Equivalence of Definitions of Limit Inferior of Sequence of Sets

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Theorem

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

Definition 1

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


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

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

Definition 2

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


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

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


Proof

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

Let:

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

Let:

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

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


Let:

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

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


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

$\ds \liminf_{n \mathop \to \infty} E_n \subseteq E$

and:

$E \subseteq \ds \liminf_{n \mathop \to \infty} E_n$


From Set Complement inverts Subsets, we have:

$\ds \map \complement E \subseteq \map \complement {\liminf_{n \mathop \to \infty} E_n} \iff \liminf_{n \mathop \to \infty} E_n \subseteq E$

The strategy will therefore be to prove:

$\ds E \subseteq \liminf_{n \mathop \to \infty} E_n$

and:

$\ds \map \complement E \subseteq \map \complement {\liminf_{n \mathop \to \infty} E_n}$


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 $x \notin E_m$.

$m$ has to exist, because the set $\set {i \in \N: x \notin E_i}$ is a finite set.


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:

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

From the definition of set union, it follows that:

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

Hence:

$E \subseteq \ds \liminf_{n \mathop \to \infty} E_n$

$\Box$


Complement of Definition 2 is contained in Complement of Definition 1

Let $x \in \map \complement E$

That is, $x \notin E$.

The definition of $E$ grants the existence of a strictly increasing sequence $\sequence {i_n}$ 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:

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

and hence $x \notin B_k$.

As $k$ was arbitrary, we have:

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

It follows that:

$\ds \map \complement E \subseteq \map \complement {\liminf_{n \mathop \to \infty} E_n}$

$\Box$


Therefore, we have established that

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

From the definition of set equality, it follows that:

$\ds \liminf_{n \mathop \to \infty} E_n = E$

$\blacksquare$


Also see


Sources

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