Characterization of Limit Superior of Sets

Theorem

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

Then:

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

where $\displaystyle \limsup_{n \to \infty} \ E_n$ is the limit superior of $E_n$.

Proof

Begin by defining $\displaystyle B_n := \bigcup_{j=n}^\infty E_j$.

Then by definition, $\displaystyle \limsup_{n \to \infty} \ E_n = \bigcap_{n=0}^\infty B_n$.

First Direction

Suppose $x$ belongs to $E_i$ for infinitely many $i \in \N$.

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

Then for any number $k$, there exists a number $a$ with $\phi (a) \geq k$.

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

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

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

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

This contradicts our assumption about $x$.

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

$\blacksquare$