# Group of Units in Unital Banach Algebra is Open

## Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra.

Let $\map G A$ be the group of units of $A$.

Then $\map G A$ is open in $A$.

## Proof

Let $x \in \map G A$.

We find an open neighborhood of $x$ contained in $\map G A$.

Clearly $x^{-1} \ne \mathbf 0_A$, so $\norm {x^{-1} } > 0$.

We have, for $y \in A$:

 $\ds \norm {1 - x^{-1} y}$ $=$ $\ds \norm {x^{-1} \paren {x - y} }$ $\ds$ $\le$ $\ds \norm {x^{-1} } \norm {x - y}$

So for $y \in A$ satisfying:

$\ds \norm {x - y} < \frac 1 {\norm {x^{-1} } }$

we have:

$\norm {1 - x^{-1} y} < 1$

From Element of Unital Banach Algebra Close to Identity is Invertible, we have that $x^{-1} y$ is invertible.

That is, $x^{-1} y \in \map G A$.

Since $\map G A$ is a group, from Group of Units is Group, we have:

$y = x \paren {x^{-1} y} \in \map G A$

So we have:

$\map {B_{\norm {x^{-1} }^{-1} } } x \subseteq \map G A$

where $\map {B_{\norm {x^{-1} }^{-1} } } x$ is the open ball centered at $x$ with radius $\norm {x^{-1} }^{-1}$.

Since $x \in \map G A$ was arbitrary, we have that $\map G A$ is open.

$\blacksquare$