Isn't this proof of a theorem about the closedness of a set wrong?
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I was reading a proof of the following theorem in my textbook: A set $A$ is closed iff $A' subseteq A$ . Proof: Suppose $A$ is closed and $x in A'$ . If $x notin A$ , then $xin A^c$ , an open set. Thus $mathcal{N}(x, delta)subseteq A^c$ for some positive $delta$ . But then $mathcal{N}(x, delta)$ can contain no points of $A$ . Thus $x$ is not an accumulation point of $A$ and so $xnotin A'$ , a contradiction. We conclude that $xin A$ . Therefore $A'subseteq A$ . Now suppose $A'subseteq A$ . To show that $A$ is closed, we show $A^c$ is open. If $A^c$ is not open, there is $xin A^c$ that is not an interior point of $A^c$ . Therefore, no $delta$ -neighborhood of $x$ is a subset of $A^c$ ; that is, each $delta$ -neighborhood of $x$ contains a point of $A$ . This point must be ...