# Inequality relating the Limit Superior and Limit Inferior of a bounded sequence

I'm doing exercise 2.4.7 in the book Understanding Analysis by Stephen Abbott. I'd like to ask, how to go about part (d) of the proof (please don't give away the complete proof). Additionally, is the proof to (a),(b),(c) rigorous and technically correct?

$$\newcommand{\absval}{\left\lvert #1 \right\rvert}$$

Let $$(a_n)$$ be a bounded sequence.

(a) Prove that the sequence defined by $$y_n = \sup \{a_k: k\ge n\}$$ converges.

(b) The limit superior of $$(a_n)$$ or $$\limsup a_n$$ is defined by

\begin{align*} \limsup{a_n} = \lim y_n \end{align*}

where $$(y_n)$$ is the sequence from part (a) of this exercise. Provide a reasonable definition for $$\liminf a_n$$ and briefly explain why it always exists for any bounded sequence.

(c) Prove that $$\liminf a_n \le \limsup{a_n}$$ for every bounded sequence, and given an example of a sequence where this inequality is strict.

(d) Show that $$\liminf{a_n} = \limsup{a_n}$$ if an only if $$\lim a_n$$ exists. In this case, all three share the same value.

I cross-reference some excellent answers on this site on the limit superior and limit inferior of a bounded sequence.

What is limit superior and limit inferior?

Limit superior and Limit inferior of a sequence of subsets of a set

Proof.

(a) The sequence $$(a_n)$$ is a bounded sequence. Hence, $$\absval{a_n} \le M$$ for all $$n \in \mathbf{N}$$.

Claim. The sequence $$(y_n)$$ is a monotonically decreasing sequence. For $$n=1$$, $$y_1 >= y_2$$ because $$y_1 = \sup \{a_k:k\ge 1\} = \max \{a_1,y_2\} \ge y_2$$. Assume that $$y_{k} \ge y_{k+1}$$. Then, $$y_{k+1} = \sup \{a_k: k+1 \ge n\} = \max \{ a_{k+1},y_{k+2} \} \ge y_{k+2}$$. Hence, $$y_{k} \ge y_{k+1}$$ for all $$k \in \mathbf{N}$$.

We can further show that $$(y_n)$$ is a bounded sequence. Since, $$\absval{a_n} \le M$$ for all $$n \in \mathbf{N}$$, the supremum of any of the subsequences $$\{a_k:k\ge n\}$$ no matter what $$k \in \mathbf{N}$$, is always less than or equal to $$M$$. Thus, $$\absval{y_n} \le M$$ for all $$n \in \mathbf{N}$$. Therefore, $$(y_n)$$ is a bounded sequence.

By the Monotone Convergence Theorem, the sequence $$(y_n)$$ is convergent.

(b) From part (a), we can deduce that, the limit superior of a sequence is, \begin{align*} \limsup{a_n} = \lim_{n\to\infty} \left(\sup \{a_k:k \ge n\}\right) = \inf \{ \sup \{a_k:k \ge n\}\} \end{align*}

whereas, the limit inferior of a sequence is, \begin{align*} \liminf{a_n} = \lim_{n\to\infty} \left(\inf \{a_k:k \ge n\}\right) = \sup \{ \inf \{a_k:k \ge n\}\} \end{align*}

The above definition is easy to verify.

Define $$z_n := \inf \{a_k:k\ge n\}$$. Let us prove that $$(z_n)$$ is a convergent sequence.

Claim. The sequence $$(z_n)$$ is a monotonically increasing sequence. For $$n=1$$, $$z_1 \le z_2$$ because $$z_1 = \inf \{a_k:k \ge 1\} = \min \{a_1,z_2\} \le z_2$$. Assume that $$z_{k} \le z_{k+1}$$. We have, $$z_{k+1} = \inf \{a_{k+1} : k+1 \ge n\} = \min \{a_{k+1},z_{k+2}\} \le z_{k+2}$$. Hence, by principle of mathematical induction, $$z_{n} \le z_{n+1}$$ for all $$n \in \mathbf{N}$$.

Claim. The sequence $$(z_n)$$ is bounded below. Since, $$(a_n)$$ is a bounded sequence, $$a_n \ge -M$$ for all $$n \in \mathbf{N}$$. So, given any $$k \in \mathbf{N}$$, the infimum of the subsequence $$\{a_{k}: k\ge n\}$$ will be greater than or equal to $$-M$$. So, $$-M$$ is a lower bound for the sequence $$(z_n)$$.

