Problem
For sequences of numbers, limit inferior and limit superior are defined as $\liminf (a_n):=\sup\{\inf\{a_k \mid k \ge n\}\}$ and $\limsup (a_n):=\inf\{\sup\{a_k \mid k \ge n\}\}$ respectively; for sequences of sets, they are defined as $\bigcup\limits_{n=1}^{\infty} \bigcap\limits_{k=n}^{\infty} A_k$ and $\bigcap\limits_{n=1}^{\infty} \bigcup\limits_{k=n}^{\infty} A_k$ respectively.
Why are they consistent?
Discussion
It suffices to find a relation between ‘<’ and ‘⊆’: ${x \le a} \subseteq {x \le b} \iff a \le b$.
Claim: $\bigcup\limits_{a \in A} \{x \le a\} = \{x \le \sup A\}$.
Proof:
The last step is due to the defintion of infimum (greatest lower bound).
With the above claim, one has
Hence, one can see that $\sup\inf \{a_k \mid k \ge n\} \le \inf\sup \{a_k \mid k \ge n\}$ and $\bigcup\limits_{n=1}^{\infty} \bigcap\limits_{k=n}^{\infty} \{x \le a_k\} \subseteq \bigcap\limits_{n=1}^{\infty} \bigcup\limits_{k=n}^{\infty} \{x \le a_k\}$ share something in common.