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:

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

- Observe that:

\begin{align} \quad x - y = x - xx^{-1}y = x(1 - x^{-1}y) \end{align}

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

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

- 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:

\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. $\blacksquare$