The title is kind of misleading because the task actually to show
Every monotonic increasing and bounded sequence (xn)n∈N 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>0∃N∈N so that |an−a|<ε for ∀n>N. (Definition of Convergence)
- A sequence (a′k)k≥1 is called a subsequence of a sequence (an)n≥1, if there is a monotonic increasing sequence (nk)k≥1∈N so that a′k=ank for ∀k≥1. (Definition of a Subsequence)
- A sequence (an)n≥1 is Cauchy, if for ∀ε>0∃N=N(ε)∈N so that |am−an|<ε for ∀m,n>N. (Definition of a Cauchy Sequence)
- (Hint) The sequence (ε⋅ℓ)ℓ∈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,q≥n 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'
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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