## Rao: Proposition 1.2.4. Superfluous section of proof?

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 $X\setminus A$?

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 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Your proof looks fine. What Rao says isn't wrong, but it can be shortened.