Sec.%2013%20Topology%20of%20the%20Reals

# Sec.%2013%20Topology%20of%20the%20Reals - Math 350 Advanced...

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Math 350 Advanced Calculus I Fall 2009 Homework Solutions Section 13 TOPOLOGY OF THE REALS 13.3 Find the interior of each set. 13.4 Find the boundary of each set. (a) Ω 1 n | n N æ Since int S S , then the only possible interior points of the set are the elements of the set itself. Suppose x 0 = 1 /n is an arbitrary element of the set S . Since every neighborhood N (1 /n ; ε ) of x 0 contains elements of R \ S , then x 0 is not an interior point. Therefore none of the elements of S are interior points so int S = . For each element x 0 = 1 /n S , every neighborhood of x 0 contains elements of S , namely x 0 itself, and elements of R \ S . Therefore each point x 0 S is a boundary point. We want to show that 0 is also a boundary point of S . To show this, we need to show that every neighborhood N (0; ε ) of 0 contains points in S and points in S . Suppose ε > 0 is arbitrary. Then N (0; ε ) contains 0, which is not in S . Since ε > 0, then the Archimedean Property implies there exists n N such that 1 /n < ε . Then 1 /n N (0; ε ). Since 1 /n S , then N (0; ε ) contains a point in S . Therefore 0 is also a boundary point. Finally, for any x 0 / S di±erent from 0, there exists a neighborhood which does not contain elements of S . Choose ε > 0 so that N ( x 0 ; ε ) misses all elements of S . Therefore bd S = S ∪{ 0 } = { 1 , 1 2 , 1 3 , . . . , 0 } . (c) { r Q | 0 < r < 2 } We want to show that bd S = £ 0 , 2 § = © x R | 0 x 2 . Suppose x 0 is an arbitrary real number in ° 0 , 2 ¢ . We want to show that x 0 is a boundary point. To show this, we need to show every neighborhood N ( x 0 ; ε ) of x 0 contains points in S and points not in S . Suppose ε > 0 is arbitrary. Then by the Density Theorems, N ( x 0 ; ε ) contains rationals in ° 0 , 2 ¢ and irrationals so contains points in S and points in R \ S . Similarly, each neighborhood of 0 or

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