Why is this zig zag function monotone increasing at $0$, but not monotone increasing in a neighborhood of $0$?

This is from The Way of Analysis by Strichartz.

I believe see a neighborhood of $0$ such that $x_1 < x_2 \implies f(x_1) \le f(x_2)$.

Why is this zig zag function monotone increasing at $0$, but not monotone increasing in a neighborhood of $0$?

• I have a couple of questions: what does it mean to be monotone increasing at a point? What is the definition of the function in question? – Theo Bendit Nov 5 '17 at 2:43
• Presumably there are to be infinitely many wiggles, say hitting the $x$-axis at $1/n$ for every $n\ne0$. – Lubin Nov 5 '17 at 2:45
• @TheoBendit I've included the defintions – Al Jebr Nov 5 '17 at 2:46
• It's not increasing on any neighbourhood of $0$ because any such neighbourhood will contain a downwards "zag". (The picture could perhaps be better - I believe it's trying to communicate a "fractal" zig-zag as Lubin described.) – Anthony Carapetis Nov 5 '17 at 2:51

I think the idea is that the function is fractal, in the sense that as we approach $0$ from either side, the function has smaller spikes with increasing frequency. Indeed, we can see that the function $f$ is such that $f\le 0$ for $x<0$, $f\ge 0$ for $x>0$, and $f=0$ at $x=0$.
$f$ is monotone at $0$, because given $x_1<x_0<x_2$ we have that $f(x_1)\le f(x_0)=0$ and $f(x_2)\ge f(x_0)=0$ so that $f(x_1)\le f(x_0)\le f(x_2)$. On the other hand, because of this spiking behavior, any neighborhood of $0$, $N=\{x\in \mathbf{R}: \lvert x\rvert<\epsilon\}$, contains an entire spike, which is clearly not a monotone function, since it increases, and then decreases. So $f$ is not monotone in a neighborhood of $0$.