How are these definitions of the limit superior and limit inferior equivalent?

Question

I have come across these three definitions of the limit superior (or upper limit) and the limit inferior (or lower limit) of a sequence of real numbers and I wonder how to establish the equivalence of these.

1. Walter Rudin: PRINCIPLES OF MATHEMATICAL ANALYSIS, 3rd edition: Definition 3.16: Given a sequence $\{s_n\}$ of real numbers, let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k} \to x$ for some subsequence $\{s_{n_k}\}$. This set contains all subsequential limits ..., plus possibly the numbers $+\infty$, $-\infty$. We now ... put $$s^{*} = \sup E,$$ $$s_{*} = \inf E.$$ The numbers $s^{*}$, $s_{*}$ are called the upper and lower limits of $\{s_n\}$; we use the notation $$ \lim_{n \to \infty} \sup s_n = s^{*}, \, \, \, \lim_{n \to \infty} \inf s_n = s_{*}.$$

2. Tom M. Apostol: MATHEMATICAL ANALYSIS, 2nd edition: Sec. 8.3: Definition 8.2: Let $\{a_n\}$ be a sequence of real numbers. Suppose there is a real number $U$ satisfying the following conditions:

i} For every $\epsilon > 0$ there exists an integer $N$ such that $n > N$ implies $$a_n < U + \epsilon.$$ ii) Given $\epsilon > 0$ and given $m > 0$, there exists an integer $n> m$ such that $$a_n > U - \epsilon.$$ Then $U$ is called the limit superior (or upper limit) of $\{a_n\}$, and we write $$U = \lim_{n \to \infty} \sup a_n.$$ Statement (i) implies that the set $\{a_1, a_2, a_3, \ldots \}$ is bounded above. If this set is not bounded above, we define $$\lim_{n\to\infty}\sup a_n = +\infty.$$ If the set is bounded above but not bounded below and if $\{a_n\}$ has no finite limit superior, then we say $\lim \sup_{n\to\infty} a_n = -\infty$. The limit inferior (or lower limit) of $\{a_n\}$ is defined as follows: $$\lim_{n\to\infty}\inf a_n = -\lim_{n\to\infty}\sup b_n,$$ where $b_n = -a_n$ for $n= 1, 2, 3, \ldots$.

3. Robert G. Bartle and Donald R. Sherbert: INTRODUCTION TO REAL ANALYSIS, 3rd edition: Exercises for Section 3.3: Problem 10: Let $(x_n)$ be a bounded sequence of real numbers, and for each $n\in \mathbb{N}$ let $s_n \colon= \sup \{x_k \colon k \geq n\}$ and let $t_n \colon= \inf \{x_k \colon k \geq n\}$. Prove that $(s_n)$ and $(t_n)$ are monotone and convergent. Also prove that if $\lim (s_n) = \lim (t_n)$, then $(x_n)$ is convergent. [One calls $\lim (s_n)$ the limit superior of $(x_n)$ and $\lim (t_n)$ the limit inferior of $(x_n)$.]

Now how can one show that the above three definitions are equivalent (i.e. these three definitions are of the same pair of numbers)?

Answer 1

I'm going to drop the boundedness condition (it is a legitimate difference) and use the extended reals. Also, I'm doing it for $\limsup$ ($\liminf$ is analogous).

Let $\{a_n\}_{n\geq 0}$ be a sequence in $\mathbb{R}$.

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Let $E$ be the set of sub-sequential limits as in $(1)$. I will show that it is closed.

Let $x$ be a limit point of $E$. Then there are $x_n\in E$ such that $x_n\to x$ as $n\to\infty$. Each $x_n$ is the limit of a subsequence $a^n_{k_j}$.

Suppose $x\in\mathbb{R}$. For all $\epsilon>0$ there is $N$ such that $\vert x-x_n\vert<\frac{\epsilon}{2}$ if $n>N$. Furthermore, there is some index $j_n$ such that $\vert x_n-a^n_{k_{j_n}}\vert<\frac{\epsilon}{2}$. We will choose the indices $j_n$ so that $k_{j_{n+1}}>k_{j_{n}}$ (they can be chosen inductively, starting with $j_1$). Now we have $a^n_{k_{j_n}}$, a subsequence of $a_k$.

Given $\epsilon>0$, we get $N$ (since $x_n$ is convergent) and if $n>N$ we have

$\vert x-a^n_{k_{j_n}}\vert\leq\vert x-x_n\vert+\vert x_n-a^n_{k_{j_n}}\vert<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

Consequently, $x$ is a sub-sequential limit and $x\in E$.

For the cases that $x=\pm\infty$ you reword the same idea into bounding below or bounding above respectively.

So $E$ contains its limit points and is closed.

(1) is equivalent to (2)

Since $E$ is closed, $\sup E\in E$ and there is a subsequence with $\sup E$ as its limit.

Let $U=\sup E$. If there were $\epsilon>0$ such that for every $N>0$ there is $n>N$ with $a_n\geq U+\epsilon$, then there would be a subsequence with limit greater than $U$, but $U$ is the supremum of the sub-sequential limits. This proves (i).

Since there is a subsequence converging to $U$ we get (ii) by letting the $n$ be the index of an element in this subsequence.

So $\sup E$ satisfies the criteria of $U$ in (2).

(2) is equivalent to (3)

Note that $s_n$ is a weakly decreasing sequence. Since it is monotone, it has a limit (it's bounded below by $-\infty$ and we are in the extended reals).

Let $U$ satisfy the criteria in (2). By (ii), we can see that for every $\epsilon>0$, $s_m>U-\epsilon$. So $\lim s_n\geq U$.

By (i) we can see that for every $\epsilon>0$ there is $N>0$ such that $s_N\leq U+\epsilon$ for some $N$. Since $s_n$ is decreasing, we actually get $s_n\leq U+\epsilon$ for all $n>N$. So $\lim s_n\leq U$.

Consequently, $\lim s_n=U$, and we see that the definitions are equivalent.

Finally, note that since $s_n$ is decreasing, $\lim_{n\to\infty} s_n=\inf_{n\geq 0} s_n=\inf_{n\geq 0}\sup_{k\geq n}a_k$. This last expression is an alternative definition.

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The only significance of the boundedness is to ensure that monotone sequences converge, working in the extended reals has the same effect with less need to separate into cases.