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 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 $\R$.

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 $r > R$:

	\(\displaystyle \)	\(\displaystyle \left\vert{x_{m_r} - \overline l}\right\vert\)	\(=\)	\(\displaystyle \left\vert{x_{m_r} - l + l - \overline l}\right\vert\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle \left\vert{x_{m_r} - l}\right\vert + \left\vert{l - \overline l}\right\vert\)	\(\displaystyle \)	Triangle Inequality
	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle \frac \epsilon 2 + \frac \epsilon 2 = \epsilon\)	\(\displaystyle \)

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

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 $\left\vert{x_n - \overline l}\right\vert < \epsilon$.

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 $\exists n_1: \left\vert{x_{n_1} - \overline l}\right\vert < 1$.

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 we can construct a subsequence $\left \langle {x_{n_r}} \right \rangle$ 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 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.

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 5.13$