# Liminf and Limsup of a sequence of sets

I am attempting to learn some measure theory and am starting with liminf and limsup of sequences of sets.

I found an example that is as follows:

$$A_n=\left\{\frac0n, \frac1n, \dots , \frac{n^2}n\right\}$$

and I am trying to find the limsup and liminf.

My understanding is that both deal with the tail sequences, and that limsup involves values that appear "infinitely often" and liminf covers values that appear "all but finitely often". Also I understand that $\liminf A_n\subset\limsup A_n$.

For the above example, if I enumerate the first few sets, it is clearly evident that ${0}$ appears i.o. It also seems (to me) that as $n\to\infty$, all of the positive rational numbers appear. I am having trouble seeing the limits. For example, no matter how large I choose $N$, there is some $n\ge N$ in which all of the rationals appear, right?

Obviously I am confused (this is all self-taught), so any explanation would be greatly appreciated. I seem to be able to make sense of liminf and limsup when the sequence is of a form similar to $[0, n/(n+1))$ and other examples, but I'm struggling with this example.

One simple question: does an event have to "not" show up sometimes to be part of the liminf, or is it just that it is allowed to be missing finitely often? Assuming for a moment that the former is true, it appears to me that {0} is definitely in both liminf and limsup: that is, no matter how large I select N, {0} is in some (in this case, all) A_n with n>N.

Moving on from there, it is clear to me that the integers begin to appear over and over again, and as n-->infinity, the rationals begin to "fill out" as well. Where I seem to be getting stuck is that the integers only show up equal to "n" and it is not clear to me how to handle the fact that the sequence is unbounded. Intuitively, all of the (positive) integers eventually show up (always), but they are far from the only values that do. {1/2} shows up always, for example. I'm not sure if {Q_+} eventually shows up, and this may be due to a lack of formal construction of Q in my past. What I recall is that Q is essentially all real numbers that can be expressed as m/n with m and n members of Z.

Thank you.

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It would be a good idea to write out a few of the sets $$A_n=\left\{\frac0n, \frac1n, \dots , \frac{n^2}n\right\}$$ explicitly. For starters, that would show that it’s not in general true that $A_n\subseteq A_{n+1}$.

\begin{align*} A_1&=\left\{\frac01,\frac11\right\}=\{0,1\}\\ A_2&=\left\{\frac02,\frac12,\frac22,\frac32,\frac42\right\}=\left\{0,\frac12,1,\frac32,2\right\}\\ A_3&=\left\{\frac03,\frac13,\frac23,\frac33,\frac43,\frac53,\frac63,\frac73,\frac83,\frac93\right\}=\left\{0,\frac13,\frac23,1,\frac43,\frac53,2,\frac73,\frac83,3\right\}\;, \end{align*}

and so on. It’s true that $A_1\subseteq A_2$, but clearly $A_2\nsubseteq A_3$.

It is easy to check that if $k\in\Bbb N$, then $k\in A_n$ for all $n\ge k$, so $\Bbb N\subseteq\liminf_n A_n$, and of course it’s very clear that $\limsup_n A_n\subseteq\Bbb Q\cap[0,\to0)$, the set of non-negative rationals.

Suppose that $a,b\in\Bbb Z^+$ and $\gcd(a,b)=1$, so that $a/b$ is a positive rational in lowest terms. Suppose further that $b>1$, so that $a/b$ is not an integer. For what $n\in\Bbb Z^+$ is it true that $a/b\in A_n$? Clearly $n$ must be a multiple of $b$. Say $n=kb$; then $$\frac{a}b=\frac{ka}{kb}=\frac{ka}n\in A_n$$ if and only if $0\le ka\le n^2$. From here you should be able to describe fairly concisely exactly which $A_n$ contain $a/b$, and that will show you exactly what $\liminf_n A_n$ and $\limsup_n A_n$ must be.

Here are the actual answers, spoiler-protected; mouse-over to see them.

$\liminf_n A_n=\Bbb N$; $\limsup_n A_n=\Bbb Q\cap[0,\to)$.

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Thank you for the thorough explanation. Truly appreciated! – Justin May 19 '12 at 21:42

As Didier pointed out, the sequence is not increasing. For example, $\frac{1}{2}$ is in $A_2$ while it is not in $A_3$. There are much more examples, as we shall see. But let's first have a closer look at the definition of limit inferior and limit superior in the case of sets.

Suppose that $\{A_n\}_{n\in\mathbb{N}}$ is a sequence of subsets of some set $X$. Then define $$\liminf_{n\to\infty}A_n:=\bigcup_{n\in\mathbb{N}}\bigcap_{m\geq n} A_n$$ and $$\limsup_{n\to\infty}A_n:=\bigcap_{n\in\mathbb{N}}\bigcup_{m\geq n} A_n$$ Suppose that $x\in\liminf_n A_n$. This means that there exists an $M\in\mathbb{N}$ such that $x\in A_m$ for all $m\geq M$. Or, stated otherwise, $x$ is an element of all but finitely many of the sets $A_n$, as you have correctly noted.

Now suppose that $x\in\limsup_nA_n$. This means that for all $n\in\mathbb{N}$, there exists an $m\in\mathbb{N}$ with the property that $x\in A_m$. Or, stated otherwise, $x$ is an element of infinitely many of the sets $A_n$. Now back to the problem.

To see what that the limit inferior of $\{A_n\}$ is $\mathbb{Z}_{\geq 0}$, suppose that $\frac{p}{q}$ is a positive non-integer rational number (i.e. $q> 1$ and $p>0$), and that $n$ is relatively prime to $q$. Then it follows easily that $\frac{p}{q}$ is not in $A_n$. Since there are infinitely many numbers relatively prime to $q$, it follows that $\frac{p}{q}$ is not in $\liminf_n A_n$. The non-negative integers, however, in $\liminf_n A_n$. To see this, note that if $n$ is an integer and $m\geq n$, then $n=\frac{mn}{m}\in A_m$. Thus, we get $$\liminf_{n\to\infty}A_n=\mathbb{Z}_{\geq 0}.$$

The limit superior of $A_n$ can be found in a similar way. Note that $\frac{p}{q}\in A_{nq}$ for any $n\in\mathbb{N}$ when $\frac{p}{q}$ is a non-negative rational. Thus, $$\limsup_{n\to\infty}A_n=\mathbb{Q}_{\geq 0}.$$

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"Limit inferior" and "Limit superior" are, I think the usual names. Limes are something else... – Arturo Magidin May 19 '12 at 21:23
@Arturo: Or even something else, if you’re an imperial Roman. – Brian M. Scott May 19 '12 at 21:26
@BrianM.Scott: Yes, I hit the same page when I just typed "limes" into the Wikipedia search box... (-: – Arturo Magidin May 19 '12 at 21:26
@Arturo: It was actually the first thing that occurred to me, since in the limit context I was already pronouncing limes as \lee-mace\ (more or less). – Brian M. Scott May 19 '12 at 21:28
I think I learned this name in when I first learned analysis. Anyway, I changed it. :) – Egbert May 19 '12 at 21:39