Limsup and Liminf are Limits of Bounds

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Theorem

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

Let $\left \langle {x_n} \right \rangle$ be bounded.

Let $\displaystyle \overline l = \limsup_{n \to \infty} x_n$ be the limit superior and $\displaystyle \liminf_{n \to \infty} x_n$ the limit inferior of $\left \langle {x_n} \right \rangle$.


Then:

$\displaystyle \overline l = \limsup_{n \to \infty} x_n = \lim_{n \to \infty} \left({\sup_{k \ge n} x_k}\right)$
$\displaystyle \underline l = \liminf_{n \to \infty} x_n = \lim_{n \to \infty} \left({\inf_{k \ge n} x_k}\right)$


Proof

First we show that:

$\displaystyle \limsup_{n \to \infty} x_n = \lim_{n \to \infty} \left({\sup_{k \ge n} x_k}\right)$


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

By Supremum of Subset, the sequence $\left \langle {M_n} \right \rangle$ decreases, for:

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


Any lower bound for $\left \langle {x_n} \right \rangle$ is clearly a lower bound for $\left \langle {M_n} \right \rangle$.

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


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

From the Bolzano-Weierstrass Theorem there exists a convergent subsequence of $\left \langle {x_n} \right \rangle$.

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

Let $l \in L$ be arbitrary.

Let $\left \langle {x_{n_r}} \right \rangle$ be a convergent subsequence of $\left \langle {x_n} \right \rangle$ 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 $\overline l = \limsup_{n \to \infty} x_n \le M$.


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 $\left\{{x_k: k \ge n}\right\}$.

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:

$\displaystyle \overline l = \limsup_{n \to \infty} x_n = \lim_{n \to \infty} \left({\sup_{k \ge n} x_k}\right)$


$\displaystyle \liminf_{n \to \infty} x_n = \lim_{n \to \infty} \left({\inf_{k \ge n} x_k}\right)$ can be proved using a similar argument.

$\blacksquare$


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