I have begun reading Rosenthal's "A First Look to Rigorous Probability Theory" and in order to reinforce my calculus background I am studying through the appendix section. Here he defined the limit of a sequence of real numbers as $\lim_{n \to \infty}x_n=x$ if for each $0 <\epsilon$ there is a natural number $N$ such that it is $|x-x_n| < \epsilon$ for $n > N$.

Then the limit infimum and supremum are defined as $\liminf_n x_n = \lim_{n \to \infty} \inf_{k\geq n} x_k$ and $\limsup_n x_n = \lim_{n \to \infty} \sup_{k\geq n} x_k$.

  1. My first question is about the interpretation of the limit infimum and limit supremum definitions. It looks to me like they are defined as the limits of sequences as well. For example, for the infimum, one can define a sequence $a_n = \inf_{k \geq n}x_k$ where $x_n$ was already a sequence of real numbers. Then the limit infimum is the limit of the sequence $a_n$ , $\lim_{n \to \infty}a_n$, so the infimums of the subsequences $x_k$ where $k \geq n$ are converging if the limit exists. It is the similar for the supremum. Is my interpretation correct here?

  2. It is said the limit infimum and supremum do always exist, though they can be infinite. I did not understand this; why it is so?

  3. In the book it is stated that the limit $\lim_{n \to \infty} x_n$ exists if and only if it is $\liminf_n x_n = \limsup_n x_n$. I could not prove this to myself. How can it be shown that this statement is true?

Thanks in advance.

  • $\begingroup$ 1. ok, 2. they are monotone sequences. $\endgroup$ Jul 23, 2014 at 7:28
  • $\begingroup$ For 3. In general the limit infimum is the least value to which a subsequence of ${a_n}$ can converge and the limit supremum is the greater value to wihch a subsequence of ${a_n}$ can converge $\endgroup$
    – Dimitri
    Jul 23, 2014 at 7:44

2 Answers 2

  1. Yes, in the book you are reading, $\liminf_n x_n$ and $\limsup_n x_n$ are defined as limits of sequences. Note that the sequence $a_n=\inf_{k\geq n} x_k$ is increasing (which means that $m\leq n$ implies $a_m\leq a_n$), while the sequence $b_n=\sup_{k\geq n} x_k$ is decreasing.

  2. The $\liminf$ and $\limsup$ are taken in the extended real line $\overline{\mathbb{R}}=\{-\infty\}\cup\mathbb{R}\cup\{\infty\}$. As an ordered set, $\overline{\mathbb{R}}$ is obtained from the ordered set $\mathbb{R}$ by adding to it the bottom (the least) element $-\infty$ and the top (the greatest) element $\infty$. As a topological space, $\overline{\mathbb{R}}$ is homeomorphic to any closed interval $[a,b]$ (with $a<b$) in $\mathbb{R}$, and is therefore compact. For example, $x\mapsto(2/\pi)\arctan x$ is a homeomorphism $\overline{\mathbb{R}}\to[-1,1]$ (where it is understood that ${-}\infty\mapsto-1$ and $\infty\mapsto1$), and at the same time it is an isomorphism of (totally) ordered sets. All this means that the limits, $\liminf$'s, and $\limsup$'s in the extended real line are just `infinitely stretched out' versions of the limits, $\liminf$'s, and $\limsup$'s in a closed interval. Since in a closed interval all increasing/decreasing sequences converge, so then do all increasing/decreasing sequences in the extended real line; in particular, $\liminf_n x_n$ and $\limsup_n x_n$ always exist. When you are thinking about the extended real line, you can imagine it as a closed interval, you can' go astray with that. If you have to prove some assertion about sequences in the extended real line, formulated in terms of the ordering and the topology (which is, after all, also determined by the ordering), then it suffices to prove the assertion for sequences in a closed interval.

  3. You can assume that $x_n$ is a sequence in a closed interval; this will make your life easier, because $\liminf_n x_n$ and $\limsup_n x_n$ are real numbers (belonging to the interval). Now prove the following: $a=\liminf_n x_n$ is the largest real number with the property that for every real number $a'<a$ there are only finitely many $n\in\mathbb{N}$ such that $x_n\leq a'$, and the analogous assertion for $b=\limsup_n x_n$ (which you will formulate yourself --- just turn everything upside-down). With these characterizations of $\liminf_n x_n$ and $\limsup_n x_n$ under the belt, you will easily prove that $\lim_{n\to\infty} x_n$ exists iff $\liminf_n x_n=\limsup_n x_n$.

