An application of absorption to teaching lim inf and lim sup (sequences)

January 10, 2009August 30, 2023Joel

Many undergraduates have difficulty understanding the notions of the lim inf and lim sup of a sequence.

[The full names for these appear to vary from author to author. In the first version of this post I called them “limit infimum” and “limit supremum”, but I think that “limit inferior” and “limit superior” are the most widely accepted names. ]

Of course, there is the basic problem that students confuse $\liminf_{n\to\infty} x_n$ with $\lim_{n\to\infty}( \inf x_n)$ (which is, strictly speaking, meaningless, but might generously be interpreted as meaning one of $\lim_{n\to\infty} x_n$ or $\inf _{n \in \mathbb{N}} x_n$ ). However, what I really mean is that the students often fail to grasp what $\liminf$ and $\limsup$ really mean. (See below for some more details of what I mean by this!)

As with epsilon and delta, we may be tempted to avoid confronting the students’ difficulties with lim inf and lim sup. For example, we can often choose between using lim inf and lim sup or using the sandwich theorem (also known as the squeeze rule). A typical example of this is the standard exercise where you have to prove the following fact at the start of a course on metric spaces.

Let $(X,d)$ be a metric space, and let $(x_n)$ and $(y_n)$ be convergent sequences in $X$ with limits $x$ and $y$ respectively. Prove that $d(x_n,y_n) \to d(x,y)$ as $n \to \infty$ .

I leave it to the reader to supply two proofs, one using the sandwich theorem, and another using $\liminf$ and $\limsup$ .

As with epsilon and delta, it may be that postponing discussion of $\liminf$ and $\limsup$ , or avoiding them altogether, is not in the best interests of the student. I have to admit that I am not sure! But I think it is worth investigating possible ways to help students to understand $\liminf$ and $\limsup$ .

In the following, for convenience, we work in the extended real line

$\overline{\mathbb{R}} =[{-\infty},{+\infty}]= \mathbb{R}\cup\{{-\infty},{+\infty}\}\,.$

This is convenient, because every subset of $\overline{\mathbb{R}}$ has a supremum and an infimum in $\overline{\mathbb{R}}$ : there is no need to worry about boundedness and non-emptiness. Those who prefer to work in $\mathbb{R}$ should add in appropriate assumptions below where necessary.

The standard approach to lim inf and lim sup

The following approach to lim inf and lim sup is entirely standard.

Let $(x_n)$ be a sequence in $\overline{\mathbb{R}}$ . Then it is standard to define sequences $(s_n)$ and $(S_n)$ in $\overline{\mathbb{R}}$ as follows: for each $n \in \mathbb{N}$ ,

$s_n = \inf \{x_n,x_{n+1},x_{n+2},\dots\}$

and

$S_n =\sup\{x_n,x_{n+1},x_{n+2},\dots\}\,.$

We may then define

$\liminf_{n \to \infty} x_n = \sup\{s_n:n\in\mathbb{N}\}$

and

$\limsup_{n \to \infty} x_n = \inf\{S_n:n\in\mathbb{N}\}\,.$

Once you have decided on an appropriate definition of convergence in $\overline{\mathbb{R}}$ , you can confirm that we also have

$\liminf_{n \to \infty} x_n = \lim_{n \to \infty} s_n$

and

$\limsup_{n \to \infty} x_n =\lim_{n \to \infty} S_n\,.$

These definitions are very clean, and are easy to apply, e.g., to prove results in the theory of measure and integration. But they do not, in themselves, give the student a very good idea of what $\liminf$ and $\limsup$ really mean for a typical sequence. In my opinion, even calculating the $\liminf$ and $\limsup$ of a few examples does not really help as much as you would expect.

One approach that can help a little is to explain that $\liminf_{n \to\infty} (x_n)$ is the minimum of all the possible limits of subsequences of the sequence $(x_n)$ , and similarly for $\limsup$ , with $\max$ in place of $\min$ .

However, in my opinion, what we should try to get across is what $\liminf$ and $\limsup$ tell us about where $x_n$ can actually be as $n$ becomes large.

The absorption approach to lim inf and lim sup

Let $(x_n)$ be a sequence of extended real numbers. Set $s = \liminf_{n \to \infty} x_n$ and $S =\limsup_{n\to\infty} x_n$ . For the rest of this post, $(x_n)$ , $s$ and $S$ will be fixed.

