Level sets of continuous functions

Detailed Solution

Since $A=f^{-1}(\{c\})$ is the preimage of the closed set $\{c\}$ under a continuous function, $A$ is always closed — but the full picture is stronger than just this. For any closed set $F\subseteq \mathbb R$, the function $f(x)=\operatorname{dist}(x,F)$ is continuous, and $f^{-1}(\{0\})=F$ exactly. So every closed subset of $\mathbb R$ — including complicated sets like a Cantor-type set, which is closed, uncountable, and has empty interior — arises as a level set of some continuous function. This makes (C) the complete characterization; (A) is true but incomplete on its own, since it doesn't capture that $A$ can be as intricate as any closed set. (B) and (D) are false: $A$ need not be a finite union of intervals (a Cantor-type level set is a counterexample), and $A$ need not be countable (a Cantor set is uncountable, and a constant function has $A=\mathbb R$). Thus the correct option is $\boxed{\mathrm{C}}$.