There is an interesting exercise on my Analysis book that I have not been able to solve:
Let $\mathbb{E,F}$ be Banach spaces, $f:\mathbb{E}\to\mathbb{F}$ of type $\mathcal{C}^k$, $k\geq1$. Asume $f'(x)$ is inversible for all $x\in\mathbb{E}$. Show that:
i) If $S=f^{-1}\{0\}$ , then $S$ does not have limit points.
ii) $Int(S)=\emptyset$ and for every compact $K\subseteq\mathbb{E}$, $K\cap S$ is finite.
iii) If $\mathbb{F}$ is finite dimentional, $S$ is enumerable, and $S$ is bounded, then $S$ is finite.
I did prove ii): Using the fact that $f$ is continuous, then $S$ is closed and $S\cap K$ is compact. So, if $S\cap K$ were not finite, then it will have a limit point by Bolzano-Weirstrass property on compact spaces. Then $S$ must be finite.
For part iii) I wanted to use the fact that if a space is totally bounded and complete, then its compact, and then use part ii). But I realized that by Riesz's Lemma one can have bounded sets that are not totally bounded.
I have no idea with part i). I think I have to use Inverse function theorem, but I really dont know how.
Any help will be very appretiated.
Advanced greetings.
For i), suppose $x\in E.$ Then $f(y) = f(x) +f'(x)(y-x) + \epsilon(y),$ where $\epsilon(y)/|y-x| \to 0$ as $y\to x.$
Now $f'(x)$ is a bounded linear invertible map from $E$ to $F.$ This implies there is a positive constant $c$ such that $|f'(x)(z)| \ge c|z|$ for $z\in E.$ Thus
$$|f(y)- f(x)| = |f'(x)(y-x)+ \epsilon(y)| \ge |f'(x)(y-x)|- |\epsilon(y)| \ge c|(y-x)|- |\epsilon(y)| = |(y-x)|(c- |\epsilon(y)|/|y-x|).$$
For $y$ close to $x, |\epsilon(y)|/|y-x| <c/2.$ For such $y$ we then get $|f(y)- f(x)|\ge (c/2)|y-x|.$
Thus there is a neighbood $U$ of $x$ such that $f(y) \ne f(x)$ for $y\in U\setminus \{x\}.$ It follows from this that the set where $f$ takes on any particular value in $F$ is discrete, in particular the value $0.$