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I'm trying to understand the proof of the following statement:

Statement: Let $X$ be a banach spase. We denote by $B(X,X)$ the set of all bounded linear operators from $X$ to $X$. Then the set of all invertible maps in $B(X,X)$ is open in $B(X,X)$.

Proof: Take invertible $T \in B(X,X)$. When $||S|| < ||T^{-1}||^{-1}, ||T^{-1}S||\leq ||T^{-1}|| \cdot||S|| < 1$.

By a proven Lemma we can say that $I+T^{-1}S$ is invertible. Then $T \cdot (I+T^{-1}S)$ is also invertible.

Question: I do not see how this proves the statement. I would have started in the following way:

Let $\Omega$ be the set of invertible functions s.t. $\Omega \subset B(X,X)$. My goal is to show that for any $T \in \Omega$ there exists a $\delta > 0$ such that $||S-T|| < \delta \implies S \in B(X,X)$.

From the question here: Open subsets of the space of linear operators I see that $||S-T|| < ||T^{-1}||^{-1}$ implies that $S$ is invertible. Using that I can conclude that $\Omega$ is open.

I don't get it how they proved this in the first proof. Following is not clear:

  • I learned that if $||T^{-1}S|| < 1 \implies I -T^{-1}S$ invertible. Is the same true for $I +T^{-1}S$?
  • The multiplication of two invertible operators is invertible. I get that. But how does the fact that $T \cdot (I+T^{-1}S)$ is invertible allow me to conclude the statement?
MyGanton
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  • $T(I+T^{-1}S)=T+S.$ Hence $T+S$ is invertible for any $|S|<\delta:=|T^{-1}|^{-1}.$ This means that the open ball $B_\delta(T)$ is contained in the set of invertible operators. – Ryszard Szwarc Dec 26 '22 at 19:54
  • So $T+S$ is invertible and since $T$ is invertible, also $S$ is invertible? – MyGanton Dec 26 '22 at 19:57
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    No. If $T$ is invertible then $T+S$ is invertible if $S$ is small enough, e.g. $|S|<\delta.$ The operator $S$ can be $0.$ Every operator $U$ close to $T$ can be represented as $U=T+S,$ where $S$ is close to $0.$ – Ryszard Szwarc Dec 26 '22 at 20:01
  • Ah, I got it! Is $I+T^{-1}S$ always invertible when $||T^{-1}S|| < 1$? Is the proof similar to the other case? – MyGanton Dec 26 '22 at 20:11
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    $|T^{-1}S|=|-T^{-1}S|$ – Ryszard Szwarc Dec 26 '22 at 20:14

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