Existence of Maximum and Minimum of Bounded Sequence

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

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:
 * $L \ne \varnothing$

From Lower and Upper Bounds for Sequences, $L$ is a bounded subset of $\R$.

Thus $L$ does have a supremum and infimum in $\R$.

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 $\dfrac \epsilon 2 > 0$.

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

Hence:
 * $\exists l \in L: \overline l \ge l > \overline l - \dfrac \epsilon 2$

Therefore:
 * $\left\vert{l - \overline l}\right\vert < \dfrac \epsilon 2$

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:
 * $\exists R: \forall r > R: \left\vert{x_{m_r} - \overline l}\right\vert < \dfrac \epsilon 2$

So, for any $r > R$:

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

So far, what has been shown is that, given any $\epsilon > 0$, there exists an infinite collection of terms of $\left \langle {x_n} \right \rangle$ which satisfy $\left\vert{x_n - \overline l}\right\vert < \epsilon$.

Next it is shown 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:
 * $\exists n_1: \left\vert{x_{n_1} - \overline l}\right\vert < 1$

Now take $\epsilon = \dfrac 1 2$ in the above.

Then:
 * $\exists n_2 > n_1: \left\vert{x_{n_2} - \overline l}\right\vert < \dfrac 1 2$

In this way a subsequence is contructed:
 * $\left \langle {x_{n_r}} \right \rangle$ satisfying $\left\vert{x_{n_r} - \overline l}\right\vert < \dfrac 1 r$

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

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

Thus $\overline l \in L$ as required.

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

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