By the Monotone Convergence Theorem, $$(z_n)$$ is convergent sequence.

(c) $$(y_n)$$ and $$(z_n)$$ are convergent sequences; their limiting value is the infimum/supremum. So, we are justified in setting, \begin{align*} y &= \inf \{y_n:n \in \mathbf{N}\}\\ z &= \sup \{z_m:m \in \mathbf{N}\} \end{align*}

Observe that, the infimum of $$k$$-tail of the sequence is always less than or equal to the supremum of the tail. Hence, $$z_k \le y_k$$ for all $$k \in \mathbf{N}$$. The visual representation makes it clear that, every $$y_n$$ is an upper bound for the sequence $$\{z_m:m\in\mathbf{N}\}$$. So, $$\inf \{y_n:n \in \mathbf{N}\}$$ is also an upper bound for the set $$Z$$. Thus, $$\liminf a_n \le \limsup a_n$$.

Consider the sequence \begin{align*} a_n := \left(\frac{n+1}{n}\right)\sin\left( \frac{n\pi}{8}\right) \end{align*}

This sequence is bounded, but not convergent, and so a strict inequality holds.

Code Snippet.

(d) Assume that $$\limsup{a_n} = \liminf{a_n} = l$$. Pick an abritrary $$\epsilon > 0$$.

import numpy as np
import matplotlib.pyplot as plt
from matplotlib import style
import math

# The sequence x_n = (n+1)/n sin(n/8)
n = np.linspace(1,500,500)
x_n = np.multiply(np.divide(n+10,n),np.sin(n/8))

# The supremum of a subsequence
def sup(x_m,n):
return max(x_m[int(n-1):])

# The infimum of a subsequence
def inf(x_m,n):
return min(x_m[int(n-1):])

limsup = [sup(x_n,m) for m in n]
liminf = [inf(x_n,m) for m in n]

plt.style.use('fivethirtyeight')
plt.figure(figsize=(20,10))
plt.xlim(0,450)
plt.text(250,1.2,r'$$\sup \{x_m: m > n\}$$')
plt.text(250,-1.2,r'$$\inf \{x_m: m > n\}$$')
plt.plot(n,x_n,'o')
plt.plot(n,limsup)
plt.plot(n,liminf)
plt.show() • Try contraposition assume that the sequence does not converge and show that lim sup and lim inf are not the same. Dec 22, 2020 at 19:38
• @FrederikRavnKlausen, let me try. Do you think (a), (b), (c) are fine? Dec 22, 2020 at 19:50
• Didn't check, but I bet you can use that a monotone bounded sequences converges somewhere. Dec 22, 2020 at 19:52
• for (d), i could be missing something but i think use the calculus 2 definition of limit of sequence exists...i don't think there's much more to it than this. math.stackexchange.com/questions/2864822/…
– BCLC
Dec 22, 2020 at 23:27

For (d) with $$U=\lim\sup a_n$$ and $$D=\lim\inf a_n:$$
Show that for any $$r>0$$ the sets $$\{n: a_n\ge U+r\}$$ and $$\{n:a_n\le D-r\}$$ are finite.
So if $$D=U=L$$ then for any $$r>0$$ the set $$\{n: |a_n-L|\ge r\}$$ is finite, so $$a_n$$ converges to $$L.$$
• There exists points $N_1,N_2$, such that, if $n \ge N_1$ \begin{align*} \sup \{a_k:k \ge n\} - l &\le \epsilon\\ a_k \le \sup \{a_k:k \ge n\} &\le l + \epsilon \end{align*} if $n \ge N_2$ \begin{align*} -\epsilon &\le \inf \{a_k:k \ge n\} - l \\ l - \epsilon &\le \inf \{a_k:k \ge n\} \le a_k \end{align*} If $n \ge \max \{N_1,N_2\}$, we have, \begin{align*} l - \epsilon \le \inf \{a_k:k \ge n\} \le &a_k \le \sup \{a_k:k \ge n\} \le l + \epsilon\\ l - \epsilon \le &a_k \le l + \epsilon \end{align*} Thus, $\lim_{n\to\infty} a_n = l$. Dec 23, 2020 at 6:12