= Solution
{c}
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, so
$$
e_xf=f(x)e_x.
$$
Suppose first that $X$ is finite, and let $J$ be an <ideal> of $R$. Define $S=\{x\in X:f(x)\ne0\text{ for some }f\in J\}$. If $x\in S$, choose such an $f$; multiplying it by $e_x/f(x)$ shows that $e_x\in J$. Every $f\in J$ satisfies
$$
f=\sum_{x\in S}f(x)e_x,
$$
where 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 <ideal>
$$
F=\{f\in\mathbb R^X:\{x:f(x)\ne0\}\text{ is finite}\}.
$$
It 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 $f_1,\ldots,f_r$ generated $F$, their <supports> would have a finite union $T$. Every element of $(f_1,\ldots,f_r)$ would vanish outside $T$, whereas $e_x\in F$ for any $x\in X\setminus T$. This contradiction proves
$$
\boxed{\mathbb R^X\text{ is Noetherian}\iff X\text{ is finite}.}
$$
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