Remark. You did not ask this, but since I have mentioned imagining the extended real line as a closed interval$\ldots$ $~$You can actually see $\overline{\mathbb{R}}$ as the closed interval $[-1,1]$, and transfer anything that's happening in $\overline{\mathbb{R}}$ to $[-1,1]$. Let $f(y)=\tan((\pi/2)y)$ for $y\in[-1,1]$, so that $f^{-1}(x)=(2/\pi)\arctan(x)$ for $x\in\overline{\mathbb{R}}$. Now define, for all $y,z\in[-1,1]$, $$ \begin{aligned} \mathit{sum}(y,z)~ &:= f^{-1}(f(y)+f(z))~,\\ \mathit{prod}(y,z)~ &:= f^{-1}(f(y)f(z))~. \end{aligned} $$ Actually the operations $\mathit{sum}$ and $\mathit{prod}$ are not defined everywhere on $[-1,1]\times[-1,1]$; for example, $\mathit{plus}$ is not defined at the two points $(1,-1)$ and $(-1,1)$ that correspond to the risque limits of the form $\infty-\infty$. Write the definitions of $\mathit{sum}$ and $\mathit{prod}$ in Mathematica (or a similar program), and then draw their $3\text{D}$ diagrams, placing on all three axes the labels $-\infty$, $-1$, $0$, $1$, $\infty$ at positions, respectively, $-1$, $-1/2$, $0$, $1/2$, $1$. You will be able to observe in the diagrams the behavior of the two operations all the way to the infinity (in either direction); at the points where the operations are not defined you will clearly see the discontinuities (at these points the diagrams contain vertical lines --- more precisely, the closures of the diagrams contain vertical lines).

  • $\begingroup$ A quick question: What exactly do we mean by a sequence in a closed interval? Do we mean that all values $x_n$ are mapped to values in an interval $[a,b]$ such that it is $a \leq x_n \leq b$? (a and b are real numbers) $\endgroup$ Jul 23, 2014 at 9:42
  • 1
    $\begingroup$ Yes, we mean exactly that. $\endgroup$
    – chizhek
    Jul 23, 2014 at 9:58
  • $\begingroup$ I have another question, it may be weird but still it bugged me. Let's assume that the sequence $x_n$ takes the values of $\pi$'s digit values after the comma. So each $x_n$ takes a value in the set $(0,1,2,3,4,5,6,7,8,9)$. If we think of the sequence $a_n = \inf_{k \geq n} x_k$ this sequence is clearly nondecreasing and bounded in $[0,9]$, but how can we know the value of limit $\liminf_n x_n$? Since we don't know exactly the values of $x_n$ for very large $n$s, we can be never sure about the value for $a_n$ for these values. $\endgroup$ Jul 23, 2014 at 12:36
  • $\begingroup$ Continued: The infimum can be zero, one or any value, we cannot observe this exactly for this almost random sequence $x_n$ at large $n$s. So, what can we say about the value $\liminf_n x_n$ for such sequences? $\endgroup$ Jul 23, 2014 at 12:39
  • $\begingroup$ For the third question, unfortunately I have failed to find a rigorous proof for that if $a$ is the largest real number such that for every $a' < a$ there is finitely many $x_n < a'$ then $a=\liminf_n x_n$. But intuitively it is clear that this is valid: If we have a plot of a bounded sequence $x_n$ and visualize a vertical line $y=a'$ then there will be finitely many $x_n$ points under that line and as soon as we increase the line over $a$, the number of $x_n$ points will be infinitely many! But I frustratingly fail to express that visualization as a mathematically valid statement... $\endgroup$ Jul 23, 2014 at 15:00

These are answers to the secondary questions in the comments after my answer to the original question; I cannot put the answers in comments, because the size of a comment is limited.

The second question. Let $(x_n)$ be a sequence taking values in the set $\{0,1,2,\ldots,9\}$, and $a:=\liminf_n x_n$. If $0$ appears infinitely many times in the sequence, then $a=0$. If $0$ appears only finitely many times, but $1$ appears infinitely often, then $a=1$. And so on. This is a slick answer. It is entirely another matter to answer the question with a definite value for a particular given sequence, such as the sequence of digits after the comma in the decimal expansion of $\pi$. Well, if you can prove that the digit $0$ appears infinitely many times in the decimal expansion of $\pi$, then you have $a=0$, and so on; in this case it takes a serious theoretical effort to determine the $\liminf$.

As for the third question, let $(x_n)$ be a sequence of real numbers such that $a:=\liminf_n x_n$ is a real number (that is, $-\infty<a<\infty$). Since the sequence of infima $a_n:=\inf_{k\geq n} x_n$ is increasing (meaning that $m\leq n$ implies $a_m\leq a_n$ -- I hate "nondecreasing"), the limit $a=\lim_{n\to\infty} a_n$ is actually the supremum: $a=\sup_n a_n$.
$\quad$Let $a_1\leq a$ and $a'<a_1$; then $a'<a$. We claim that there exists $m$ such that $x_n\geq a'$ for every $n\geq m$: since $a=\sup_n a_n$, there exists $m$ such that $a_m\geq a'$; but then $x_n\geq a_m\geq a'$ for every $n\geq m$.
$\quad$Now let $a_1>a$, and set $a':=\tfrac{1}{2}(a+a_1)$; we have $a<a'<a_1$. In this case we claim that given any $m$ there exists $n\geq m$ such that $x_n<a'$: since $a_m=\inf_{k\geq m} x_k\leq a<a'$, there exists $n\geq m$ such that $x_n<a'$.
$\quad$We have proved that $a$ is the largest of all real numbers $a_1$ with the property that for every $a'<a_1$ there are only finitely many $n$ such that $x_n<a'$.


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