Assume $f(x)$ is a smooth function on $\mathbb{R}$ and $f$ does not vanish on any interval. In other words, $f$ can have zero points but we cannot find any interval $(a, b)$ such that $f(x)=0$ for all $x \in (a, b)$. Denote by $\mathcal{Z} = \{x \in \mathbb{R}, f(x) = 0\}$ the zero point set of $f$. Assume $\mathcal{Z}$ is non-empty.


**Question:** Is it possible that every point in $\mathcal{Z}$ is an accumulation point of $\mathcal{Z}$?


Here $x$ is an accumulation point of $\mathcal{Z}$ means there exists a sequence $\{x\_n\}$ such that $x\_n \in \mathcal{Z}$, $x\_n \neq x$ and $\lim\_{n\to \infty}x\_n = x$ under the usual Euclidean topology.



Zero points of a smooth function on $\mathbb{R}$
The answer is yes.


We can make a smooth function whose zero set is the Cantor set. Simply place a smooth function on each middle third segment as you build the Cantor set, so that on each of these segments, the function arches from 0 up to a height that vanishes quickly as the segments become small, and then back down to 0, but smooth, and with all derivatives vanishing on the endpoints.


The zero set of this function will be the Cantor set itself, which is a closed nowhere dense set in which every point is an accumulation point. The function is smooth on each middle-third segment, and it is smooth at the points of the Cantor set, since the heights of the pieces on the segments approaching it vanish quickly enough.


In fact, this method can show that every closed set in the reals is the zero set of a smooth function. One places a little smooth bump into each open interval of the complement, and by making the heights sufficiently low as the intervals become small, you can get smoothness at the limits of these bumps.


