If [tex]A\subseteq B[/tex] are both subsets of a topological space [tex](X,\tau)[/tex], is it true that any closed subset of A is also a closed subset of B?
Ahh i see. Thanks. But is that the only problem here? If we insist that A is compact in B, then that fixes the problem and the statement is true, right?Hurkyl's point is that any topological space is both open and closed as a subset of itself. If A is NOT closed as a subset of topological space B, since it IS closed as a subset of itself, the statement "any closed subset of A is also a closed subset of B" is false.