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A finite, nonempty set always contains its supremum |
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Definition
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If a< L for every element a in the set A, then supA |
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Definition
False, let A = (0,1) and L = 1. Then supA=1=L. |
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If A and B are sets with the property that a<b for every a element of A, and every b element of B, then it follows that supA<infB. |
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Definition
False: let A=(0,1) and B=(1,2). Then supA=1=infB. |
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Term
If sup A = s and sup B = t, the sup(A+B) = s+t. The set A+B is defined as A+B = {a+b: a elem. A and b elem. B}. |
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Definition
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If sup A <= sup B, then there exists an element b elem. B that is an upper bound for A. |
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Definition
False, Let A=B=(0,1). Then supA=supB but no element of B is an upper bound for A. |
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sequences (xn) and (yn), which both diverge, but whose sum (xn + yn) converges; |
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Definition
Let xn = (−1)^n and yn = (−1)^(n+1); then xn + yn = 0 for all n. |
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sequences (xn) and (yn), where (xn) converges, (yn) diverges, and (xn + yn) converges; |
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Definition
This is impossible by the Algebraic Limit Theorem, part (ii). If (xn + yn) converges and (xn) converges, then so does (xn + yn − xn) = (yn). |
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a convergent sequence (bn) with bn != 0 for all n such that (1/bn) diverges; |
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Definition
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an unbounded sequence (an) and a convergent sequence (bn) with (an − bn) bounded; |
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Definition
This is impossible. By Theorem 2.3.2, (bn) is bounded since it converges. Since (bn) and (an−bn) are both bounded, so is (bn + an − bn) = (an). |
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two sequences (an) and (bn), where (an*bn) and (an) converge but (bn) does not. |
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Definition
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A sequence that does not contain 0 or 1 as a term but contains subsequences converging to each of these values. |
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Definition
(1/3, 2/3, 1/4, 3/4, 1/5, 4/5, . . . ). |
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Term
A monotone sequence that diverges but has a convergent subsequence. |
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Definition
A monotone sequence that diverges must be unbounded. Since it is monotone, such a sequence must increase or decrease without bound, and any subsequence will also either increase or decrease without bound, therefore diverging. |
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Term
A sequence that contains subsequences converging to every point in the infinite set {1, 1/2, 1/3, 1/4, 1/5, . . . }. |
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Definition
(1, 1, 1/2, 1, 1/2, 1/3, 1, 1/2, 1/3, 1/4, 1, 1/2, 1/3, 1/4, 1/5, . . . ). |
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Term
An unbounded sequence with a convergent subsequence. |
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Definition
(1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, . . . ). |
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Term
A sequence that has a subsequence that is bounded but contains no subsequence that converges. |
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Definition
This is impossible; the bounded subsequence is itself a bounded sequence and must have a convergent subsequence. |
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A Cauchy sequence that is not monotone. |
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Definition
(1, −1/2, 1/3, −1/4, 1/5, . . . ). |
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A monotone sequence that is not Cauchy. |
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Definition
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A Cauchy sequence with a divergent subsequence. |
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Definition
This is impossible since a Cauchy sequence converges, and all subsequences of a convergent sequence also converge. |
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An unbounded sequence containing a subsequence that is Cauchy. |
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Definition
(1, 2, 1, 3, 1, 4, 1, 5, . . . ). |
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An arbitrary intersection of compact sets is compact. |
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Definition
Compact sets are closed and bounded. The intersection of closed sets is closed. The intersection of bounded sets is bounded. Therefore the intersection of compact sets is compact. |
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Let A C R be arbitrary, and let K C R be compact. Then, the intersection A n K is compact. |
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Definition
False: let A = (0, 1) and K = [0, 1]. Then A n K = (0, 1) is not closed and therefore is not compact. |
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Term
If F1 ≥ F2 ≥ F3 ≥ F4 ≥ . . . is a nested sequence of nonempty closed sets, then the intersection n(n=1 to inf)Fn != empty set. |
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Definition
False: let Fn = [n,inf). Then n(n=1 to inf)Fn = empty set |
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A finite set is always compact. |
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Definition
A finite set is a finite union of closed sets (each point is a closed interval) and is closed. A finite set is bounded. Therefore a finite set is compact. |
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A countable set is always compact. |
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Definition
False: N is countable but not bounded and thus not compact. |
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