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.


2 Answers 2


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 $A$


$G(A)$ 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}||}. $

  • 2
    $\begingroup$ Very nice @RenjiRodrigo $\endgroup$ Jan 20, 2018 at 16:40
  • 1
    $\begingroup$ Do you need the factor of $\frac12$? $\endgroup$
    – mr_e_man
    Nov 21, 2019 at 3:42
  • $\begingroup$ Not really @mr_e_man, any <1 constant is sufficient. $\endgroup$ May 11 at 2:42
  • $\begingroup$ @prikarsartam - I think you don't need any constant; it's sufficient that $|h|<1/|x^{-1}|$. $\endgroup$
    – mr_e_man
    May 11 at 2:49
  • $\begingroup$ Right, makes sense. $\endgroup$ May 11 at 4:01

It's because if $a$ is invertible and $\|x\|<\frac1{\|a^{-1}\|}$, then$$(a+x)\left(a^{-1}-(a^{-1})^2x+(a^{-1})^3x^2-(a^{-1})^4x^3+\cdots\right)=1.$$


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