# Limits of sequences of sets

I am starting to learn about the liminf and limsup of a sequence of sets. However, I do not understand the very basics, such as why $\lim\sup A_n=\bigcap_{n\in N}\bigcup_{k>n} A_k,$ and why $\lim\sup A_n=\bigcup_{n\in N}\bigcap_{k>n}A_k.$ Can someone break these definitions down for me?

Also, why is $\lim\inf\subseteq\lim\sup$?

Let $\langle A_n:n\in\Bbb N\rangle$ be a sequence of sets. Imagine, for instance, that $A_0=[2,3]$, $A_1=\{4\}$, and $A_n=\left[-1,\frac1n\right]$ for $n\ge 2$. Then $\bigcap_{n\in\Bbb N}A_n=\varnothing$, but that’s because $A_0$ and $A_1$ don’t really ‘fit’; if we ignored them, we’d have a nice decreasing sequence of closed intervals whose intersection is $[-1,0]$. The idea of the $\liminf$ is to throw away any peculiarities caused by finitely many initial terms of the sequence of sets and get at what we might call the ‘essential’ intersection. Specifically, instead of looking at the set of points that are in all of the $A_n$’s, we look at the points that are in every $A_n$ from some point on. Such points are said to be eventually in the sets in the sequence.

Specifically, $\bigcap_{n\ge m}A_n$ is the set of points that are in $A_m,A_{m+1},A_{m+2},\ldots$, i.e., in every $A_n$ from $m$ on. Temporarily say that a point is $m$-good if it belongs to every $A_n$ with $n\ge m$; then $\bigcap_{n\ge m}A_n$ is the set of points that are $m$-good. Thus, $\bigcup_{m\in\Bbb N}\bigcap_{n\ge m}A_n$ is the set of points that are $m$-good for at least one $m\in\Bbb N$. These ‘good’ points are precisely the points that are eventually in the terms of the sequence $\langle A_n:n\in\Bbb N\rangle$, so they are precisely the ones that we want in $\liminf_{n\in\Bbb N}A_n$, and we define

$$\liminf_{n\in\Bbb N}A_n=\bigcup_{m\in\Bbb N}\bigcap_{n\ge m}A_n\;.$$

Now let $A_0=\{2\}$, $A_1=[0,1]$, $A_2=[-1,0]$, and $A_n=\{3(-1)^n\}$ for $n\ge 3$, so that $A_3=A_5=A_7=\ldots=\{-3\}$, and $A_4=A_6=\ldots=\{3\}$. Then

$$\bigcup_{n\in\Bbb N}A_n=[-1,1]\cup\{-3,2,3\}\;,$$

but again this is a little misleading: points of $[-1,1]$ are in the union only because of $A_0,A_1$, and $A_2$, and without those three sets the union would be just $\{-3,3\}$. The $\limsup$ is designed to get at the ‘essential’ part of the union by including only those points that are in infinitely many of the sets $A_n$. Such points are said to be frequently in the sets in the sequence.

Specifically, $\bigcup_{n\ge m}A_n$ contains every point that is in at least one $A_n$ with $n\ge m$, so if we define

$$\limsup_{n\in\Bbb N}A_n=\bigcap_{m\in\Bbb N}\bigcup_{n\ge m}A_n\;,$$

we’re saying that $\limsup_{n\in\Bbb N}A_n$ is the set of points $x$ that have the following property:

no matter how big an $m\in\Bbb N$ you pick, there is an $n\ge m$ such that $x\in A_n$.

A little thought should convince you that this really just says that $\{n\in\Bbb N:x\in A_n\}$ is infinite, which is precisely the idea that we wanted to capture.

• Thanks for the explanation. It is really clear. – user85362 Aug 26 '13 at 1:05
• @Abhishek: You’re very welcome; glad it helped. – Brian M. Scott Aug 26 '13 at 1:06
• This is a amazing explanation, thank you!! – user77404 Aug 26 '13 at 2:37
• @user77404: You’re welcome! – Brian M. Scott Aug 26 '13 at 2:44
• Excellent answer. Just one thing, should $$\bigcup_{n\in\Bbb N}A_n=[-1,1]\cup\{-3,3\}\;,$$ be $$\bigcup_{n\in\Bbb N}A_n=[-1,1]\cup\{-3,2,3\}\;,$$? – sonicboom Sep 25 '13 at 13:34

There are two notions of limits

$$\lim\sup A_n=\{x\in X\mid \forall n\in N\exists k>n:x\in A_k\}=\bigcap_{n\in N}\bigcup_{k>n} A_k$$ In other words, $\lim\sup$ is the set of $x$ that are in infinitely many $A_k$.

$$\lim\inf A_n=\{x\in X \mid \exists n\in N\forall k>n: x\in A_k\}=\bigcup_{n\in N}\bigcap_{k>n}A_k$$ This is the set of $x$ that are in each $A_k$ starting from an index $n$. Comparing these descriptions it follows immediatly that $\lim\inf\subseteq\lim\sup$

If $\lim\sup A_n=\lim\inf A_n$, then this is defined as the limit $\lim A_n$. They are equal, for example, if the sequence is increasing or decreasing.