real_analysisii_homework3-1

real_analysisii_homework3-1 - Homework 3 Hctor Guillermo...

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ector Guillermo Cu´ ellar R´ ıos February 9, 2006 13.9 Prove the folowing. (a) An accumulation point of a set S is either an interior point of S or a boundary point of S . Proof. Suppose that an accumulation point x of a set S is neither an interior point of S nor a boundary point of S , then x is an interior point of R \ S , thus there is a neighborhood N of x such that N R \ S it follow that N S = and N * S = but this is contradicts the definition of an accumulation point. (b) A boundary point of a set S is either an accumulation point of S or an isolated point of S . Proof. Lets consider several cases: Case 1: The boundary point x S . Now we consider two subcases: Subcase a: There is a deleted neighborhood N * of x such that N * S = then by definition 13.14 x is an isolated point. Subcase b: For every deleted neighborhood N * of x, N * S 6 = then by definition 13.14 x is an accumulation point. Case 2: The boundary point x 6∈ S . By definition 13.3 we know that for every neighborhood N of x , N S 6 = . Since we know that x 6∈ S it follows that for every deleted neighborhood N * of x , N * S 6 = thus by definition 13.14
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This note was uploaded on 03/05/2009 for the course MATH 117 taught by Professor Akhmedov,a during the Spring '08 term at UCSB.

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real_analysisii_homework3-1 - Homework 3 Hctor Guillermo...

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