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 $$\overline l$$.


 * 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 $$\underline l$$.