Complex Analysis/Appendix/Proofs/Theorem 1.1

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We prove that a set is closed if and only if it contains all of its limit points.

We assume that contains all of its limit points and we show that its complement is open. Let . Then, since is not a limit point of , there is a ball that contains no point of , that is . Since this is true for all , it follows that is closed.

We now assume that there is a limit point of in and show that is not closed. Let be a limit point of . Then There is no neighborhood of that is contained in the complement of , and therefore is not closed.