Liminf of Set Sequence

Last updated Nov 1, 2022

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

1. Since $\bigcap\limits_{k \geq n} A_{k} \subset \bigcap\limits_{k \geq n+1} A_{k}$ for any $n \in \mathbb{N}$, $(\bigcap\limits_{k \geq n} A_{k}){n=1}^{\infty}$ is Non-Decreasing Function and we can write $$\liminf\limits{n \to \infty} A_{n} = \lim\limits_{n \to \infty} \bigcap\limits_{k \geq n} A_{k}.$$
2. $x \in \liminf\limits_{n \to \infty} A_{n}$ If and Only If there exists an $n \in \mathbb{N}$ so that $\forall k \geq n$, $x \in A_{k}$ by definition of Set Union and Set Intersection. Thus $$\liminf\limits_{n \to \infty} A_{n} = {x \in \bigcup\limits_{n \in \mathbb{N}} A_{n} : x \not\in A_{n} \text{ for finitely many } n \in \mathbb{N}}$$

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