Exercise 5.21

I’m going to show this for the infinitely differentiable case (and even that won’t be complete). Cauchy’s function $F(x)=e^{-1∕x^2}$ for $x\ne 0$, $F(0)=0$, is the classic counterexample of a non-constant, infinitely differentiable real function such that $f^{(n)}(0)=0$ for all orders $n$ (you can easily find a proof of this, or you can try it yourself). We can use this to define a function $\tilde{F}(x)=F(x)$ for $x>0$, $\tilde{F}(x)=0$ for $x\le 0$, which is infinitely differentiable everywhere on $R$ and whose zero set is $(-\infty ,0]$. Finally, for $a<b$ let $F_{a,b}(x)=\tilde{F}(x-a)\tilde{F}(b-x)$, an infinitely differentiable function on $R$ whose zero set is $(-\infty ,a]\cup [b,\infty )$.

The complement of a closed set $E$ of $R$ is an open set consisting of a (possibly infinite) collection of open intervals $(a_n,b_n)$. The function $f(x)=\sum <!--\; FUNCTION\; APPLICATION\; -->F_{a_n,b_n}(x)$ is well-defined since at most one of the terms in the sum is non-zero for any given $x$, infinitely differentiable everywhere on $R$, and has zero set $E$.