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${\mathbb{Z}}^{+}$,

   ${\displaystyle \bigcup _I}{\displaystyle \bigcap _{i\in I}}{A}_i={\displaystyle \bigcap _{n=1}^{\mathrm{\infty }}}{\displaystyle \bigcup _{i=n}^{\mathrm{\infty }}}{A}_k.$(1)

   Let $x$ be an element of theLHS, 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 canfind an $l\in I$ such that $l>k$. Being a member of $I$, we havethat $x\in A_l\subseteq {\bigcup }_{i=k+1}^{\mathrm{\infty }}{A}_i$. 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. Thisproves 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 theleast $n_1$ 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,

   ${\displaystyle \bigcup _I}{\displaystyle \bigcap _{i\in I}}{A}_i={\displaystyle \bigcup _{n=1}^{\mathrm{\infty }}}{\displaystyle \bigcap _{i=n}^{\mathrm{\infty }}}{A}_k.$(2)

   Suppose first that $x$ isan 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 ofthe 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 finitecomplement is a subset of the set of all infinite subsets. Thethird assertion is now clear from the previous two propositions. QED

Corollary. If $\{A_i\}$ is a decreasingsequence of sets, then

$$lim\; inf{A}_i=lim\; sup{A}_i=lim{A}_i=\bigcap A_i.$$

Similarly, if $\{A_i\}$ is an increasingsequence 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 adescending chain of sets, since the other case is completelyanalogous. Let $A_1\supseteq A_2\supseteq \mathrm{\dots }$ be a descendingchain of sets. Set $A={\bigcap }_{i=1}^{\mathrm{\infty }}{A}_i$. We shall showthat

$$lim\; sup{A}_i=lim\; inf{A}_i=lim{A}_i=A.$$

First, by thedefinition 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 aboveTheorem, $lim\; inf{A}_i\subseteq lim\; sup{A}_i=A$, so we only need toshow that $A\subseteq lim\; inf{A}_i$. But this is immediate from thedefinition of $A$, being the intersection of all $A_i$ with subscripts $i$ taking on all values of ${\mathbb{Z}}^{+}$. Its complement is the emptyset, clearly finite. Having shown both the existence and equalityof the and of the $A_i$’s, we concludethat the limit of $A_i$’s exist as well and it is equal to $A$.QED