Solution
= Solution
{c}
Because the image is integer-valued, the open interval $(n-\tfrac12,n+\tfrac12)$ contains exactly one possible image value, namely $n$. Thus
$$
f^{-1}(\{n\})=f^{-1}\bigl((n-\tfrac12,n+\tfrac12)\bigr).
$$
The right side is open in $A$ by <continuity>, so
$$
\boxed{f^{-1}(\{n\})\text{ is open in }A.}
$$
This includes the possibility of an empty <fiber>. The source notation $f^{-1}(n)$ is understood as the <preimage> of the singleton $\{n\}$.