limit of sequence of sets
Recall that $lim\; sup$ and $lim\; inf$ of a sequence^{} of sets $\{{A}_{i}\}$ denote the and the of $\{{A}_{i}\}$, respectively. Please click here (http://planetmath.org/LimitSuperiorOfSets) to see the definitions and here (http://planetmath.org/LimitSuperior) to see the specialized definitions when they are applied to the real numbers.
Theorem. Let $\{{A}_{i}\}$ be a sequence of sets with $i\in {\mathbb{Z}}^{+}=\{1,2,\mathrm{\dots}\}$. Then

1.
for $I$ ranging over all infinite subsets of ${\mathbb{Z}}^{+}$,
$$lim\; sup{A}_{i}=\bigcup _{I}\bigcap _{i\in I}{A}_{i},$$ 
2.
for $I$ ranging over all subsets of ${\mathbb{Z}}^{+}$ with finite compliment,
$$lim\; inf{A}_{i}=\bigcup _{I}\bigcap _{i\in I}{A}_{i},$$ 
3.
$lim\; inf{A}_{i}\subseteq lim\; sup{A}_{i}$.
Proof.

1.
We need to show, for $I$ ranging over all infinite subsets of ${\mathbb{Z}}^{+}$,
$\bigcup _{I}}{\displaystyle \bigcap _{i\in I}}{A}_{i}={\displaystyle \bigcap _{n=1}^{\mathrm{\infty}}}{\displaystyle \bigcup _{i=n}^{\mathrm{\infty}}}{A}_{k}.$ (1) Let $x$ be an element of the LHS, the left hand side of Equation (1). Then $x\in {\bigcap}_{i\in I}{A}_{i}$ for some infinite subset $I\subseteq {\mathbb{Z}}^{+}$. Certainly, $x\in {\bigcup}_{i=1}^{\mathrm{\infty}}{A}_{i}$. Now, suppose $x\in {\bigcup}_{i=k}^{\mathrm{\infty}}{A}_{i}$. Since $I$ is infinite^{}, we can find an $l\in I$ such that $l>k$. Being a member of $I$, we have that $x\in {A}_{l}\subseteq {\bigcup}_{i=k+1}^{\mathrm{\infty}}{A}_{i}$. By induction^{}, we have $x\in {\bigcup}_{i=n}^{\mathrm{\infty}}{A}_{i}$ for all $n\in {\mathbb{Z}}^{+}$. Thus $x$ is an element of the RHS. This proves one side of the inclusion ($\subseteq $) in (1).
To show the other inclusion, let $x$ be an element of the RHS. So $x\in {\bigcup}_{i=n}^{\mathrm{\infty}}{A}_{i}$ for all $n\in {\mathbb{Z}}^{+}$ In ${\bigcup}_{i=1}^{\mathrm{\infty}}{A}_{i}$, pick the least element ${n}_{0}$ such that $x\in {A}_{{n}_{0}}$. Next, in ${\bigcup}_{i={n}_{0}+1}^{\mathrm{\infty}}{A}_{i}$, pick the least ${n}_{1}$ such that $x\in {A}_{{n}_{1}}$. Then the set $I=\{{n}_{0},{n}_{1},\mathrm{\dots}\}$ fulfills the requirement $x\in {\bigcap}_{i\in I}{A}_{i}$, showing the other inclusion ($\supseteq $).

2.
Here we have to show, for $I$ ranging over all subsets of ${\mathbb{Z}}^{+}$ with ${\mathbb{Z}}^{+}I$ finite,
$\bigcup _{I}}{\displaystyle \bigcap _{i\in I}}{A}_{i}={\displaystyle \bigcup _{n=1}^{\mathrm{\infty}}}{\displaystyle \bigcap _{i=n}^{\mathrm{\infty}}}{A}_{k}.$ (2) Suppose first that $x$ is an element of the LHS so that $x\in {\bigcap}_{i\in I}{A}_{i}$ for some $I$ with ${\mathbb{Z}}^{+}I$ finite. Let ${n}_{0}$ be a upper bound^{} of the finite set^{} ${\mathbb{Z}}^{+}I$ such that for any $n\in {\mathbb{Z}}^{+}I$, $$. This means that any $m\ge {n}_{0}$, we have $m\in I$. Therefore, $x\in {\bigcap}_{i={n}_{0}}^{\mathrm{\infty}}{A}_{i}$ and $x$ is an element of the RHS.
Next, suppose $x$ is an element of the RHS so that $x\in {\bigcap}_{k=n}^{\mathrm{\infty}}{A}_{k}$ for some $n$. Then the set $I=\{{n}_{0},{n}_{0}+1,\mathrm{\dots}\}$ is a subset of ${\mathbb{Z}}^{+}$ with finite complement that does the job for the LHS.

3.
The set of all subsets (of ${\mathbb{Z}}^{+}$) with finite complement is a subset of the set of all infinite subsets. The third assertion is now clear from the previous two propositions^{}. QED
Corollary. If $\{{A}_{i}\}$ is a decreasing sequence of sets, then
$$lim\; inf{A}_{i}=lim\; sup{A}_{i}=lim{A}_{i}=\bigcap {A}_{i}.$$ 
Similarly, if $\{{A}_{i}\}$ is an increasing sequence of sets, then
$$lim\; inf{A}_{i}=lim\; sup{A}_{i}=lim{A}_{i}=\bigcup {A}_{i}.$$ 
Proof. We shall only show the case when we have a descending chain of sets, since the other case is completely analogous. Let ${A}_{1}\supseteq {A}_{2}\supseteq \mathrm{\dots}$ be a descending chain of sets. Set $A={\bigcap}_{i=1}^{\mathrm{\infty}}{A}_{i}$. We shall show that
$$lim\; sup{A}_{i}=lim\; inf{A}_{i}=lim{A}_{i}=A.$$ 
First, by the definition of of a sequence of sets:
$$lim\; sup{A}_{i}=\bigcap _{n=1}^{\mathrm{\infty}}\bigcup _{i=n}^{\mathrm{\infty}}{A}_{k}=\bigcap _{n=1}^{\mathrm{\infty}}{A}_{n}=A.$$ 
Now, by Assertion 3 of the above Theorem, $lim\; inf{A}_{i}\subseteq lim\; sup{A}_{i}=A$, so we only need to show that $A\subseteq lim\; inf{A}_{i}$. But this is immediate from the definition of $A$, being the intersection^{} of all ${A}_{i}$ with subscripts $i$ taking on all values of ${\mathbb{Z}}^{+}$. Its complement is the empty set^{}, clearly finite. Having shown both the existence and equality of the and of the ${A}_{i}$’s, we conclude that the limit of ${A}_{i}$’s exist as well and it is equal to $A$. QED
Title  limit of sequence of sets 

Canonical name  LimitOfSequenceOfSets 
Date of creation  20130322 15:00:34 
Last modified on  20130322 15:00:34 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  8 
Author  CWoo (3771) 
Entry type  Theorem 
Classification  msc 03E20 
Classification  msc 28A05 
Classification  msc 60A99 
Related topic  LimitSuperior 