# Equivalence of Definitions of Limit Inferior of Sequence of Sets

## Theorem

The following definitions of the concept of Limit Inferior of Sequence of Sets are equivalent:

### Definition 1

Let $\set {E_n : n \in \N}$ be a sequence of sets.

Then the limit inferior of the sequence, denoted $\displaystyle \liminf_{n \mathop \to \infty} \ E_n$, is defined as:

 $\displaystyle \liminf_{n \mathop \to \infty} \ E_n$ $:=$ $\displaystyle \bigcup_{n \mathop = 0}^\infty \ \bigcap_{i \mathop = n}^\infty E_n$ $\displaystyle$ $=$ $\displaystyle \paren {E_0 \cap E_1 \cap E_2 \cap \ldots} \cup \paren {E_1 \cap E_2 \cap E_3 \cap \ldots} \cup \cdots$

### Definition 2

Let $\left\{{E_n : n \in \N}\right\}$ be a sequence of sets.

Then the limit inferior of the sequence, denoted $\displaystyle \liminf_{n \mathop \to \infty} \ E_n$, is defined as:

$\displaystyle \liminf_{n \mathop \to \infty} \ E_n := \set {x: x \in E_i \text{ for all but finitely many } i}$

## Proof

Let $\left\{{E_n : n \in \N}\right\}$ be a sequence of sets.

Let:

$\displaystyle B_n := \bigcap_{j \mathop = n}^\infty E_j$

Let:

$\displaystyle \liminf_{n \to \infty} \ E_n := \bigcup_{n \mathop = 0}^\infty B_n$

that is, $\displaystyle \liminf_{n \to \infty} \ E_n$ according to definition 1.

Let:

$E := \left\{{x : x \in E_i \text{ for all but finitely many} \ i}\right\}$

that is, $\displaystyle \liminf_{n \to \infty} \ E_n$ according to definition 2.

By definition of set equality, it is enough to prove:

$\displaystyle \liminf_{n \to \infty} \ E_n \subseteq E$

and:

$E \subseteq \displaystyle \liminf_{n \to \infty} \ E_n$

From Set Complement inverts Subsets, we have:

$\displaystyle \complement \left({E}\right) \subseteq \complement \left({\liminf_{n \to \infty} \ E_n}\right) \iff \liminf_{n \to \infty} \ E_n \subseteq E$

The strategy will therefore be to prove:

$\displaystyle E \subseteq \liminf_{n \to \infty} \ E_n$

and:

$\displaystyle \complement \left({E}\right) \subseteq \complement \left({\liminf_{n \to \infty} \ E_n}\right)$

### Definition 2 is contained in Definition 1

Let $x \in E$.

By definition $x$ is in all but a finite number of $E_i$.

Let $m \in \Z_{\ge 0}$ be the largest integer such that $x \notin E_m$.

Let $M \in \Z$ such that $m < M$.

Then as $x \in E_j$ for all $j > m$ it follows by definition of set intersection that:

$x \in B_M = \displaystyle \bigcap_{j \mathop = M}^\infty E_j$

From the definition of set union, it follows that:

$x \in \displaystyle \bigcup_{n \mathop = 0}^\infty B_n = \liminf_{n \to \infty} \ E_n$

Hence:

$E \subseteq \displaystyle \liminf_{n \to \infty} \ E_n$

$\Box$

### Complement of Definition 2 is contained in Complement of Definition 1

Let $x \in \complement \left({E}\right)$

That is, $x \notin E$.

The definition of $E$ grants the existence of a strictly increasing sequence $\left \langle {i_n}\right \rangle$ of natural numbers, such that:

$\forall n \in \N: x \notin E_{i_n}$

It follows that for every $k \in \N$, there exists an $n \in \Z$ with $i_n > k$.

Subsequently, from the definition of set intersection:

$B_k = \displaystyle \bigcap_{j \mathop = k}^\infty E_j \subseteq E_{i_n}$

and hence $x \notin B_k$.

As $k$ was arbitrary, we have:

$x \notin \displaystyle \bigcup_{k \mathop = 0}^\infty B_k = \liminf_{n \to \infty} \ E_n$

It follows that:

$\displaystyle \complement \left({E}\right) \subseteq \complement \left({\liminf_{n \to \infty} \ E_n}\right)$

$\Box$

Therefore, we have established that

$x \in \displaystyle \liminf_{n \to \infty} \ E_n \iff x \in E$

From the definition of set equality, it follows that:

$\displaystyle \liminf_{n \to \infty} \ E_n = E$

$\blacksquare$