# Thread: lim sup and lim inf of a countable collection of sets

1. ## lim sup and lim inf of a countable collection of sets

Hi all,

Problem statement:
To build a case of a countable collection of sets of real numbers for which the lim inf and lim sup are not equal.

My Attempt:
(I normally associate lim sup and lim inf with sequences so I'm finding it a bit hard to think in terms of sets (intervals))

Anyway, let me call my countable collection An.
Suppose that I have two countable collection of sets, Bm and Cm where:

- Bm is a countable collection of non-empty decreasing closed sets (intervals) of reals with
B1 = [x, y-e]
- Cm is a countable collection of non-empty decreasing closed sets (intervals) of reals with C1 = [y+e, z]
(with x < y < z; e > 0)

And Ai = Bj if i is even and Ai = Cj if i is odd.

As n goes to infinity, lim inf An = L1 is going to lie somewhere between x and y-e
and lim sup An = L2 is going to lie somewhere between y+e and z
So, L1 cannot be equal to L2.

Is my example valid?

2. ## Re: lim sup and lim inf of a countable collection of sets

Originally Posted by director
Hi all,

Problem statement:
To build a case of a countable collection of sets of real numbers for which the lim inf and lim sup are not equal.

My Attempt:
(I normally associate lim sup and lim inf with sequences so I'm finding it a bit hard to think in terms of sets (intervals))

Anyway, let me call my countable collection An.
Suppose that I have two countable collection of sets, Bm and Cm where:

- Bm is a countable collection of non-empty decreasing closed sets (intervals) of reals with
B1 = [x, y-e]
- Cm is a countable collection of non-empty decreasing closed sets (intervals) of reals with C1 = [y+e, z]
(with x < y < z; e > 0)

And Ai = Bj if i is even and Ai = Cj if i is odd.

As n goes to infinity, lim inf An = L1 is going to lie somewhere between x and y-e
and lim sup An = L2 is going to lie somewhere between y+e and z
So, L1 cannot be equal to L2.

Is my example valid?

I don't think this example is right.
Try something like this.
Let the countable family be $A_{m,n}=\left(\frac{2}{m+n},\frac{3}{m}\right)$, $m,n$ naturals.

Then $\limsup A_{m,n} = \max A_{m,n}= A_{1,1}=(1,3)$ but $\liminf A_{m,n} =\lim_{n\rightarrow\infty,m\rightarrow\infty}A_{m, n}=\{0\}.$

3. ## Re: lim sup and lim inf of a countable collection of sets

the definitions of liminf/limsup for a sequence of set is not exactly the same you are used for a sequence of natural numbers.

$\liminf_{n\to\infty} A_n = \bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n$ and $\limsup_{n\to\infty} A_n = \bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n$

If you think about this in these terms the most simple exemple of a countable set with different limits should be An = [-1/n, 1,n], where limsup is (-1,1), liminf is {0}

4. ## Re: lim sup and lim inf of a countable collection of sets

A point $x$ is in $\liminf_{n\to \infty}A_n$ if there exists $N \in \mathbb{N}$ such that for all $n\ge N$, $x \in A_n$. That same point $x$ will be in $\limsup_{n\to \infty}A_n$ if for all $N\in \mathbb{N}$, there exists $n\ge N$ such that $x \in A_n$.

In the OP example, if $B_m = [x,y-e/m]$ and $C_m = [y+e/m,z]$, then $\liminf_{n\to \infty} A_n = \emptyset, \limsup_{n\to \infty}A_n = [x,z]\setminus {y}$, so that could be what you are looking for.