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 $\ds E \subseteq \liminf_{n \mathop \to \infty} E_n$.
$\Box$
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 $n \notin E_{n + 1}$.
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:
- $\ds \limsup_{n \mathop \to \infty} E_n \subseteq E$
$\Box$
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:
- $\ds \lim_{n \mathop \to \infty} E_n = \bigcup_{k \mathop \in \N} {E_k}$
$\blacksquare$
Sources
- 1951: J.C. Burkill: The Lebesgue Integral ... (previous) ... (next): Chapter $\text {I}$: Sets of Points: $1 \cdot 1$. The algebra of sets