What I think we would like students to understand is that, for large $n$ , $x_n$ is “almost” in $\null[s,S]$ , and there is no strictly smaller closed interval for which this is true.

Recall, in my terminology, a set absorbs a sequence if at most finitely many terms of the sequence lie outside the set.

In terms of absorption we can say various things about the relationships between $(x_n)$ , $s$ and $S$ . These standard facts are usually expressed using more standard terminology, e.g., in terms of a sequence “eventually lying within a set” or, for non-absorption, infinitely many terms of the sequence lying outside a set.

Let $a$ and $b$ be extended real numbers.

If $a<s$ , then $(a,{+\infty}]$ absorbs the sequence $(x_n)$ .
If $a > s$ , then $\null[a,{+\infty}]$ does not absorb $(x_n)$ .
If $b>S$ , then $\null[{-\infty},b)$ absorbs $(x_n)$ .
If $b < S$ , then $\null[{-\infty},b]$ does not absorb $(x_n)$ .
If $\null [s,S] \subseteq (a,b)$ then $(a,b)$ absorbs $(x_n)$ .
If $\null[a,b]$ is a proper subset of $\null[s,S]$ , then $\null[a,b]$ does not absorb $(x_n)$ .

Of course, we do not know whether or not one or both of $\null[s,S]$ and $(s,S)$ absorb $(x_n)$ . However, it is true that $\null[s,S]$ is equal to the intersection of all closed extended-real intervals which absorb the sequence $(x_n)$ .

Note added 4/12/09: Note that condition 6 in the above list
is not strong enough to recover much information in situations where $\null [s,S]$ itself fails to absorb the sequence. For example,for the sequence $x_n = 1/n$ , the interval $\null[{-1},0]$ satisfies conditions 5 and 6 above, without being equal to $\null[s,S]$ .

As mentioned above, all of these statements may be expressed using more standard terminology. Is the language of absorption helpful here?

In due course, I plan to return to this topic in the setting of function limits rather than sequences. This will then connect up with continuity and semicontinuity of functions.

Joel Feinstein

Tommi January 11, 2009 / 8:54 am

Greetings.

I am an undergraduate student. For me, the following explanation is useful: “What I think we would like students to understand is that, for large n, x_n is “almost” in [s, S], and there is no strictly smaller closed interval for which this is true.”

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vlorbik January 13, 2009 / 4:47 pm

effective use of “almost always” and “infinitely often”
(probability language) did a lot for me when i first encountered these.

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gowers January 21, 2009 / 4:26 pm

I notice that even with the language of absorption it is not wholly easy to say what the lim sup of a sequence is, at least if you regard quantifiers as difficult. You have to say that $(-\infty,a]$ absorbs for every $a>s$ and fails to absorb for any $a<s$ .

One could imagine getting round this problem as follows. First let us say that an interval $\null[-a,b]$ almost absorbs a sequence if $\null[a-\delta,b+\delta]$ absorbs the sequence for every $\delta>0$ . (Note that this introduces just one quantifier once they are happy with the absorption concept.) Next, let us say that an interval (or more general set) $S$ tempts a sequence $(a_n)$ if $a_n\in S$ for infinitely many $n$ . Finally, let us say that an interval $\null[a,b]$ almost tempts a sequence if $\null[a-\delta,b+\delta]$ tempts that sequence for every $\delta>0$ .

Then $a$ is the lim sup of a sequence if $(-\infty,a]$ almost absorbs the sequence and $\null[a,\infty)$ almost tempts it.

The advantage of the tempts concept is that it doesn’t force you to think what it means for a set not to absorb a sequence. Of course, one would prove a lemma to the effect that $S$ absorbs a sequence if and only if the complement of $S$ does not tempt it.

The disadvantage is of course that it introduces yet more nonstandard terminology.

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Joel January 21, 2009 / 5:01 pm

Thanks for those suggestions Tim!
It is not easy to balance the advantages and disadvantages of introducing non-standard terminology. I’m trying not to introduce too much.
Of course, a set $A$ tempts a sequence $(x_n)$ if and only if some subsequence of $(x_n)$ lies entirely within the set $A$ , and this is true if and only if the complement of $A$ does not absorb $(x_n)$ . (I set the last part of that as an exercise for the second-year students.) So, it may be possible to get the students to understand “does not absorb” without using extra terminology. I do find tempting tempting though!
Joel

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Joel February 5, 2009 / 2:03 pm

Working in the real line:
If you want to do without the ‘almost’ concepts, then $a$ is the lim sup of the sequence if and only if, for all $\varepsilon>0$ , the interval $({-\infty},a+\varepsilon)$ absorbs the sequence and the interval $(a-\varepsilon,+\infty)$ tempts it.

