Why is the space of surjective operators open? Suppose $E$ and $F$ are given Banach spaces. Let $A$ be a continuous surjective map. Why is there a small ball around $A$ in the operator topology, such that all elements in this ball are surjective?
 A: Since this question bugged me, I decided to write down the proof (I don't have access to Lang's book, so I hope my argument is not much more complicated than necessary). The idea is the same as in the proof of the Banach-Schauder theorem.
By the open mapping theorem we may scale the norm on $E$ in such a way that $A$ maps the unit ball of $E$ onto the unit ball of $F$, that is $B_{\leq 1} F \subset A(B_{\leq 1}E)$. Since $A$ is linear we have $B_{\leq r} F \subset A(B_{\leq r}F)$ for all $r > 0$.
Claim. If $B: E \to F$ is such that $\alpha := \|A - B\| < 1$ then $B$ is onto.
Proof. Let $f \in F$. We want to show that there is $e$ such that $f = Be$. For convenience, we put $f_{0} = f$ and assume $\|f_{0}\| \leq 1$.
Choose $e_{0}$ with $\|e_{0}\| \leq 1$ such that $Ae_{0} = f_{0}$. Define $f_{1} = f_{0} - Be_{0}$ and observe $\|f_{1}\| = \|(A - B) e_{0}\| \leq \alpha$, so we may choose $e_{1}$ with $\|e_{1}\| \leq \alpha$ such that $Ae_{1} = f_{1}$. Now $f_{2} = f_{1} - Be_1$ has norm $\|f_{2}\| = \|(A - B)e_{1}\| \leq \alpha^{2}$, so we obtain by induction two sequences $\{f_{n}\}_{n=0}^{\infty}$ and $\{e_{n}\}_{n=0}^{\infty}$ having the following properties:


*

*$\|e_{n}\|, \|f_{n}\| \leq \alpha^{n}$ for all $n$.

*$e_{n}$ is such that $f_{n} = A(e_{n})$,

*$f_{n+1} = f_{n} - B(e_{n}) = (A-B)(e_{n})$.


Finally $e = \sum_{n = 0}^{\infty} e_{n}$ has norm $\|e\| \leq \sum_{n=0}^{\infty} \alpha^{n} = \frac{1}{1-\alpha}$ and, moreover,
\[
B(e) = \sum_{n=0}^{\infty} B(e_{n}) = \sum_{n=0}^{\infty} (f_{n} - f_{n+1}) = f_{0} = f,
\]
as we wanted to show.
A: A bounded linear operator $A: E \to F$, where $E$ and $F$ are Banach spaces, is surjective if and only if there is $c > 0$ such that for all $\phi \in F^*$,
$\|A^* \phi\| \ge c \|\phi\|$ (see e.g. Rudin, "Functional Analysis", Theorem 4.15).
Now if $A$ is such an operator, so is $B$ for $\|A-B\| < c$, since 
 $$\|B^* \phi\| \ge \|A^* \phi\| - \|A^* - B^*\| \|\phi\| \ge (c - \|A-B\|) \|\phi\|$$
