# limit of sequence of sets

Recall that $\limsup$ and $\liminf$ 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 $i\in\mathbb{Z}^{+}=\{1,2,\ldots\}$. Then

1. 1.

for $I$ ranging over all infinite subsets of $\mathbb{Z}^{+}$,

 $\limsup A_{i}=\bigcup_{I}\bigcap_{i\in I}A_{i},$
2. 2.

for $I$ ranging over all subsets of $\mathbb{Z}^{+}$ with finite compliment,

 $\liminf A_{i}=\bigcup_{I}\bigcap_{i\in I}A_{i},$
3. 3.

$\liminf A_{i}\subseteq\limsup A_{i}$.

Proof.

1. 1.

We need to show, for $I$ ranging over all infinite subsets of $\mathbb{Z}^{+}$,

 $\displaystyle\bigcup_{I}\bigcap_{i\in I}A_{i}=\bigcap_{n=1}^{\infty}\bigcup_{i% =n}^{\infty}A_{k}.$ (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, $x\in\bigcup_{i=1}^{\infty}A_{i}$. Now, suppose $x\in\bigcup_{i=k}^{\infty}A_{i}$. Since $I$ is infinite, we can find an $l\in I$ such that $l>k$. Being a member of $I$, we have that $x\in A_{l}\subseteq\bigcup_{i=k+1}^{\infty}A_{i}$. By induction, we have $x\in\bigcup_{i=n}^{\infty}A_{i}$ 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 $x\in\bigcup_{i=n}^{\infty}A_{i}$ for all $n\in\mathbb{Z}^{+}$ In $\bigcup_{i=1}^{\infty}A_{i}$, pick the least element $n_{0}$ such that $x\in A_{n_{0}}$. Next, in $\bigcup_{i=n_{0}+1}^{\infty}A_{i}$, pick the least $n_{1}$ such that $x\in A_{n_{1}}$. Then the set $I=\{n_{0},n_{1},\ldots\}$ fulfills the requirement $x\in\bigcap_{i\in I}A_{i}$, showing the other inclusion ($\supseteq$).

2. 2.

Here we have to show, for $I$ ranging over all subsets of $\mathbb{Z}^{+}$ with $\mathbb{Z}^{+}-I$ finite,

 $\displaystyle\bigcup_{I}\bigcap_{i\in I}A_{i}=\bigcup_{n=1}^{\infty}\bigcap_{i% =n}^{\infty}A_{k}.$ (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 $n\in\mathbb{Z}^{+}-I$, $n. This means that any $m\geq n_{0}$, we have $m\in I$. Therefore, $x\in\bigcap_{i=n_{0}}^{\infty}A_{i}$ and $x$ is an element of the RHS.

Next, suppose $x$ is an element of the RHS so that $x\in\bigcap_{k=n}^{\infty}A_{k}$ for some $n$. Then the set $I=\{n_{0},n_{0}+1,\ldots\}$ is a subset of $\mathbb{Z}^{+}$ with finite complement that does the job for the LHS.

3. 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

 $\liminf A_{i}=\limsup A_{i}=\lim A_{i}=\bigcap A_{i}.$

Similarly, if $\{A_{i}\}$ is an increasing sequence of sets, then

 $\liminf A_{i}=\limsup A_{i}=\lim A_{i}=\bigcup A_{i}.$

Proof. We shall only show the case when we have a descending chain of sets, since the other case is completely analogous. Let $A_{1}\supseteq A_{2}\supseteq\ldots$ be a descending chain of sets. Set $A=\bigcap_{i=1}^{\infty}A_{i}$. We shall show that

 $\limsup A_{i}=\liminf A_{i}=\lim A_{i}=A.$

First, by the definition of of a sequence of sets:

 $\limsup A_{i}=\bigcap_{n=1}^{\infty}\bigcup_{i=n}^{\infty}A_{k}=\bigcap_{n=1}^% {\infty}A_{n}=A.$

Now, by Assertion 3 of the above Theorem, $\liminf A_{i}\subseteq\limsup A_{i}=A$, so we only need to show that $A\subseteq\liminf A_{i}$. 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 LimitOfSequenceOfSets 2013-03-22 15:00:34 2013-03-22 15:00:34 CWoo (3771) CWoo (3771) 8 CWoo (3771) Theorem msc 03E20 msc 28A05 msc 60A99 LimitSuperior