# Equivalence of Definitions of Limit Inferior of Sequence of Sets

## Contents

## 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:

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

$\blacksquare$

## Also see

- Inner Limit of Sequence of Sets in Normed Space described via the point-to-set distance function induced by the norm of the space
- Inner Limit in Hausdorff Space by Open Neighborhoods

## Sources

- 1970: Avner Friedman:
*Foundations of Modern Analysis*... (previous) ... (next): $\S 1.1$: Rings and Algebras: Problem $1.1.2$