Monotone Convergence Theorem (Real Analysis)/Increasing Sequence

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

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

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

By the Continuum Property, it has a supremum, $B$.

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. As $B$ is still an upper bound for $\left \langle {x_n} \right \rangle$. thus we know that $B\geq x_n \implies x_n\leq B$ Hence $\forall n \geq N: B-\epsilon<x_n\leq x_N<\epsilon $.

Then:

Hence the result.