Prove that the set of invertible elements in a Banach algebra is open I am reading Banach algebras. There is a question in the text that show that set of all invertible elements is an open set.
I have thought a lot about it, considering it as a normed space but am not exactly sure how to show that it has an open ball around any invertible element.
 A: I tried to explain step by step, because I studied the subject, here what I wrote at the time
Definition: Let $A$ be  Banach algebra and G(A) the set of it's invertible elements.
Theorem:
 If $x$ is invertible, $h \in A$ and $|h| < \frac{1}{2||x^{-1}||}$.  Then,
$x+h \in G(A)$
Proof:
We have  $||x^{-1}h|| \leq ||x^{-1}|| \;||h|| < \frac{1}{2}<1$. So, $1_A +x^{-1}h$ is invertible by the series
$$(1_A +x^{-1}h)^{-1} =\sum^{\infty}_{k=0} (-x^{-1}h)^k, $$
which converges because $||x^{-1}h||<1.$ So,
$$x+h=x(1_A+x^{-1}h) $$ is invertible,
because it's the product of two invertible elements, $x$ and $(1_A+x^{-1}h).$
Theorem: 
For every Banach algebra $A$, $G(A)$ is a open subset of $A$
Proof:
$G(A)$ is open, because given $x \in G(A)$, we have
 $$B(x, \frac{1}{2||x^{-1}||}) \subset G(A). $$
 it follows from
 $$ B(x, \frac{1}{2||x^{-1}||}) =\{ y \in A\;|\; ||y-x||\leq \frac{1}{2||x^{-1}||}  \}$$
 we take $h=y-x,$ so $y=x+h$. By the last theorem we have $x+h \in G(A)$, given $x \in A$ and $||h|| \leq \frac{1}{2||x^{-1}||}. $
A: It's because if $a$ is invertible and $\|x\|<\frac1{\|a^{-1}\|}$, then$$(a+x)\left(a^{-1}-(a^{-1})^2x+(a^{-1})^3x^2-(a^{-1})^4x^3+\cdots\right)=1.$$
