Monotone Convergence Theorem (Real Analysis)/Increasing Sequence

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Theorem

Let $\sequence {x_n}$ be increasing and bounded above.


Then $\sequence {x_n}$ converges to its supremum.


Proof

Suppose $\sequence {x_n}$ 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 $\epsilon > 0$.

By the definition of supremum, $B - \epsilon$ is not an upper bound.

Thus:

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

But $\sequence {x_n}$ is increasing.

Hence:

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

But $B$ is still an upper bound for $\sequence {x_n}$.


Then:

\(\displaystyle \forall n > N: B - \epsilon\) \(<\) \(\displaystyle x_n \le B\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \forall n > N: B - \epsilon\) \(<\) \(\displaystyle x_n < B + \epsilon\) Real Plus Epsilon
\(\displaystyle \leadsto \ \ \) \(\displaystyle \forall n > N: \size {x_n - B}\) \(<\) \(\displaystyle \epsilon\) Negative of Absolute Value: Corollary 1


Hence the result.

$\blacksquare$


Sources