Exercise 17.2

Proof. $A$ is closed in $Y$ iff there exists $B\subseteq X$ closed in $X$ such that $A=Y\cap B$ by Theorem $17.2$. But then, $A$ is the intersection of closed sets, and so is closed. □

Proof. Since $A$ is closed in $Y$, it follows from Theorem 17.2 that $A=B\cap Y$ where $B$ is some closed set in $X$. Hence by definition, $X-B$ is open in $X$. Also, since $Y$ is closed in $X$, we have that $X-Y$ is open in $X$ by definition. We then have

by DeMorgan’s law. Since both $X-B$ and $X-Y$ are open in $X$, clearly their union must also be open since we are in a topological space. Hence $X-A$ is open in $X$ so that $A$ is closed in $X$ by definition. □