Recall that $\limsup$ and $\liminf$ of a sequence of sets $\lbrace A_i \rbrace$ denote the limit superior and the limit inferior of $\lbrace A_i \rbrace$ , respectively. Please click here to see the definitions and here to see the specialized definitions when they are applied to the real numbers.

Theorem. Let $\lbrace A_i \rbrace$ be a sequence of sets with $i\in\mathbb{Z}^{+}=\lbrace 1,2,\ldots \rbrace$ . Then

1. for $I$ ranging over all infinite subsets of $\mathbb{Z}^{+}$ , $$\limsup A_i=\bigcup_{I}\bigcap_{i\in I}A_i,$$
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. $\liminf A_i\subseteq\limsup A_i$ .

Proof.

1. We need to show, for $I$ ranging over all infinite subsets of $\mathbb{Z}^{+}$ , \begin{eqnarray} \bigcup_{I}\bigcap_{i\in I} A_i=\bigcap_{n=1}^\infty \bigcup_{i=n}^\infty A_k.\end{eqnarray}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 $I=\lbrace n_0,n_1,\ldots \rbrace$ 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, \begin{eqnarray} \bigcup_{I}\bigcap_{i\in I} A_i=\bigcup_{n=1}^\infty \bigcap_{i=n}^\infty A_k.\end{eqnarray}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 $n\in\mathbb{Z}^{+}-I$ , $n<n_0$ . 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 $I=\lbrace n_0,n_0+1,\ldots\rbrace$ 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 $\lbrace A_i \rbrace$ is a decreasing sequence of sets, then $$\liminf A_i=\limsup A_i=\lim A_i=\bigcap A_i.$$ Similarly, if $\lbrace A_i \rbrace$ 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 limit superior 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 limit superior and limit inferior of the $A_i$ 's, we conclude that the limit of $A_i$ 's exist as well and it is equal to $A$ . QED