lim sup and lim inf of sequence of sets.

Question

I was wondering if someone would be so kind to provide a very simple explanation of lim sup and lim inf of s sequence of sets. For a sequence of subsets $A_n$ of a set $X$, the $\limsup A_n= \cap_{N=1}^\infty ( \cup_{n\ge N} A_n )$ and $\liminf A_n = \cup_{N=1}^\infty (\cap_{n \ge N} A_n)$. But i am having a hard time imagining what that really means unions of intersections and intersections of unions i think maybe causing the trouble. I read the version on wikipedia but that did n't resolve this either. Any help would be much appreciated.

Answer 1

A member of $$ \bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n $$ is a member of at least one of the sets $$ \bigcap_{n\ge N} A_n, $$ meaning it's a member of either $A_1\cap A_2 \cap A_3 \cap \cdots$ or $A_2\cap A_3 \cap A_4 \cap \cdots$ or $A_3\cap A_4 \cap A_5 \cap \cdots$ or $A_4\cap A_5 \cap A_6 \cap \cdots$ or $\ldots$ etc. That means it's a member of all except finitely many of the $A$.

A member of $$ \bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n $$ is a member of all of the sets $$ \bigcup_{n\ge N} A_n, $$ so it's a member of $A_1\cup A_2 \cup A_3 \cup \cdots$ and of $A_2\cup A_3 \cup A_4 \cup \cdots$ and of $A_3\cup A_4 \cup A_5 \cup \cdots$ and of $A_4\cup A_5 \cup A_6 \cup \cdots$ and of $\ldots$ etc. That means no matter how far down the sequence you go, it's a member of at least one of the sets that come later. That means it's a member of infinitely many of them, but there might also be infinitely many that it does not belong to.

Answer 2

In terms of sets, we have the following interpretations:

• $\displaystyle x\in\bigcup_{i\in I} A_i$ means that $x$ is in at least one of the $A_i$ sets.
• $\displaystyle x\in\bigcap_{i\in I} A_i$ means that $x$ is in all of the $A_i$ sets.

So this means that

1. $\bigcap_{N\ge1}\bigcup_{n\ge N} A_n$ are all elements somewhere in $A_N,A_{N+1},A_{N+2},\dots$, no matter how large N is, which is logically equivalent to "being in infinitely many of the $A_i$ sets."
2. $\bigcup_{N\ge1}\bigcap_{n\ge N} A_n$ are all elements in every single one of $A_{N},A_{N+1},A_{N+2},\dots$ for some $N$. This is logically equivalent to every element being "in all but finitely many the $A_i$ sets."

Answer 3

Are you familiar with the real analysis definition of $$\limsup_{n\to\infty} x_n = \inf_{m\geq 0} \sup_{n\geq m} x_n~?$$

The same definition can be applied to any sequence of elements in a complete lattice. Now apply it to the power set $2^X$ of some base set $X$ with set inclusion as the partial order.