Proof that a converging increasing sequence converges to a number greater than any term of the sequence using Cauchy Criterion

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 $\sup{T}$.

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I assume when you say increasing that is to be interpreted as strictly increasing? :) – cardinal Feb 24 '12 at 13:45
Hint: Assume the sequence $a_1 > a_2 > \cdots > a_n > \cdots$ has a limit $U$. Now suppose for some $m$ that $U \leq a_m$. What can you say about $a_{m+1}$? Hence? – cardinal Feb 24 '12 at 13:47
The term "increasing sequence" is not strong enough, since it may be that $a_n=a$ constantly. You may want strictly increasing. – Ragib Zaman Feb 24 '12 at 13:48
I can say that $a_{m+1}>U$ and generaly if $n>m$ then $a_{n}>a_{m}$ – nick Feb 24 '12 at 13:50
Done. Thaks cardinal – nick Feb 24 '12 at 13:52

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 $n \in \mathbb{N}.$

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Correct Thank you – nick Feb 24 '12 at 13:53

You don't need any version of completeness: the desired result holds in any ordered field.

See e.g. Theorem 22 in $\S 2.1$ of these notes. It says, in particular, that a weakly increasing sequence in any ordered field converges iff the set of terms has a least upper bound $L$, and if so $L$ is the limit of the sequence.

If you now assume that the sequence is strictly increasing, then the least upper bound $L$ cannot be less than or equal to any term of the sequence: if $L \leq x_n$, then $x_{n+1} > x_n \geq L$, contradicting the fact that $L$ is an upper bound for the set of terms of the sequence.

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As many have pointed out, we need the hypothesis that the sequence is strictly increasing.

Suppose there was some $m\in \mathbb{N}$ such that $a_m>a$. Then for $\epsilon=a_m-a$, there is an $N>0$ such that $|a_n-a| < \epsilon$ for all $n>N$. I.e. $-a_m+a<a_n-a<a_m-a$. We deduce that $a_n<a_m$ for all $n>N$. This is a contradiction, since we just proved that all terms after $a_N$(and consequently terms that occur after $a_m$) are smaller than $a_m$, which cannot happen in an increasing sequence.

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