Limit of Decreasing Sequence of Sets is Intersection

From ProofWiki
Jump to navigation Jump to search


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:

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


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

Aiming for a contradiction, suppose $\exists n \in \N: x \notin E_n$.

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


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

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


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

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:

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


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

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:

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