just want to see if i got these: 1.let U be open in X and A closed in X then U-A is open in X and A-U is closed in X. 2. if A is closed in X and B is closed in Y then AxB is closed in XxY. my proof: 1. A'=X-A which is open in X X-(A-U)=Xn(A'U U)=A'U(U) but this is a union of open sets and thus it's open thus: A-U is closed. U'=X-U which is closed in X X-(U-A)=Xn(U'UA) but this is an intersection of closed sets and thus it's also closed. which means that U-A is open. 2. let's look at: XxY-AxB=((X-A)xY)U(Xx(Y-B)) now X-A is open in X and Y is open in Y so (X-A)xY is open in XxY the same for Xx(Y-B) so we have here that it equals an arbitrary union of open sets so it's also an open set, thus AxB.