Proof. Let S be a closed set and x&isinX a limit point thereof. Then, x must be in S, for otherwise the complement of S would constitute an open neighborhood of x that does not intersect S.
Conversely, suppose that S contains all of its limit points. We shall show that the complement of S is an open set. Let x¬inS be given. By assumption x is not a limit point, and hence there exists an open neighborhood U of x that contains only finitely many points of S, call them
<math>y_1,\ldots y_n</math>
Using the T_{0} assumption, we may choose open neighborhoods
<math>U_1,\ldots, U_n</math>
of x, such that U_{i} avoids y_{i}. The intersection of U with all the U_{i} produces an open neighborhood of x that avoids S altogether. This proves that the complement of S is open, and therefore that S is closed. Q.E.D.
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