1. Let's clarify the problem: We want to understand why the set difference $B \setminus C$ is closed, given that $B$ and $C$ are subsets of a topological space. 2. Recall that $B \setminus C = B \cap C^c$, where $C^c$ is the complement of $C$. 3. If $B$ is closed and $C$ is open, then $C^c$ is closed because the complement of an open set is closed. 4. The intersection of two closed sets is closed. Since $B$ is closed and $C^c$ is closed, their intersection $B \cap C^c = B \setminus C$ is closed. 5. Therefore, $B \setminus C$ is closed if $B$ is closed and $C$ is open. 6. If $C$ is not open, $B \setminus C$ may not be closed. The key condition is the openness of $C$ and closedness of $B$. Final answer: $B \setminus C$ is closed because it is the intersection of the closed set $B$ and the closed set $C^c$ (the complement of $C$), assuming $C$ is open and $B$ is closed.