Equivalence of Definitions of Limit Superior of Sequence of Sets

Theorem
The following definitions of a limit superior of a sequence of sets are equivalent:

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.