Proposition

Suppose $S$ is a subspace of $X$. Prove that a subset $B \subset S$ is closed in $S$ if and only if it is equal to the intersection of $S$ with some closed subset of $X$.

Solution

Let $B \subset S$.

Suppose that $B$ is closed in $S$. Then $S \setminus B$ is open in $S$. By the definition of a subspace topology, $S \setminus B = S \cap U$ for some open $U \subset X$.

We claim that $(X \setminus U) \cap S = B$.

$(X \setminus U) \cap S$ and $B$ are subsets of $S$. Thus it suffices to check if each point in $S$ is either in both of them or in neither of them. For any $x \in S$,

Therefore, $B = (X \setminus U) \cap S$ where $X \setminus U$ is closed in $X$.

Suppose that $B = S \cap C$ where $C$ is a closed subset of $X$.

Then $X \setminus C$ is open in $X$. Then $S \cap (X \setminus C)$ is open in $S$. We claim that $S \cap (X \setminus C) = S \setminus B$. Note that each of them is a subset of $S$. Let $x \in s$.

Therefore, $S \setminus B$ is open in $S$, so $B$ is closed in $S$.