Why is the lim inf the union of intersections For my statistics class we had elementary set theory.
It was stated that:
$$\inf_{k\geq n } A_k = \bigcap\limits_{k=n}^{\infty} A_k$$
and 
$$\sup_{k\geq n } A_k = \bigcup\limits_{k=n}^{\infty} A_k$$
From this was deduced that:
$$\lim\limits_{n\to\infty} \inf A_k = \bigcup\limits_{n=1}^{\infty} \bigcap\limits_{k=n}^{\infty} A_k$$
and 
$$\lim\limits_{n\to\infty} \sup A_k = \bigcap\limits_{n=1}^{\infty} \bigcup\limits_{k=n}^{\infty} A_k$$
I absolutely have no idea why. Could someone explain it to me in the least technical way possible? I neither get why the intersection of Ak from n onwards should be the infimum nor why the union of all intersections should be the limit of that infimum.

It is clear that for real numbers, e.g. $3$ and $8$, that the minimum is the smaller one, i.e. $3$. There is a way to talk about smaller sets: a set $A$ is "smaller" than a set $B$ if it is completely contained in it, i.e. $A\subseteq B$. This is natural somehow, as e.g. $\{1,2,3\}$ is smaller than $\{1,2,3,4,5,6\}$. In this case, check that the intersection gives
$$A\cap B=A.$$
This means that the intersection somehow acts like the minimum function on sets. By the same reasoning, you can argue that the union gives $A\cup B=B$ and acts like a maximum function. It is a bit less natural if neither $A\subseteq B$ nor $B\subseteq A$, but above should deal as a motivation.
Because $\inf$ and $\sup$ generalizes minimum and maximum for infinitely many numbers (and sets in our case), it is natural to define
$$\inf \,\{A_k\}=\bigcap A_k,\qquad \sup \,\{A_k\}=\bigcup A_k.$$
Now remember that $\liminf_{n\to\infty}$ was defined as $\sup_{n\in\Bbb N}\inf_{k\ge n}$. This hopefully explains why the formula for $\liminf$ contains both a union and an intersection.
