Limit of Sets Exists iff Limit Inferior contains Limit Superior

Theorem

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

Then $\Bbb S$ converges to a limit if and only if:

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

Proof

Sufficient Condition

Let $\Bbb S$ converge to a limit.

Then by definition:

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

and so by definition of set equality:

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

$\Box$

Necessary Condition

Suppose that:

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

From Limit Superior includes Limit Inferior:

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

whether or not $\Bbb S$ converges to a limit.

By definition of set equality:

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

and so by definition $\Bbb S$ converges to a limit.

$\blacksquare$