Rao: Proposition 1.2.4. Superfluous section of proof?

[Rasalhague]

Rao: Topology: 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: A'\subseteq A.

Rao's proof of (A'\subseteq A) \Rightarrow (X\setminus A \in T) goes like this:

Suppose A' \subseteq A. Then for all x \in X \setminus A, x \notin A'. Hence, there exists a neighborhood U of x such that U \cap A = \varnothing. In other words, x \in U \subseteq X\setminus A.

To me, this looks like enough to show that (A'\subseteq A) \Rightarrow (X\setminus A \in T), since a set A is open iff each of its points belongs to a neighborhood which is a subset of A. So X\setminus A is open. In other words, A is closed.[/QUOTE]

But Rao goes on:

Since U is a neighborhood of x and U\subseteq X\setminus A, X\setminus A is also a neighborhood of X. So each point x of X\setminus A has a neighborhood X\setminus A which is contained in X\setminus A. Hence X\setminus A is open. Therefore, A is closed.

This seems superfluous to me. Am I missing something? Why not just say U is the neighborhood of x that's a subset of X\setminus A?

 

[micromass]

Your proof looks fine. What Rao says isn't wrong, but it can be shortened.