# Limsup and Liminf are Limits of Bounds

## Theorem

Let $\sequence {x_n}$ be a sequence in $\R$.

Let $\sequence {x_n}$ be bounded.

Let $\ds \overline l = \limsup_{n \mathop \to \infty} x_n$ be the limit superior and $\ds \liminf_{n \mathop \to \infty} x_n$ the limit inferior of $\sequence {x_n}$.

Then:

$\ds \overline l = \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \mathop \ge n} x_k}$
$\ds \underline l = \liminf_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\inf_{k \mathop \ge n} x_k}$

## Proof

First we show that:

$\ds \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \mathop \ge n} x_k}$

Let $M_n = \ds \sup_{k \mathop \ge n} x_k$.

By Supremum of Subset, the sequence $\sequence {M_n}$ decreases, for:

$m \ge n \implies \set {k \in \N: k \ge m} \subseteq \set {k \in \N: k \ge n}$

A lower bound for $\sequence {x_n}$ is also a lower bound for $\sequence {M_n}$.

So from the Monotone Convergence Theorem (Real Analysis) it follows that $\sequence {M_n}$ converges.

Suppose $M_n \to M$ as $n \to \infty$.

From the Bolzano-Weierstrass Theorem there exists a convergent subsequence of $\sequence {x_n}$.

Let $L$ be the set of all numbers which are the limit of some subsequence of $\sequence {x_n}$.

Let $l \in L$ be arbitrary.

Let $\sequence {x_{n_r} }$ be a convergent subsequence of $\sequence {x_n}$ such that $x_{n_r} \to l$ as $r \to \infty$.

Then $\forall n_r \ge n: x_{n_r} \le M_n$.

Hence $l \le M_n$ by Lower and Upper Bounds for Sequences and hence (from the same theorem) $l \le M$.

This is true for all $l \in L$, so $\overline l = \limsup_{n \mathop \to \infty} x_n \le M$.

Now, from Terms of Bounded Sequence Within Bounds, we have that:

$\forall \epsilon > 0: \exists n: \forall k \ge n: x_k < \overline l + \epsilon$

Thus $\overline l + \epsilon$ is an upper bound for $\set {x_k: k \ge n}$.

So $M \le M_n \le \overline l + \epsilon$.

Thus from Real Plus Epsilon, $M \le \overline l$.

Thus we conclude that $M = \overline l$ and hence:

$\ds \overline l = \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \ge n} x_k}$

$\ds \liminf_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop to \infty} } {\inf_{k \mathop \ge n} x_k}$

can be proved using a similar argument.

$\blacksquare$