Motivation to define $\limsup$ and $\liminf$ of sets

Let $$\{A_n\}$$ a collection of set. We define $$\limsup_{n\to \infty }A_n:=\bigcap_{n\in\mathbb N}\bigcup_{m\geq n}A_m\quad \text{and}\quad \liminf_{n\to \infty }A_n:=\bigcup_{n\in\mathbb N}\bigcap_{m\geq n}A_m.$$

These definition looks a bit coming from nowhere.

• Is there any reason to define the limsup and liminf of sets like that?
• Is there any connection with limsup and liminf defined for a real sequence for example ?

If $$(x_n)$$ is a sequence in a lattice $$(P,\leq)$$, $$\liminf_{n\to \infty }a_n:=\sup_{n\in\mathbb N} \inf_{k\geq n}x_k\quad \text{and}\quad \limsup_{n\to \infty }x_n=\inf_{n\in\mathbb N}\sup_{k\geq n}x_k.$$

If we use standard notation, we use$$a\wedge b:=\inf\{a,b\}\quad \text{and}\quad a\vee b:=\sup\{a,b\}.$$ So, $$\limsup_{n\to \infty }x_n:=\bigvee_{n\in\mathbb N}\bigwedge_{m\geq n}x_m\quad \text{and}\quad \liminf_{n\to \infty }x_n:=\bigwedge_{n\in\mathbb R}\bigvee_{m\geq n}x_m.$$

Take the lattice $$(2^{\mathbb R},\subset )$$ where the order is the usual inclusion. Then, $$A\wedge B$$ is the biggest set contain in $$A$$ and $$B$$ and $$A\vee B$$ is the smallest set where $$A$$ and $$B$$ are contained. Therefore,

$$A\wedge B:=A\cap B\quad \text{and}\quad A\vee B:=A\cup B.$$

So, $$\inf_{k\geq n}A_k:=\bigwedge_{k\geq n}A_k=\bigcap_{k\geq n}A_k,$$ and thus $$\sup_{n\in\mathbb N}\inf_{k\geq n}A_k=\bigvee_{n\in\mathbb N}\bigwedge_{k\geq n}A_k=\bigcup_{n\in\mathbb N}\bigcap_{k\geq n}A_k=:\liminf_{n\to \infty }A_k.$$ Same idea with the $$\limsup$$.

• Thanks, that makes a lot of sense. – Bruce Sep 28 '20 at 17:39

For a set $$A$$, define the characteristic function $$\chi_A(x)=\begin{cases} 1 \text{ if } x\in A \\ 0 \text{ otherwise}\end{cases}$$

Then the definition of $$\liminf$$ and $$\limsup$$ of sets are such that $$\liminf \chi_{A_i}=\chi_{\liminf A_i}$$ and $$\limsup \chi_{A_i}=\chi_{\limsup A_i}$$. Even if you ignore the actual interpretation of the sets, this is enough to make the definitions useful.

First of all there is a typo, the second set should be $$\bigcup$$ $$\bigcap$$. With that in mind, see if you can spot an analogy between statements 1 and 2 here:

1. The lim sup of a sequence $$\{ a_n \}$$ is a number $$M$$ such that $$M$$ is "nearly" larger than every term of the sequence "after a while," while being the smallest such $$M$$.

2. The lim sup of a sequence $$\{ A_n \}$$ is a set $$M$$ which contains every $$A_n$$ "after a while," and is the smallest such set.

In these statements, "... after a while" can be interpreted to "there exists $$N$$ such that for all $$n \geq N$$ ... For the real sequences, "a is 'nearly' greater than b" means that, for a given $$\varepsilon > 0$$, $$a > b - \varepsilon$$ . Similarly, spot the analogy between statements 3 and 4:

1. the lim inf of a sequence $$\{ a_n \}$$ is a number $$m$$ such that $$m$$ is "nearly" smaller than each $$a_n$$ "after a while," while being the largest such $$m$$.

2. The lim inf of a sequence of sets $$\{ A_n \}$$ is a set $$M$$ which is contained in every $$A_n$$ "after a while," while being the largest such $$M$$.