Limsup of Set Sequence

Last updated Nov 1, 2022

Let $({A}{n}){n=1}^{\infty}$ be a Sequence of Sets. Then $$\limsup\limits_{n \to \infty} A_{n} = \bigcap\limits_{n \in \mathbb{N}} \bigcup\limits_{k \geq n} A_{k}.$$

1. Since $\bigcup\limits_{k \geq n} A_{k} \supset \bigcup\limits_{k \geq n+1} A_{k}$ for any $n \in \mathbb{N}$, $(\bigcup\limits_{k \geq n} A_{k}){n=1}^{\infty}$ is Non-Increasing Function and we can write $$\limsup\limits{n \to \infty} A_{n} = \lim\limits_{n \to \infty} \bigcup\limits_{k \geq n} A_{k}.$$
2. If $x \in \limsup\limits_{n \to \infty} A_{n}$ If and Only If for every $n \in \mathbb{N}$ there exists $k \geq n$, s.t. $x \in A_{k}$ by definition of Set Union and Set Intersection. Thus $$\limsup\limits_{n \to \infty} A_{n} = {x \in \bigcup\limits_{n \in \mathbb{N}} A_{n} : x \in A_{n} \text{ for infinitely many } n \in \mathbb{N}}.$$

• Set Union
• Set Intersection
• Limit of a Monotonic Set Sequence