(X)$ is closed in $X$.
• Proof: By Theorem 1, $\mathrm{Inv}(X)$ is open. And so $\mathrm{Sing}(X) = X \setminus \mathrm{Inv}(X)$ is closed. $\blacksquare$