Monotone Convergence Theorem

This article was Proof of the Week between 30 June 2009 and 6 July 2009.

[edit] Theorem

Let $\left \langle {x_n} \right \rangle$ be a sequence in $\mathbb{R}$ .

Let $\left \langle {x_n} \right \rangle$ be increasing and bounded above.

Then $\left \langle {x_n} \right \rangle$ converges to its supremum.

Let $\left \langle {x_n} \right \rangle$ be decreasing and bounded below.

Then $\left \langle {x_n} \right \rangle$ converges to its infimum.

Suppose $\left \langle {x_n} \right \rangle$ is increasing and bounded above.

Let its supremum be $B$ .

We need to show that $x_n \to B$ as $n \to \infty$ .

Let $ε > 0$ .

Since $B - ε$ is not an upper bound, by the definition of supremum.

Thus $\exists x_N: x_N > B - \epsilon$ .

But $\left \langle {x_n} \right \rangle$ is increasing.

Hence $\forall n > N: x_N \ge x_N > B - \epsilon$ .

But $B$ is still an upper bound for $\left \langle {x_n} \right \rangle$ .

	$\forall n > N: B - \epsilon$	$<$	$x_n \le B$
$\Longrightarrow$	$\forall n > N: B - \epsilon$	$<$	$x n < B + ε$	Real Plus Epsilon
$\Longrightarrow$	$\forall n > N: \left\|{x_n - B}\right\|$	$<$	$ε$	Negative of Absolute Value: Corollary

Hence the result.

$\blacksquare$

If $\left \langle {x_n} \right \rangle$ is decreasing and bounded below then $\left \langle {-x_n} \right \rangle$ is increasing and bounded above.

Thus the above result applies and the proof follows.

$\blacksquare$