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:
- $\ds \overline l = \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \ge n} x_k}$
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- $\ds \liminf_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop to \infty} } {\inf_{k \mathop \ge n} x_k}$
can be proved using a similar argument.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 5$: Subsequences: Exercise $\S 5.15 \ (6)$