 Solution

For $x\in X$, let $e_x$ be the indicator function of $\{x\}$. In the ring of functions $R={\mathbb{R}}^X$, multiplication is pointwise, soSuppose first that $X$ is finite, and let $J$ be an ideal of $R$. Define $S=\{x\in X:f(x)\ne 0\text{ for some }f\in J\}$. If $x\in S$, choose such an $f$; multiplying it by $e_x/f(x)$ shows that $e_x\in J$. Every $f\in J$ satisfieswhere the real scalars are constant functions. Thus $J=(e_x:x\in S)$ has a finite generating set of an ideal, so $R$ is a Noetherian ring. If $X$ is empty, $R$ is the zero ring, and the conclusion still holds.

If $X$ is infinite, consider the idealIt is an ideal because a sum has support contained in the union of the two supports, and multiplication by any function cannot enlarge support. If $f_1,\dots ,f_r$ generated $F$, their supports would have a finite union $T$. Every element of $(f_1,\dots ,f_r)$ would vanish outside $T$, whereas $e_x\in F$ for any $x\in X\setminus T$. This contradiction proves