In Walter Rudin's Principles of mathematical analysis Exercise 5.21, it is proved that for any closed subset $E\subseteq \mathbb{R}$, there exists a smooth function $f$ on $\mathbb{R}$ such that $E=\{x\in \mathbb{R}\mid f(x)=0\}$.
Since $E^c$ is open in $\mathbb{R}$, $E^c$ is a countable disjoint union of open intervals $(a_i,b_i)$. On an interval $(a,b)$, if we let $f(x)=\exp(\frac{1}{(x-a)(x-b)})$, $x\in(a,b)$; $f(x)=0$, $x\notin(a,b)$, ($f(x)=\exp(\frac{1}{(x-a)})$ or $f(x)=\exp(\frac{1}{(x-b)})$ on $(a,\infty)$ and $(-\infty,b)$ resp.), then $f$ is smooth. Hence we can get a smooth function on $\mathbb{R}$ such that $E$ is the zero point sets of $f$.
(1). How about a closed subset $E$ of $\mathbb{R}^n$? is it true that for any closed subset $E\subseteq \mathbb{R}^n$, there exists a smooth function $f$ on $\mathbb{R}^n$ such that $E=\{x\in \mathbb{R}^n\mid f(x)=0\}$?
An open set in $\mathbb{R}^n$ cannot be written as a disjoint union of countable open balls hence the proof above is not valid. The partition of unity only claims that for an open set $U$, there exists open $V$ in $U$ such that $\bar V\subseteq U$ and a smooth function $f$ on $\mathbb{R}^n$ such that $supp f\subseteq U$, $f|_V=1$.
(2). The taylor series of $f(x)=\exp(\frac{1}{(x-a)(x-b)})$ does not converge around points $a,b$ hence $f$ is not analytic at $a,b$. Is it true that for any closed subset $E\subseteq \mathbb{R}$, there exists an analytic function $f$ on $\mathbb{R}$ such that $E=\{x\in \mathbb{R}\mid f(x)=0\}$?
Best Answer
1) Let $\{B_{r_j}(a_j)\}$ be a countable collection of open balls (radius $r_j$, centre $a_j$) whose union is the complement of $E$. Take $f(x) = \sum_j c_j g(\|x - a_j\|/r_j)$ for some smooth function $g$ which is nonzero on $[0,1)$ and $0$ on $[1,\infty)$ and a suitable sequence of positive numbers $c_j$.
2) The zeros of a nonconstant analytic function form a discrete set.