Let $\langle A_n:n\in\Bbb N\rangle$ be a sequence in $X$,$\lim\sup A_n=\bigcap_{n\in N}\bigcup_{k>n} A_k,$,can we conclude that $A_n\subset lim\sup A_n$ when n tends to $\infty$? If not,what,s the relationship between $A_n$ and $lim\sup A_n$ when $n$ is large enough.
Upper limit of a sequence of sets
analysiscalculuslimitslimsup-and-liminfreal-analysis
Related Solutions
(1) Your proof that $\liminf A_n\subseteq \limsup A_n$ is correct.
Exercise 1: Under what conditions does equality hold in the above inclusion, i.e., under what conditions is it true that $\liminf A_n=\limsup A_n$?
(2) Note that $x\in \liminf A_n$ if and only if there exists a positive integer $N$ such that $x\in A_n$ for all $n\geq N$ if and only if there exists a positive integer $N$ such that $x\in \bigcap_{k=N}^{\infty} A_k$ if and only if $x\in \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} A_k$.
I will leave the remaining questions as easy exercises (with similar solutions).
Exercise 2: Let $\{A_n\}$ be a sequence of measurable subsets of a measurable space $(X,\mu)$. Assume that $\Sigma_{n=1}^{\infty} \mu(A_n)<\infty$. Let $A=\limsup A_n$. Prove that $\mu(A)=0$. (Hint: Use the fourth assertion in your question, namely, use the characterization of $\limsup A_n$ in your question.)
Exercise 3: Give an example (in the context of Exercise 2) where $\lim_{n\to\infty} \mu(A_n)=0$ but that $\mu(A)>0$. Do not assume that $\Sigma_{n=1}^{\infty} \mu(A_n)<\infty$. (Hint: let $\{A_n\}$ be an appropriate sequence of intervals in $[0,1]$, for example.)
Sure, these all have relatively simple interpretations:
- $\sup A_k$: Union. It's the set of all points that are in $A_k$ for at least one $k$.
- $\inf A_k$: Intersection. It's the set of all points that are $A_k$ for all $k$.
- $\limsup A_k$: This is the set of points $x$ such that $x$ is in infinitely many $A_k$.
- $\liminf A_k$: This is the set of points $x$ such that $x$ is in all but finitely many $A_k$.
In general,
$$\liminf A_k \subseteq \limsup A_k.$$
This is because if $x \in A_k$ for all but finitely many $k$, then it must be in $A_k$ for infinitely many $k$. However, $x$ could be in $A_k$ for all even $k$ (infinitely many $k$) but not in $A_k$ for all odd $k$. Then $x \in \limsup A_k$ but $x \notin \liminf A_k$.
$$A := \lim A_k \text{ if } \liminf A_k = A = \limsup A_k.$$
This is a strong statement. Suppose $\lim A_k$ exists. Then if $x \in A_k$ for infinitely many $k$, there exist only finitely many $k'$ such that $x \notin A_{k'}$. In other words, there exists a $K$ such that $x \in A_k$ for all $k \geq K$.
Best Answer
Put $A_{2n}:= \{0,n\}$ and $A_{2n+1}:=\{n\}$. Then $\limsup A_n = \{0\}$ but there is no relation between $A_n$ and $\limsup A_n$ in the sense of inclusion. In $\limsup A_n$ you get exactly those elements which are contained in infinitely many $A_n$'s so you have enough freedom to destroy any strict relation in this direction.