Show that if \(U\) is open in \(X\) and \(A\) is closed in \(X\), then \(U - A\) is open in \(X\), and \(A - U\) is closed in \(X\).
Solution
1. \(U-A\) is open in \(X\).
\(U-A=U\cap(X-A)\), and \(X-A\) is open because \(A\) is closed; a finite intersection of open sets is open.
2. \(A-U\) is closed in \(X\).
\(A-U=A\cap(X-U)\), and \(X-U\) is closed because \(U\) is open; an intersection of closed sets is closed.
3. Q.E.D.
Steps 1 and 2.