# Zero points of a smooth function on $\mathbb{R}$

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.

• mathoverflow.net/questions/196167/… Aug 16, 2021 at 5:56
• It has been known for almost a century that any closed set in a manifild is the zero set of an infinitely smooth function (Whitney). Aug 16, 2021 at 6:20
• Another account: mathoverflow.net/a/24037/1946 Aug 16, 2021 at 6:22
• @bathalf15320 I find your comment unnecessarily dismissive. Are you claiming to know everything that was known a century ago? This xkcd comic seems relevant: xkcd.com/1053. Aug 16, 2021 at 17:32
• It seems to me that if an answer on MO fails to acknowledge that it is a special case of known results, especially of such a celebrated one, either out of ignorance or for some other reason, then it is a matter of common courtesy to point this out--precedence is one of the most valued treasures of a mathematician, or indeed any creative artist. Jun 24 at 7:52