(Or you can use $({-\infty},a+\varepsilon]$ and $\null[a-\varepsilon,+\infty)$ if you wish.)

If we are working in the extended real line, the definitions proposed by Tim need some minor modifications. For a start, we need to work with intervals $\null[{-\infty},a]$ instead of $({-\infty},a]$ (etc.). But also, in the case where $a={-\infty}$ , we need an appropriate definition of what it means for the interval $\null[{-\infty},a]=\{{-\infty}\}$ to almost absorb (or almost tempt) our sequence. We can’t use ${-\infty}+\delta$ , because that is still ${-\infty}$ . For ‘almost absorbs’, we need to say instead that, for all positive real numbers M, $\null[-\infty,-M]$ absorbs the sequence (or some equivalent formulation of this). Obviously we need to do something similar for the interval $\null[+\infty,+\infty] = \{+\infty\}$ .

For my next comment, let’s assume instead that we are working with bounded sequences of real numbers, so that we can safely work in the real line.

It might be nice to define what it means for a general subset $E$ of the real line to almost absorb or almost tempt a sequence $(x_n)$ .

Let $E$ be a non-empty subset of the real line. Which of the following is the best definition of $E$ almost absorbs $(x_n)$ ?

– Every open superset of $E$ absorbs the sequence $(x_n)$ .

– For all $\varepsilon>0$ , the set $\{y \in \mathbb{R}: \textrm{dist}(y,E)<\varepsilon\}$ absorbs the sequence $(x_n)$ .

– For all $\varepsilon>0$ , the set $\{y \in \mathbb{R}: \textrm{dist}(y,E)\leq\varepsilon\}$ absorbs the sequence $(x_n)$ .

The second and third of these are easily seen to be equivalent.
For closed intervals (which are, perhaps, what we care most about), all three are equivalent. Otherwise, they differ (even for closed sets).

My initial instinct was to prefer the first definition, in view of its topological nature. But:
(a) I suspect that the others are easier for students to think about and to check;
(b) for open sets $E$ , the first definition is probably not what we want. Perhaps ‘every open superset of the closure of $E$ …’ would be a better attempt, but this is getting messy.

Whichever version you go for, if you set $s=\liminf_{n\to \infty}x_n$ and $S=\limsup_{n\to \infty}x_n$ (as in the original post), then $\null[s,S]$ is the minimum closed interval which almost absorbs the sequence. This also works for general sequences in the extended real line, provided that you use an appropriate definition of ‘almost absorbs’. As mentioned above, some care is needed with intervals which have one (or both) endpoints equal to $\pm \infty$ .

Joel Feinstein 5/2/09

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- Joel February 3, 2010 / 9:14 pm
  
  I don’t understand why this post and its comments keep getting corrupted 😦
  Joel
  3/2/10
  
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rose August 8, 2009 / 4:02 pm

iam an ug student.the explation about the limsup and liminf is very useful.
the language is simple and clear.but we need more examples.if the examples are published then it will be fantastic

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- Joel December 4, 2009 / 10:02 am
  
  Examples are, of course, crucial!
  Let’s look at some specific illustrative examples here.
  I will give each example in a separate comment, and I will work with bounded sequences in the real line, so that there are no issues with $\pm\infty$ .
  
  First let’s look at one of the most standard oscillating sequences,
  
  $x_n=(-1)^n$ ,
  
  i.e. the sequence ${-1},1,{-1},1,\dots$ .
  
  Since every term of this sequence is in the closed interval
  $\null[{-1},1]$ , that closed interval already absorbs the sequence (not required), and hence also almost absorbs it (as required).
  
  However, since both ${-1}$ and $1$ occur infinitely often, no closed interval strictly smaller than $\null[{-1},1]$ almost absorbs the sequence.
  
  In terms of “tempts” and “almost tempts”, both
  $(-\infty,{-1}]$ and $\null[1,\infty)$ tempt the sequence (not required) and hence also almost tempt it (as required).
  
  So we have found that the closed interval $\null[{-1},1]$ has the required properties: it is the minimum closed interval which almost absorbs the sequence.
  Thus $\liminf_{n\to\infty} x_n = {-1}$ and
  $\limsup_{n\to\infty} x_n = 1$ .
  
