Question 3.T.7: A monotonic sequence is convergent if, and only if, it is bo...

If $(x_{n})$ is convergent then, by Theorem 3.2, it must be bounded.

(i) Assume $(x_{n})$ is increasing and bounded. The set $A = \left\{x_{n} : n ∈ \mathbb{N}\right\}$ will then be non-empty and bounded, so, by the completeness axiom, it has a least upper bound. Let sup A = x. By Definition 2.6, given ε > 0 there is an $x_{N} ∈ A$ such that

$x_{N} > x − ε$.

Since $x_{n}$ is increasing,

$x_{n} ≥ x_{N}$  for all n ≥ N.        (3.6)

But x is an upper bound for A, hence

$x ≥ x_{n}$  for all $n ∈ \mathbb{N}$.        (3.7)

From (3.6) and (3.7) we arrive at

$|x_{n} − x| = x − x_{n} ≤ x − x_{N} < ε$  for all n ≥ N,

which means $x_{n} → x$.

(ii) If $(x_{n})$ is decreasing and bounded, then $(−x_{n})$ is increasing and bounded, and, from (i), we have

lim$(−x_{n}) =$ sup(−A).

But since sup(−A) = − inf A, it follows that lim $x_{n} =$ inf A.