Deforming a linear map a little preserves surjectivity Let $X$ be a Banach space, and $A: X\to X$ be a surjective linear map. Define
$$\eta(A) = \{\lambda\in \mathbb{C}: A - \lambda I \text{ is surjective}\}$$
where $I: X\to X$ is the identity. Show that $\eta(A)$ is an open subset of $\mathbb{C}$. 
This is a problem from a graduate qualifying exam and I'm a bit stuck on it—it suffices to show that a neighborhood of $0$ in $\mathbb{C}$ is contained in $\eta(A)$. My initial idea was something like "given $Ax \in X$, consider $y\in A^{-1}(Ax + \lambda x)$. This $y$ should under $A - \lambda I$ map to someone close to $Ax$, if $\lambda$ is small. Then we use the difference between $Ay - \lambda y$ and $Ax$ to adjust $y$. We repeat and hope we get convergence". And that's about all I've got so far, but it requires the open mapping theorem, which requires $A$ to be continuous. Can the problem even be done without the continuity of $A$? Any help is appreciated, regarding whether the problem is possible without the continuity of $A$ and how actually to prove the result. 
Edit: this is Problem 6 of UCLA’s analysis qual from Spring 2007. 
 A: I got the solution assuming $A$ continuous (which is essentially a working out of my original idea), so I will add it here for posterity’s sake. 
By the open mapping theorem, $A$ maps the unit ball to a neighborhood of 0. Since the truth of the problem is invariant under scalar multiplication of the map $A$, we can without loss of generality assume 
$$A(\{||x||\leq 1\})\supseteq\{||x||\leq1\}$$
So then, for any $x\in X$, we can find an element, which we will somewhat abusively denote $A^{-1}x$, with norm less than that of $x$. If then $\lambda <1/2$, consider for $x\in X$ the element
$$y= \sum_{n=1}^\infty \lambda^{n-1}A^{-n}x$$
Where $A^{-n}x$ denotes $A^{-1}A^{-1}\dots A^{-1}x$. 
Since $||A^{-n}||\leq ||x||$, the infinite sum $y$ is dominated by a geometric series and so does indeed converge to a well-defined element of $X$. Applying $A-\lambda I$ to $y$ and using the continuity of $A$, we get after some telescoping 
$$(A-\lambda I) y=(\lim_{n\to\infty} x-\lambda^nA^{-n}x) =x$$
Since again, the norm of $\lambda^nA^{-n}x$ goes to 0. $x$ was arbitrary, so that $A-\lambda I$ is indeed surjective. 
A: No, it's not true in general; you need the continuity of $A$.
I don't have an explicit counterexample, but start with a Hamel basis $B$ for $X$ (infinite dimensional), and partition it into countably many non-empty sets $(B_n)_{n=1}^\infty$. Define a (necessarily unbounded) linear operator $T$ on $A$ by its action on $B$, by $T(x) = \frac{1}{n} x$, where $x \in B_n$.
It's not hard to see that $T$ is surjective. Pick any $y \in X$, and we may find basis vectors $x_1, \ldots, x_n \in B$ and scalars $a_1, \ldots, a_n$ such that
$$y = \sum_{i=1}^n a_i x_i.$$
For every $i = 1, \ldots, n$, we have that $x_i \in B_{m_i}$ for some $m_i$, and so
$$T\left(\sum_{i=1}^n a_i m_i x_i\right) = \sum_{i=1}^n a_i m_i T x_i = \sum_{i=1}^n a_i m_i \frac{1}{m_i} x_i = \sum_{i=1}^n a_i x_i = y.$$
Now, I claim that $\frac{1}{n} \notin \eta(T)$, even though $0 \in \eta(T)$. In particular,
$$\left(T - \frac{1}{n}I\right)^{-1}B_n = \emptyset.$$
Fix $y \in B_n$. If there were some $x \in X$ such that $\left(T - \frac{1}{n}I\right)x = y$, then we could express $x$ by
$$x = \sum_{i=1}^m a_i x_i$$
for some $x_i \in B$. Applying $T - \frac{1}{n} I$ to this linear combination will annihilate any terms with $x_i \in B_n$. Every term without a $0$ coefficient will not belong to $B_n$, but still, by assumption, will sum to $y \in B_n$. This violates the linear independence of the basis. Thus, $T - \frac{1}{n} I$ is not surjective, as required.
