Limit superior and limit inferior of sets

Question

I have searched for the answer in wikipedia and math stackexchange. However, I do not have any background in real analysis and all the answers seem very complicated to me to understand. I am wondering if anyone can provide an intuitive answer or a graphical answer to visualize the concept easily.

Answer 1

If $(\mathcal P,\leq)$ is a lattice (i.e. a poset s.t. all pair of element $(a,b)$ has an infimum and a supremum), then for a sequence $(a_n)\in \mathcal P^{\mathbb N}$, $$\limsup_{n\to \infty }a_n:=\inf_{n\in\mathbb N}\sup_{k\geq n}a_ n\quad \text{and}\quad \liminf_{n\to \infty }a_n:=\sup_{n\in\mathbb N}\inf_{k\geq n}a_k.$$

Now, consider $(\mathcal P(\mathbb R),\subset )$ is a lattice and $$\inf(A,B):=A\cap B\quad \text{and}\quad \sup(A,B):=A\cup B,$$ for all $A,B\in \mathcal P(\mathbb R)$.

So, for a sequence $(A_n)\in \mathcal P(\mathbb R)^{\mathbb N}$, $$\inf_{k\geq n}A_k=\bigcap_{k\geq n}A_k,$$ and thus $$\liminf_{n\to \infty }A_n:=\sup_{n\in\mathbb N}\inf_{k\geq n}A_k=\bigcup_{n\in\mathbb N}\bigcap_{k\geq n}A_k.$$ And similarly, $$\limsup_{n\to \infty }A_n=\bigcap_{n\in\mathbb N}\bigcup_{k\geq n}A_k.$$