  Exercise:
  Investigate this sequence using the standard definitions of $\liminf$ and $\limsup$ .
  
  Perhaps it is better to work out the $\liminf$ and $\limsup$ separately, rather than focussing on the closed interval $\null[s,S]$ as I have here?
  
  In this case, following Tim’s suggestions, you could note (as above) that $({-\infty},{-1}]$ almost tempts the sequence (in this case it actually does tempt the sequence), and that $\null [{-1},\infty)$ almost absorbs the sequence (in this case, it actually does absorb the sequence), so that tells us that
  $\liminf_{n\to\infty} x_n = {-1}$ .
  
  Similarly, $\null[1,\infty)$ almost tempts the sequence (in this case it actually does tempt the sequence), and $({-\infty},{1}]$ almost absorbs the sequence (in this case, it actually does absorb the sequence), so that tells us that $\limsup_{n\to\infty} x_n = 1$ .
  
  Now that sequence is not very interesting. So in my next comment I’ll look at a slightly more interesting sequence.
  
  Joel
  December 4 2009
  
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Joel December 4, 2009 / 10:27 am
Second specific example on $\liminf$ and $\limsup$

For our second example, let us modify the previous example slightly, and consider
$x_n = (-1)^n (n+1)/n = (-1)^n(1 + \frac{1}{n}).$

Note here that $|x_n| = 1 + \frac{1}{n} > 1$ , and that $|x_n|\to 1$ as $n \to \infty$ .
So, $x_n \notin [{-1},1]$ , but, for large $n$ , $x_n$ is close to that closed interval. In fact, for all $n$ , we have $x_n \in [{-(1 + \frac{1}{n})},1 + \frac{1}{n}]$ .

We may suspect that $\null [{-1},1]$ is the “important” closed interval here.
We could really do with a good name for the interval $\null[\liminf_{n\to\infty}x_n,\limsup_{n\to\infty}x_n]$ . For now, let me call that interval the limmy closed interval for the sequence.

OK, so we suspect that our limmy closed interval is $\null [{-1},1]$ . Let’s check this carefully.

Suppose that $\null [{-1},1] \subseteq (a,b)$ for some real numbers $a$ and $b$ .
Of course, this just means that $a< -1$ and $b > 1$ . Then, for large $n$ we have both
${-(1 + \frac{1}{n})} > a$ and $1 + \frac{1}{n} < b$ , and hence we certainly have
$x_n \in (a,b)$ . This shows that $(a,b)$ absorbs the sequence $(x_n)$ .

We have now established that $\null [{-1},1]$ almost absorbs the sequence $(x_n)$ . However, as in the previous example, both $({-\infty},{-1}]$ and $[1,\infty)$ almost tempt the sequence (in this example, they both actually tempt the sequence). It follows, as before, that $\null [{-1},1]$ really is the minimum closed interval which almost absorbs the sequence. Thus $\null [{-1},1]$ is the limmy closed interval, and we (again) have $\liminf_{n\to\infty}x_n = -1$ and $\limsup_{n\to\infty}x_n = 1$ .

Again, you can calculate $\liminf_{n\to\infty} x_n$ and $\limsup_{n\to\infty} x_n$ separately as in the previous example, either directly from the standard definition , or by showing that:
- $({-\infty},{-1}]$ almost tempts the sequence and $\null[{-1},\infty)$ almost absorbs the sequence, so $\liminf_{n\to\infty}x_n = -1$ ;
- $\null[1,\infty)$ almost tempts the sequence and $({-\infty},1]$ almost absorbs the sequence, so $\limsup_{n\to\infty}x_n = 1$ .
Here are some further comments which may help students to understand these examples.

Here, $1$ is the least real number $b$ such that $({-\infty},b]$ almost absorbs the sequence.

Equivalently, $1$ is the infimum of all the real numbers $b$ such that $({-\infty},b]$ absorbs the sequence.

Similarly, $-1$ is the greatest real number $a$ such that $\null[a,\infty)$ almost absorbs the sequence.

Equivalently, $-1$ is the supremum of all the real numbers $a$ such that $\null[a,\infty)$ absorbs the sequence.

