The set where a derivative vanishes is G-delta

Question

If $f:I\to R$ ($I$ - interval) is differentiable, then $\{x\colon f'(x)=0\}$ is a $G_{\delta}$ set.

The lecturer didn't prove this fact and I found no proof in my books. How it can be proven?

Answer 1

For positive integers $M,N$, let $$\eqalign{L(M,N) &= \bigcap_{n > N} \{x \; : \; n (f(x + 1/n) - f(x)) \ge 1/M \}\cr U(M,N) &= \bigcap_{n > N} \{x \; : \; n (f(x + 1/n) - f(x)) \le - 1/M \}}$$ Each $L(M,N)$ and $U(M,N)$ is the intersection of closed sets, therefore closed.
Given that $f$ is everywhere differentiable, $f'(x) = 0$ if and only if for all $M$ and $N$, $x \notin L(M,N) \cup U(M,N)$. Thus $$ \{x: f'(x) = 0\} = \bigcap_{M,N \in \mathbb N} (L(M,N) \cup U(M,N))^c$$ is a $G_\delta$.