Equivalence of Definitions of Limit Superior of Sequence of Sets

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Theorem

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

Definition 1

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


Then the limit superior of $\set {E_n: n \in \N}$, denoted $\displaystyle \limsup_{n \mathop \to \infty} \ E_n$, is defined as:

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

Definition 2

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


Then the limit superior of $\set {E_n: n \in \N}$, denoted $\displaystyle \limsup_{n \mathop \to \infty} \ E_n$, is defined as:

$\displaystyle \limsup_{n \mathop \to \infty} \ E_n = \set {x : x \in E_i \text { for infinitely many } i}$


Proof

Begin by defining:

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

Then by definition 1:

$\displaystyle \limsup_{n \to \infty} \ E_n = \bigcap_{n \mathop = 0}^\infty B_n$


First Direction

Let $x$ belong to $E_i$ for infinitely many $i \in \N$.

Let $\phi \left({n}\right)$ be the sequence consisting of these numbers in increasing order.

Then for any number $k$, there exists a number $a$ with $\phi \left({a}\right) \ge k$.

Hence:

$\displaystyle x \in E_{\phi \left({a}\right)} \subseteq \bigcup_{j \mathop = k}^\infty E_j = B_k$

Since $k$ was arbitrary, it follows that $x\in B_n$ for each $n$.

So:

$\displaystyle x \in \limsup_{n \to \infty} \ E_n$


Second Direction

Let:

$\displaystyle x \in \bigcap_{n \mathop = 0}^\infty B_n$

If $x$ occurs in only finitely many $E_i$'s, then there is a largest value of $i$ (call it $i_0$) for which the membership holds.

Hence:

$x \notin \left({E_{i_0 + 1} \cup E_{i_0 + 2} \cup \ldots}\right) = B_{i_0 + 1}$

Therefore:

$\displaystyle x \notin \bigcap_{n \mathop = 0}^\infty B_n$

This contradicts our assumption about $x$.

Hence $x$ belongs to infinitely many members of the sequence.

$\blacksquare$


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