b S 0 3 1 4 1 5 S 1 3 1 5 int S 1 3 bd S 1 1 3 5 S 1 3 isolated points 1 5 c S

B s 0 3 1 4 1 5 s 1 3 1 5 int s 1 3 bd s 1 1 3 5 s 1

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(b) S = ( [0 , 3) (1 , 4) ) ∪ {- 1 , 5 } S = (1 , 3) ∪ {- 1 , 5 } : int S = (1 , 3) , bd S = {- 1 , 1 , 3 , 5 } , S 0 = [1 , 3], isolated points: {- 1 , 5 } . (c) S = { r Q : 0 < r < 2 } int S = , bd S = [0 , 2] , S 0 = [0 , 2], no isolated points. (d) S = n n 2 + 1 : n N S = { 1 / 2 , 2 / 5 , 3 / 10 , 4 / 17 , 5 / 26 , . . . } . int S = , bd S = S ∪ { 0 } , S 0 = { 0 } , isolated points: { 1 / 2 , 2 / 5 , 3 / 10 , 4 / 17 , 5 / 26 , . . . } = S . (e) S = { x R : 0 < | x - 3 | ≤ 2 } S = [1 , 3) (3 , 5] : int S = (1 , 3) (3 , 5) , bd S = { 1 , 3 , 5 } , S 0 = [1 , 5], no isolated points. 4
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(f) S = 1 n : n N [1 , ) S = 1 , 1 2 , 1 3 , 1 4 , 1 5 , · · · [1 , ) int S = (1 , ) , bd S = { 0 } ∪ 1 , 1 2 , 1 3 , 1 4 , 1 5 , · · · , S 0 = [1 , ) ∪ { 0 } , isolated points: 1 2 , 1 3 , 1 4 , 1 5 , · · · . 5
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3. Prove that if α = sup S and α / S , then (a) α is an accumulation point of S . Let N * ( α, ) be a deleted neighborhood of α . Since α = sup S , there is an element s S such that α - < s α but, since α / S, α - < s < α. Thus, every deleted neighborhood of α contains a point of S which implies that m is an accumulation point of S . (b) S is an infinite set. Let s 1 be any element of S . By Theorem 3 (lecture notes), there exists s 2 S such that s 1 < s 2 < α. Now assume we have selected s 1 , s 2 , . . . , s n S such that s 1 < s 2 < . . . < s n < α. By Theorem 3, there is an element s n +1 S such that s n < s n +1 < α .
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  • Fall '08
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  • Math, Sets, accumulation point, Closed set, General topology

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