# How can I understand that the set of invertible maps in $B(X,X)$ is open in $B(X,X)$?

Let $$X$$ be a complete normed space. I define $$B(X,X)=\{T:X\rightarrow X|~ T ~\text{linear and bounded}\}$$ where we have $$||T||=\sup \{||T(x)||: ||x||\leq 1\}$$I say that $$T\in B(X,X)$$ is invertible if there exists $$S:X\rightarrow X$$ linear, such that

1. $$STx=x$$ for all $$x$$
2. $$TSx=x$$ for all $$x$$
3. $$||S||<\infty$$

Now we have the theorem that the set of invertible maps in $$B(X,X)$$ is open in $$B(X,X)$$

Our prof told us that this mean that if we take a small ball around an operator, then the operators in this ball are still invertible.

But I somehow don't see how to get from the statement to this conclusion.

Could maybe someone help me further and explain this a bit in detail?

• Does this answer your question? Prove that the set of invertible elements in a Banach algebra is open Oct 20, 2022 at 18:28
• @AnneBauval no not quite, because I know how to prove this, what isn't clear to me is why this is equivalent to say that in a small ball around an operator all the other operators are also invertible. What does open have in common with this? Oct 20, 2022 at 18:30
• HINT: Start by showing that every operator in the (open) ball of radius $1$ centered at the identity is invertible. Oct 20, 2022 at 18:32
• @TedShifrin okey thanks I will try it! Oct 20, 2022 at 18:33
• By your answer to me, I understand that Ted Shifrin's hint is of no more use than my link, since you said you already know how to prove that in a small ball around an invertible operator, all the other operators are also invertible. If your only concern is to understand why this proves that the set of invertible operators is open, it will be very easy to solve! It is simply the definition of an open set in a normed vector space or more generally in a metric space. Oct 20, 2022 at 18:42