Confusion about definition of limit of infimum and supremum of a sequence of sets I'm confused by the definition of $\lim \inf A_n$ and $\lim \sup A_n$. From the definitions of $\inf A_k$ and $\sup A_k$, shouldn't I have 
$$\lim_{n\to\infty} \inf A_n = \bigcap_{n=1}^{\infty}A_n \quad\text{and}\quad\lim_{n\to\infty} \sup A_n = \bigcup_{n=1}^{\infty}A_n\;?$$
These formulas are taken from Section 1 of Sidney Resnick's A Probability Path textbook.

 A: That is not correct. Suppose that $A_1=\emptyset$ and that $A_n=\{1\}$ if $n>1$. Then$$\bigcap_{k=n}^\infty A_k=\begin{cases}\emptyset&\text{ if }n=1\\\{1\}&\text{ otherwise}\end{cases}$$and therefore$$\liminf_{n\to\infty}A_n=\{1\}.$$But$$\bigcap_{n=1}^\infty A_n=\emptyset.$$
A: Obviously, $$\inf_{k\geq n}A_k=\bigcap_{k=n}^\infty A_k$$ is an increasing sequence of sets, ordered by inclusion.  (As $n$ increases, we take the intersection of fewer sets.)  With your proposed definition, we would take the smallest of those sets as the lim inf.  Doesn't seem reasonable, does it?  
A: No, you are misunderstanding this definition (which, admittedly, could benefit from some more explanation). Note that for real sequences $(a_n)$ you have
$$ \liminf_{n\to\infty} a_n= \sup_{n\ge1}\inf_{k\ge n} a_k$$
and
$$ \limsup_{n\to\infty} a_n= \inf_{n\ge1}\sup_{k\ge n} a_k.$$
Using these formulas analogously with the definitions of infima and suprema of sequences of sets as in the picture leads to the desired identities, and should also make clear why your initial idea is wrong.
