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By Allan L. Scherr

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We will prove this direction by proving the contrapositive. Assume E ⊆ R is disconnected. We want to find two non-empty, disjoint sets A, B satisfying E = A∪B such that there never exists a convergent sequence (xn ) → x with (xn ) contained in one of A or B, and x an element of the other. Because E is disconnected, there exist separated sets A and B satisfying E = A ∪ B. Now suppose (xn ) is contained in A and (xn ) → x. Then either x ∈ A or x is a limit point of A, and in either case x ∈ A.

Thus, ∞ k=N |ak | converges by the Comparison Test and part (b). Because ∞ N −1 |ak | = k=1 and N −1 k=1 ∞ |ak | + k=1 |ak | k=N |ak | is just a finite sum, the series ∞ k=1 |ak | converges. 10. (a) The idea here is that eventually the terms an “look like” a non-zero constant times 1/n, and we know that any series of this form diverges. To make this precise, let 0 = l/2 > 0. Because (nan ) → l, there exists N ∈ N such that nan ∈ V 0 (l) for all n ≥ N . A little algebra shows that this implies we must have nan > l/2, or an > (l/2) (1/n) for all n ≥ N.

Where x + y = s intersects an nth stage square). Moving to the n + 1th stage means removing the open middle third of this shaded region. 1, implying that the line x + y = s must intersect a (n + 1)st stage square. This shows that there exist xn+1 , yn+1 ∈ Cn+1 where xn+1 + yn+1 = s. (b) We have (xn ) and (yn ) with xn , yn ∈ Cn and xn + yn = s for all n. The sequence (xn ) doesn’t converge, but (xn ) is bounded so by the BolzanoWeierstrass Theorem there exists a convergent subsequence (xnk ).

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An analysis of time-shared computer systems by Allan L. Scherr


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