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: [itex]A'\subseteq A[/itex].
Rao's proof of [itex](A'\subseteq A) \Rightarrow (X\setminus A \in T)[/itex] goes like this:
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Suppose [itex]A' \subseteq A[/itex]. Then for all [itex]x \in X \setminus A[/itex], [itex]x \notin A'[/itex]. Hence, there exists a neighborhood U of x such that [itex]U \cap A = \varnothing[/itex]. In other words, [itex]x \in U \subseteq X\setminus A[/itex].
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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:
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Since U is a neighborhood of x and [itex]U\subseteq X\setminus A[/itex], [itex]X\setminus A[/itex] is also a neighborhood of X. So each point x of [itex]X\setminus A[/itex] has a neighborhood [itex]X\setminus A[/itex] which is contained in [itex]X\setminus A[/itex]. Hence [itex]X\setminus A[/itex] is open. Therefore, A is closed.
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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]?
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