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Daniel J Smith Custom TMUA Set A Paper 2 · Q3

Paper 2Logic & proofSequences & seriesNumber & counting

We say a sequence of real numbers xnx_n is a Cauchy sequence if

for all ε>0\varepsilon>0 there exists a positive integer N∈NN\in\mathbb N such that for all m,n≥Nm,n\geq N we have ∣xm−xn∣<ε|x_m-x_n|<\varepsilon

Which one of the following statements is true if and only if xnx_n is not a Cauchy sequence?

AThere exists ε>0\varepsilon>0 such that for every N∈NN\in\mathbb N there exist m,n≥Nm,n\geq N with ∣xm−xn∣≥ε|x_m-x_n|\geq\varepsilon
BFor every ε>0\varepsilon>0 there exists N∈NN\in\mathbb N and there exists m,n≥Nm,n\geq N with ∣xm−xn∣≥ε|x_m-x_n|\geq\varepsilon
CThere exists N∈NN\in\mathbb N such that for all ε>0\varepsilon>0 and all m,n≥Nm,n\geq N we have ∣xm−xn∣≥ε|x_m-x_n|\geq\varepsilon
DFor every ε>0\varepsilon>0 and every N∈NN\in\mathbb N there exists m,n≥Nm,n\geq N with ∣xm−xn∣<ε|x_m-x_n|<\varepsilon
EThere exists ε>0\varepsilon>0 and there exists N∈NN\in\mathbb N such that for all m,n≥Nm,n\geq N we have ∣xm−xn∣≥ε|x_m-x_n|\geq\varepsilon
FFor every N∈NN\in\mathbb N there exists ε>0\varepsilon>0 such that for all m,n≥Nm,n\geq N we have ∣xm−xn∣≥ε|x_m-x_n|\geq\varepsilon
GThere exists ε>0\varepsilon>0 such that for every N∈NN\in\mathbb N and for all m,n≤Nm,n\leq N we have ∣xm−xn∣≥ε|x_m-x_n|\geq\varepsilon
HThere exists ε>0\varepsilon>0 such that for every N∈NN\in\mathbb N there exist m,n≥Nm,n\geq N with ∣xm−xn∣<ε|x_m-x_n|<\varepsilon

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