Existence of Maximum and Minimum of Bounded Sequence

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

Let $$L$$ be the set of all real numbers which are the limit of some subsequence of $$\left \langle {x_n} \right \rangle$$.

Then $$L$$ has both a maximum and a minimum.

Proof
From the Bolzano-Weierstrass Theorem we know that $$L \ne \varnothing$$.

From Lower and Upper Bounds for Sequences, $$L$$ is a bounded set.

Thus $$L$$ does in fact have a supremum and infimum in $$\mathbb{R}$$.

The object of this proof is to confirm that $$\overline l = \sup \left({L}\right) \in L$$ and $$\underline l = \inf \left({L}\right) \in L$$, that is, that these points do actually belong to $$L$$.


 * First we show that $$\overline l \in L$$.

To do this, we show that $$\exists \left \langle {x_{n_r}} \right \rangle: x_{n_r} \to \overline l$$ as $$n \to \infty$$, where $$\left \langle {x_{n_r}} \right \rangle$$ is a subsequence of $$\left \langle {x_n} \right \rangle$$.

Let $$\epsilon > 0$$. Then $$\frac \epsilon 2 > 0$$.

Since $$\overline l = \sup \left({L}\right)$$, and therefore by definition the smallest upper bound of $$L$$, $$\overline l - \frac \epsilon 2$$ is not an upper bound of $$L$$.

Hence $$\exists l \in L: \overline l \ge l > \overline l - \frac \epsilon 2$$.

Therefore $$\left|{l - \overline l}\right| < \frac \epsilon 2$$.

Now because $$l \in L$$, we can find $$\left \langle {x_{m_r}} \right \rangle$$, a subsequence of $$\left \langle {x_n} \right \rangle$$, such that $$x_{m_r} \to l$$ as $$n \to \infty$$.

So $$\exists R: \forall r > R: \left|{x_{m_r} - \overline l}\right| < \frac \epsilon 2$$.

So, for any $$r > R$$:

$$ $$ $$

Thus we have shown that $$\forall r > R: \left|{x_{m_r} - \overline l}\right| < \epsilon$$.

We have not finished yet. All we have done so far is show that, given any $$\epsilon > 0$$, there exists an infinite collection of terms of $$\left \langle {x_n} \right \rangle$$ which satisfy $$\left|{x_n - \overline l}\right| < \epsilon$$.

Now what we need to do is to show how to construct a subsequence $$\left \langle {x_{n_r}} \right \rangle$$ such that $$x_{n_r} \to \overline l$$ as $$n \to \infty$$.

Take $$\epsilon = 1$$ in the above: then $$\exists n_1: \left|{x_{n_1} - \overline l}\right| < 1$$.

Now take $$\epsilon = \frac 1 2$$ in the above: then $$\exists n_2 > n_1: \left|{x_{n_2} - \overline l}\right| < \frac 1 2$$.

In this way we can construct a subsequence $$\left \langle {x_{n_r}} \right \rangle$$ satisfying $$\left|{x_{n_r} - \overline l}\right| < \frac 1 r$$.

But $$\frac 1 r \to 0$$ as $$r \to \infty$$ from the corollary to Power of Reciprocal.

From the Squeeze Theorem, it follows that $$\left|{x_{n_r} - \overline l}\right| \to 0$$ as $$r \to \infty$$.

Thus $$\overline l \in L$$ as we were to prove.


 * A similar argument shows that the infimum $$\underline l$$ of $$L$$ is also in $$L$$.

Note
From Limit of a Subsequence we know that if $$\left \langle {x_n} \right \rangle$$ is convergent then $$L$$ has exactly one element.