# invertible elements in a Banach algebra form an open set

Theorem - Let $\mathcal{A}$ be a Banach algebra with identity element $e$ and $G(\mathcal{A})$ be the set of invertible elements in $\mathcal{A}$. Let $B_{r}(x)$ denote the open ball of radius $r$ centered in $x$.

Then, for all $x\in G(\mathcal{A})$ we have that

 $B_{\|x^{-1}\|^{-1}}(x)\subseteq G(\mathcal{A})$

and therefore $G(\mathcal{A})$ is open in $\mathcal{A}$.

Proof : Let $x\in G(\mathcal{A})$ and $y\in B_{\|x^{-1}\|^{-1}}(x)$. We have that

 $\|e-x^{-1}y\|=\|x^{-1}x-x^{-1}y\|=\|x^{-1}(x-y)\|\leq\|x^{-1}\|\|x-y\|<\|x^{-1% }\|\|x^{-1}\|^{-1}=1$

So, by the Neumann series (http://planetmath.org/NeumannSeriesInBanachAlgebras) we conclude that $e-(e-x^{-1}y)$ is invertible, i.e. $x^{-1}y\in G(\mathcal{A})$.

As $G(\mathcal{A})$ is a group we must have $y\in G(\mathcal{A})$.

So $B_{\|x^{-1}\|^{-1}}(x)\subseteq G(\mathcal{A})$ and the theorem follows. $\square$

Title invertible elements in a Banach algebra form an open set InvertibleElementsInABanachAlgebraFormAnOpenSet 2013-03-22 17:23:22 2013-03-22 17:23:22 asteroid (17536) asteroid (17536) 6 asteroid (17536) Theorem msc 46H05