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 $\S 5.15 \ (4)$