Sunday, October 6, 2019

calculus - Show that every monotonic increasing and bounded sequence is Cauchy.



The title is kind of misleading because the task actually to show




Every monotonic increasing and bounded sequence (xn)nN is Cauchy



without knowing that:




  • Every bounded non-empty set of real numbers has a least upper
    bound. (Supremum/Completeness Axiom)

  • A sequence converges if and only if it is Cauchy. (Cauchy
    Criterion)

  • Every monotonic increasing/decreasing, bounded and real
    sequence converges to the supremum/infimum of the codomain (not sure

    if this is the right word).



However, what is allowed to use listed as well:




  • A sequence is called covergent, if for ε>0NN so that |ana|<ε for n>N. (Definition of Convergence)

  • A sequence (ak)k1 is called a subsequence of a sequence (an)n1, if there is a monotonic increasing sequence (nk)k1N so that ak=ank for k1. (Definition of a Subsequence)

  • A sequence (an)n1 is Cauchy, if for ε>0N=N(ε)N so that |aman|<ε for m,n>N. (Definition of a Cauchy Sequence)

  • (Hint) The sequence (ε)N is unbounded for ε>0. (Archimedes Principle)




Would appreciate any help.


Answer



If xn is not Cauchy then an ε>0 can be chosen (fixed in the rest) for which, given any arbitrarily large N there are p,qn for which $p\varepsilon.$



Now start with N=1 and choose xn1, xn2 for which the difference of these is at least ε. Next use some N beyond either index n1, n2 and pick $N'\varepsilon.$ Continue in this way to construct a subsequence.



That this subsequence diverges to + can be shown using the Archimedes principle, which you say can be used, since all the differences are nonnegative and there are infinitely many differences each greater than ε, a fixed positive number.


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