Limit of Decreasing Sequence of Sets is Intersection

Definition
Let $\sequence {E_k}_{k \mathop \in \N}$ be an decreasing sequence:


 * $\forall k \in \N: E_{k + 1} \subseteq E_k$

Then $\sequence {E_k}_{k \mathop \in \N}$ has a limit such that:
 * $\ds \lim_{n \mathop \to \infty} E_n = \bigcap_{k \mathop \in \N} {E_k}$

Proof
Let $E = \ds \bigcap_{k \mathop \in \N} {E_k}$.

Let $x \in E$.

By definition of set intersection:
 * $\forall E_n \in \sequence {E_k}_{k \mathop \in \N}: x \in E_n$

Thus $x \in E_n$ for all but finitely many (that is, zero) terms of $\sequence {E_k}_{k \mathop \in \N}$.

That is:
 * $x \in \ds \liminf_{n \mathop \to \infty} E_n$

where $\ds \liminf_{n \mathop \to \infty} E_n$ denotes the limit inferior of $\sequence {E_k}_{k \mathop \in \N}$.

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

Let $x \in \ds \limsup_{n \mathop \to \infty} E_n$.

$\exists n \in \N: x \notin E_n$.

Then as $E_{n + 1} \subseteq E_n$ it follows that $n \notin E_{n + 1}$.

Hence:
 * $\forall m \in \N: m > n: x \notin E_m$

and so there is only a finite number of $i \in \N$ for which $x \in E_i$, that is, where $m < n$.

That is, by definition of limit superior:
 * $x \notin \ds \limsup_{n \mathop \to \infty} E_n$

But this contradicts our assertion that $x \in \ds \limsup_{n \mathop \to \infty} E_n$.

Hence:
 * $\forall n \in \N: x \in E_n$

and so by definition of set intersection:
 * $x \in E$

So we have shown that:
 * $\ds \limsup_{n \mathop \to \infty} E_n \subseteq E$

Hence we have:
 * $\ds \limsup_{n \mathop \to \infty} E_n \subseteq E$

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

Hence by Subset Relation is Transitive:


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

and it follows from Limit of Sets Exists iff Limit Inferior contains Limit Superior that:


 * $\ds \lim_{n \mathop \to \infty} E_n = \bigcup_{k \mathop \in \N} {E_k}$