Given $A_n$ $=$ {$w$$|$$0$ $\le$$w$$\le$$1$$-$$\frac{1}{n}$}
Find
$\limsup_{n \to \infty}$$A_n$ $\text{and}$ $\liminf_{n \to \infty}$$A_n$.
Can anyone guide me on how to solve this question?
I know that $\liminf_{n \to \infty}$$A_n$$=$$$
\bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n$$
and
$\limsup_{n \to \infty}$$A_n$$=$ $$
\bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n
$$
$A_1$$=$${0}$
$A_2$$=$$[$$0$,$\frac{1}{2}$$]$
$A_3$$=$$[$$0$,$\frac{2}{3}$$]$
$A_4$$=$$[$$0$,$\frac{3}{4}$$]$
and when n goes to infinity, $A_n$ is approaching 1.
is
$\liminf_{n \to \infty}$$A_n$ equals to the union of all the intersection of $A_m$ from m equal to n till infinity? which means I will get [0,1)?
Please correct me, thanks.
Best Answer
Note that $\lim \inf_{n \rightarrow \infty} A_n = \bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n$ can also be interpreted as follows: this is the set of $x$ such that $x$ is eventually in every $A_n$ (if the left $N$ is that $x$ is in the union of the intersections of all $A_n$ with $n \ge N$, this defines this tail (eventually = all but finitely many)).
Also $\lim \sup_{n \rightarrow \infty} A_n = \bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n$ can be seen as all $x$ such that $x$ is in infinitely many $A_n$ (i.e. the set $\{n \in \mathbb{N}: x \in A_n \}$ is infinite). This makes it clear that the lim inf is smaller than the limsup in general (being in almost all $A_n$ implies being in infinitely many of them).
Now look at your $A_n = \{x: 0 \le x \le 1-\frac{1}{n} \}$. Its clear that these sets are increasing for increasing $n$, so if a set is in one $A_n$, it will be in all later ones. So here the lim sup and lim inf will be equal to each other (being in infinitely many is equivalent to being in one, which is equivalent to being in almost all of them). And as every $x < 1$ will eventually be smaller than some $1 - \frac{1}{n}$, all of $[0,1)$ is in these coinciding limits, and no other points are even in one of them.