Is $A + zB$ invertible for sufficiently small $z>0$, when $A$ is a invertible operator, and $B$ is a bounded linear operator? I have a question on the existence of an inverse for a particular operator. A similar question in the case of a matrix a matrix was asked previously on m.se. In this question a users asks if the matrix $A-\varepsilon I$ is invertible for sufficiently small $\varepsilon > 0$.
The answer is yes, and here is one of the answers given to that question.

Yes, if $\mathbf A$ is any $n \times n$ matrix, then $\mathbf  A+\epsilon \mathbf I$ is invertible for sufficiently small $\epsilon > 0$. This is because $\det (\mathbf A + \epsilon \mathbf I)$ is a
  polynomial in $\epsilon$ of degree $n$, and so it has a finite number
  of zeroes.

My question is as follows. Is $A - zB$ invertible for sufficiently small $z>0$, where $A$ is an invertible operator, and $B$ is a bounded linear operator. Is so, why is this the case?
If these were $n \times n$ matrices I think it would be true as once we take $z$ small enough we can be sure that subtracting $zB$ from $A$ would not lead to any zero rows in the resulting $A-zB$. But I don't know if anything similar can be justified in the case of bounded linear operators?
 A: Yes. The set of invertible operators is open. 
More concretely, the geometric series implies that $X$ is invertible if $\|I-X\|<1$. Now suppose that $\|A-C\|<\frac1{\|A^{-1}\|}$. Then
$$
\|I-A^{-1}C\|=\|A^{-1}(A-C)\|\leq\|A^{-1}\|\,\|A-C\|<1.
$$
So $A^{-1}C$ is invertible, and thus $C$ is invertible. 
In the end, if $|z|<\frac1{\|B\|\,\|A^{-1}\|}$, we have 
$$
\|A-(A+zB)\|=\|zB\|=|z|\,\|B\|<\frac1{\|A^{-1}\|}
$$
and then $A+zB$ is invertible. 
A: One standard proof is as follows: first, note that $A + zB$ is invertible if and only if $A^{-1}(A + z B) = I + zA^{-1}B$ is invertible.  To that end, it suffices to note that $I + zB$ is invertible if $z$ is sufficiently small and $B$ is a bounded operator.
Now, consider the infinite sum
$$
C = \sum_{k=0}^\infty (-zB)^k
$$
where $B^0 = I$, the identity operator.  Show that if this infinite sum converges (in operator norm), then the result is a $2$-sided inverse to the operator $I + zB$. Then show that for sufficiently small $z$, the sum converges.
