The fiber $f^{-1}(a)$ is a discrete (and countable) set

Question

Let $f:ℝ→ℝ$ be a real analytic function. Then my question is: Show that for a real number $a$, the fiber $f^{-1}(a)$ is a discrete (and countable) set unless $f = a$.

Answer 1

This should be trivial if you have the identity theorem for ananlytic functions, but here's how to show it directly:

Assume $f^{-1}(a)$ has a limit point $c$. Then there is a strictly monotonic sequence $x_n\to c$ with $f(x_n)=a$. Hence between $x_n$ and $x_{n+1}$ there exists a $\xi_n$ with $f'(\xi_n)=0$ and by continuity $f'(c)=f'(\lim \xi_n)=\lim f'(\xi_n)=0$. That is, $c$ is a limit point of $(f')^{-1}(0)$. By induction, $c$ is limit point of $(f^{(n)})^{-1}(0)$ for all $n\ge 1$, hence $f^{(n)}(c)=0$ for all $n\ge 1$ and the Taylor expansion around $c$ is constant. Let $u=\inf\{x\in\mathbb R\colon f|_{[x,c]}=a\}$. If $u\ne-\infty$, then $u$ is a limit point of $f^{-1}(a)$ and by what we just saw, $f=a$ on an open neighbourhood of $c$, contradicting the definition of $u$ as $\inf$; we conclude $u0-\infty$. Similarly, $\sup\{x\in\mathbb R\colon f|_{[c,x]}=a\}=+\infty$ and ultimately $f=a$.

Therefore, unless $f$ is constant, $f^{-1}(a)$ is discrete. As a discrete subset of $\mathbb R$, it is at most countable.