# Terms of Bounded Sequence Within Bounds

## Theorem

Let $\sequence {x_n}$ be a sequence in $\R$.

Let $\sequence {x_n}$ be bounded.

Let the limit superior of $\sequence {x_n}$ be $\overline l$.

Let the limit inferior of $\sequence {x_n}$ be $\underline l$.

Then:

- $\forall \epsilon > 0: \exists N: \forall n > N: x_n < \overline l + \epsilon$

- $\forall \epsilon > 0: \exists N: \forall n > N: x_n > \underline l - \epsilon$

## Proof

### Upper Bound

First we show that:

- $\forall \epsilon > 0: \exists N: \forall n > N: x_n < \overline l + \epsilon$

Aiming for a contradiction, suppose this proposition were to be false.

That would mean that for some $\epsilon > 0$ it would be true that for each $N$ we would be able to find $n > N$ such that $x_n \ge \overline l + \epsilon$.

From Limit of Subsequence of Bounded Sequence it would follow that there exists a convergent subsequence whose limit was $l \ge \overline l + \epsilon$.

This would contradict the definition of $\overline l$.

$\Box$

### Lower Bound

Next, in the same way, we show that:

- $\forall \epsilon > 0: \exists N: \forall n > N: x_n > \underline l - \epsilon$

Aiming for a contradiction, suppose this proposition were to be false.

That would mean that for some $\epsilon > 0$ it would be true that for each $N$ we would be able to find $n > N$ such that $x_n \le \underline l + \epsilon$.

From Limit of Subsequence of Bounded Sequence it would follow that there exists a convergent subsequence whose limit was $l \le \underline l + \epsilon$.

This would contradict the definition of $\underline l$.

$\blacksquare$

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 5$: Subsequences: Exercise $\S 5.15 \ (4)$