Terms of Bounded Sequence Within Bounds

Theorem

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

Let $\left \langle {x_n} \right \rangle$ be bounded.

Let the limit superior of $\left \langle {x_n} \right \rangle$ be $\overline l$.

Let the limit inferior of $\left \langle {x_n} \right \rangle$ 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

• First we show that $\forall \epsilon > 0: \exists N: \forall n > N: x_n < \overline l + \epsilon$:

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 $\overline l$.

$\blacksquare$

• Next, in the same way, we show that $\forall \epsilon > 0: \exists N: \forall n > N: x_n > \underline l - \epsilon$:

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

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 5.15 \ (4)$