If we have an strictly increasing sequence $a_{n}$ and we have proved that it converges to a number $a$ can we prove that $a_{n}<a\:\: \forall n\in \mathbb{N}$ without using the Least Upper Bound Property or the Monotone Convergence Theorem? The Cauchy Criterion (Every Cauchy sequence converges) is the only version of Completeness that can be used.
I encountered this problem while trying to prove that the Cauchy Criterion implies the Least Upper Bound Property using the hints in the second answer here Equivalence of Completeness Axioms of Real Numbers. I have proved that $\{x_n\} $ and $\{y_n\}$ converge to the same limit but I am unable to prove that that limit is $\sup{T}$.
Best Answer
I will assume the sequence is strictly increasing, since if the sequence is merely increasing there is an obvious counterexample.
Suppose there exists $N \in \mathbb{N}$ such that $a_N \geq a.$ Let $ \epsilon = a_{N+1} -a > 0.$ Then for all $ n > N$ we have $ |a_n-a| \geq \epsilon$, which contradicts the assumption that $a_n \to a.$ Thus no such $N$ can exist, and $a_n < a$ for all $n \in \mathbb{N}.$