Suppose $f^{(k)}(0) = 0$ for all $k \ge 0$, where $f$ is $C^\infty$. Is 0 then an accumulation point of a set where $f$ vanishes?

Question

Question in title. I'm wondering to what extent some form of the identity theorem for analytic functions still holds for smooth functions. For an analytic function on an open connected set, all derivatives vanishing at a point imply that the function is identically zero. Does a weaker statement hold for $C^{\infty}$ functions hold, namely that $f^{(k)}(0) = 0$ for all $k \ge 0$ implies that 0 is a non-isolated root of $f$? If so, can this statement be strengthened at all?

Answer 1

I actually just came up with a counterexample: let $f(x) = e^{-\frac{1}{|x|}}$ for $x \neq 0$, and $f(0) = 0$. All derivatives of $f$ vanish at $0$, but clearly $f$ has only one root.

My intuition was that for $a$ in an open interval $I$ about zero, we can write $f(a) = O(|a|^n)$ for all $n \in \Bbb{N}$ using Taylor's theorem. This is not enough to conclude that $f$ vanishes sufficiently close to zero, if $f$ decays exponentially as you approach zero.