Monotone Convergence Theorem (Real Analysis)

Theorem
Every bounded monotone sequence is convergent.

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

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

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

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

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

Proof for Increasing Sequence
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 $\epsilon > 0$.

Since $B - \epsilon$ 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 $\left \langle {x_n} \right \rangle$.

Then:

Hence the result.

Proof for Decreasing Sequence
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.