Divergent Sequence may be Bounded

Theorem
While every Convergent Sequence is Bounded, it does not follow that every bounded sequence is convergent.

There exist bounded divergent sequences.

Proof
Let $$\left \langle {x_n} \right \rangle$$ be the sequence in $\R$ defined as $$x_n = \left({-1}\right)^n$$.

It is clear that $$\left \langle {x_n} \right \rangle$$ is bounded: above by $$1$$ and below by $$-1$$.

Suppose $$x_n \to l$$ as $$n \to \infty$$.

Let $$\epsilon > 0$$.

Then $$\exists N \in \R: \forall n > N: \left|{\left({-1}\right)n - l}\right| < \epsilon$$.

But there are values of $$n > N$$ for which $$\left({-1}\right)^n = \pm 1$$.

Thus $$\left|{1 - l}\right| < \epsilon$$ and $$\left|{-1 - l}\right| < \epsilon$$.

This can not be true for every $$\epsilon > 0$$ and so we have the contradiction that $$l = 1$$ and $$l = -1$$.

This can not be the case as a Sequence has One Limit at Most.

Thus $$\left \langle {x_n} \right \rangle$$ has no limit and, while definitely bounded, is unmistakably divergent.

Alternatively, note the subsequences of $$\left \langle {x_n} \right \rangle$$:
 * $$\left \langle {x_{n_r}} \right \rangle$$ where $$\left \langle {n_r} \right \rangle$$ is the sequence defined as $$n_r = 2r$$;
 * $$\left \langle {x_{n_s}} \right \rangle$$ where $$\left \langle {n_s} \right \rangle$$ is the sequence defined as $$n_s = 2s+1$$.

The first is $$1, 1, 1, 1, \ldots$$ and the second is $$-1, -1, -1, -1, \ldots$$

So $$\left \langle {x_n} \right \rangle$$ has are two subsequences with different limits.

From Limit of a Subsequence, that means $$\left \langle {x_n} \right \rangle$$ can not be convergent.