Joel
December 4 2009

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Joel January 3, 2010 / 6:13 pm

At this point, let me remind the reader that, for a bounded sequence of real numbers $(x_n)$ , $(x_n)$ converges if and only if
$\liminf_{n\to\infty} x_n = \limsup_{n\to\infty} x_n$ .
In this case, we also have
$\lim_{n \to \infty} x_n = \liminf_{n\to\infty} x_n = \limsup_{n\to\infty} x_n$ .
This means that you can take any of your favourite convergent sequences, and write down their $\liminf$ and $\limsup$ immediately. (Given that I mentioned boundedness above, you may wish to recall that every convergent sequence of real numbers is bounded.)
For example, with $x_n = \frac{2n^2+4n+1}{3n^2+7}$ , it is a standard exercise to check that $\lim_{n\to\infty} x_n = 2/3$ .
Thus we also have
$\liminf_{n\to\infty} x_n = \limsup_{n\to\infty} x_n = 2/3$
here.
If you work in the extended real line instead, with appropriate definitions, then you no longer need to worry about boundedness.
Here I think that it is amusing to note that, in the extended real line, the bizarre statement
$n \to \infty$ as $n \to \infty$
actually has some content!
Joel Feinstein
January 3 2010

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Joel January 8, 2010 / 10:26 am

Elementary remarks and exercises

Let ${\bf x}=(x_n)$ and ${\bf y}=(y_n)$ be bounded sequences of real numbers. We can use obvious (coordinatewise) algebraic operations to define $-{\bf x}$ , ${\bf x} + {\bf y}$ , etc. The following four facts are then standard (but we shall discuss them further below):

$\limsup_{n\to\infty} (-x_n) = - \liminf_{n\to\infty} x_n\,;$
$\liminf_{n\to\infty} (-x_n) = - \limsup_{n\to\infty} x_n\,;$
$\limsup_{n\to\infty} (x_n+y_n) \leq \limsup_{n\to\infty} x_n + \limsup_{n\to\infty} y_n\,;$
$\liminf_{n\to\infty} (x_n+y_n) \geq \liminf_{n\to\infty} x_n + \liminf_{n\to\infty} y_n\,.$

In order to discuss these further, let us introduce some (temporary?) notation for the limmy interval of the (bounded) sequences ${\bf x}=(x_n)$ , say
$L[{\bf x}] = L[(x_n)] = [\liminf_{n\to \infty} x_n,\limsup_{n\to\infty} x_n]$ .
Recall our discussion from earlier comments:

$L[{\bf x}] = L[(x_n)]$ is the minimum closed interval which almost absorbs the sequence $(x_n)$ .

For sets $A$ and $B$ of real numbers, we define $-A$ and $A+B$ by
${-A}= \{-x:x\in A\}$
and
$A+B =\{a+b:a \in A,b \in B\}\,.$
Then it is a very easy exercise to check that the set $A$ absorbs/almost absorbs/tempts/almost tempts the sequence ${\bf x}$ if and only if the set $-A$ absorbs/almost absorbs/tempts/almost tempts the sequence $-{\bf x}$ .
Similarly, with a little more work, you can see that if $A$ absorbs/almost absorbs ${\bf x}$ and $B$ absorbs/almost absorbs ${\bf y}$ then $A+B$ absorbs/almost absorbs ${\bf x} +{\bf y}$ .

Note, however, that the corresponding statement for tempting is false for the sum.

Exercise: give a counterexample.

Armed with these facts, we can now establish the four standard facts mentioned at the start of this comment. In terms of limmy intervals, these take the following form.
$L[{-\bf x}] = {-L[{\bf x}]}$
and
$L[{\bf x + y}] \subseteq L[{\bf x}] + L[{\bf y}]\,.$

(This latter inclusion may be strict.)

I am NOT claiming that this is the quickest way to prove these standard facts! However, it may help to provide greater understanding of them.

Warning: The results for sums need more care when working in the extended real line, where sums are not always defined.
Exercise: How much of the above remains valid when you work in the extended real line?

Joel
January 8 2010

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ipk January 20, 2010 / 8:59 am

Thanks this slowing starting to make sense. how does limsup and liminf for a sequence of sets, apply to probability theory. any examples would be much appreciated

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- Joel January 20, 2010 / 9:41 am
  
  The two measure theory/probability theory results that spring to mind here are Fatou’s lemma and the Borel-Cantelli lemma. Because you are dealing with sequences of sets instead of sequences of points, there are some subtle differences. However, you can make a connection if you look at the indicator functions of the sets involved and take the pointwise limsup/liminf.
  I’ll come back to say more about this when I have finished my large pile of marking!
  Joel
  January 20 2010
  
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ipk January 20, 2010 / 9:46 am

great thanks look forward to it

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