THEOREM

An increasing (or nondecreasing) sequence $\{ s_{n}\}$ that is bounded from above converges. A decreasing (or nonincreasing) sequence $\{ s_{n}\}$ that is bounded from below converges.

To prove this theorem, we need the following property of real numbers. The set of real numbers is defined by a collection of axioms. One of these axioms is the Completeness Axiom.

As an example, consider the set $S$: $\{x\,|\,x^{2} < 2,\,\,x>0\}$. The set of upper bounds to $S$ is the set $\{x\,|\,x^{2}\geq 2,\,\,x>0\}$.

We prove the theorem for a nondecreasing sequence $\{ s_{n}\}$. The proofs for the other three cases are similar.

Proof

Suppose $\{ s_{n}\}$ is a nondecreasing sequence that is bounded from above. Since $\{ s_{n}\}$ is bounded from above, there is a positive number $K$ (an upper bound), so that $s_{n}\leq K$ for every $n$. From the Completeness Axiom, the set $\{ s_{n}\}$ has a least upper bound $L.$ That is, $s_{n}\leq L$ for every $n.$

Then for any $\varepsilon >0,$ $L-\varepsilon$ is not an upper bound of $\{ s_{n}\} .$ That is, $L- \varepsilon < s_{N}$ for some integer $N$. Since $\{ s_{n}\}$ is nondecreasing, $s_{N}\leq s_{n}$ for all $n > N.$ Then for all $n > N,$ $$\begin{equation*} L - \varepsilon < s_{n}\leq L < L + \varepsilon \end{equation*}$$

That is, $\vert s_{n}-L\vert < \varepsilon$ for all $n > N,$ so the sequence $\{ s_{n}\}$ converges to $L$.