Let $A$ be a commutative ring with a norm $∥·∥\geq0$ such that $∥ab∥\leq ∥a∥∥b∥$, $∥a+b∥\leq \text{max}\{∥a∥,∥b∥\}$, $∥1∥=1$ and $∥a∥=0$ if and only if $a=0$. We assume $A$ is complete for this norm.
If $∥x∥<1$ and $y$ is an invertible element, then how to prove $x+y$ is invertible?
This question is motivated by the proof of lemma 2 in the answer of this question, and user 10676 said that it is a well-known result, but I can't see this, because we can't deduce $∥y^{-1}∥<1$. Am I right?
Thanks!
In the answer you link to, there are additional conditions without which your statement does not necessarily hold, as Robert Israel notes in his comment.
Namely, what we have in the linked answer on top of what you write is that the norm is multiplicative (i.e. $\lvert \lvert ab \rvert \rvert = \lvert \lvert a \rvert \rvert \lvert \lvert b \rvert \rvert $), and that the invertible element you call $y$ has norm $\lvert \lvert y \rvert \rvert = \mathbf{1}$. Then it is indeed clear that
$\lvert \lvert xy^{-1} \rvert \rvert < 1$, hence
$1+xy^{-1}$ is invertible (geometric sum, ring complete), hence
$y+x = y(1+xy^{-1})$ is invertible.