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Unformatted text preview: 1.79 1. Let int . Thus is a neighborhood of . Therefore, is a neighborhood of , so that is an interior point of . 2. Clearly, if , then . Therefore, assume which implies that is a boundary point of . Every neighborhood of contains other points of . Hence . 1.80 Assume that is open. Every has a neighborhood which is disjoint from . Hence no is a closure point of . contains all its closure points and is therefore closed. Conversely, assume that is closed. Let be a point its complement . Since is closed and / , is not a boundary point of . This implies that has a neighborhood which is disjoint from , that is . Hence, is an interior point of . This implies that contains only interior points, and hence is open.contains only interior points, and hence is open....
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- Fall '10