: Proposition 1.2.4. If (X,T) is a topological space, a subset A of X is closed iff the the derived set of A is a subset of A: [itex]A'\subseteq A[/itex].
Rao's proof of [itex](A'\subseteq A) \Rightarrow (X\setminus A \in T)[/itex] goes like this:
To me, this looks like enough to show that [itex](A'\subseteq A) \Rightarrow (X\setminus A \in T)[/itex], since a set A is open iff each of its points belongs to a neighborhood which is a subset of A. So [itex]X\setminus A[/itex] is open. In other words, A is closed.[/QUOTE]
But Rao goes on:
This seems superfluous to me. Am I missing something? Why not just say U is the neighborhood of x that's a subset of [itex]X\setminus A[/itex]?