This page is for testing MathJax in my blog. I wrote some custom shorthand like $\zeros \in \R^n$.

## Problem

For sequences of numbers, limit inferior and limit superior are defined as $\liminf (a_n):=\sup\{\inf\{a_k:k \ge n\}\}$ and $\limsup (a_n):=\inf\{\sup\{a_k:k \ge n\}\}$ respectively; for sequences of sets, they are defined as $\displaystyle \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} A_k$ and $\displaystyle \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} A_k$ 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: $\displaystyle \bigcup_{a \in A} \{x \le a\} = \{x \le \sup A\}$.

Proof:

\begin{aligned} & x \in \bigcup_{a \in A} \{x \le a\} \\ \iff& x \le a \;\forall a \in A \\ \iff& x \text{ is a lower bound of } A \\ \iff& x \le \inf A \end{aligned}

The last step is due to the defintion of infimum (greatest lower bound).

With the above claim, one has

\begin{aligned} & \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} \{x \le a_k\} \\ =& \bigcap_{n=1}^{\infty} \bigcup_{a \in \{a_k:k \ge n\}} \{x \le a\} \\ =& \bigcap_{n=1}^{\infty} \{ x \le \sup \{a_k:k \ge n\}\} \\ =& \{x \le \inf\sup \{a_k:k \ge n\}\} \end{aligned}

Hence, one can see that $\sup\inf \{a_k:k \ge n\} \le \inf\sup \{a_k:k \ge n\}$ and $\displaystyle \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} \{x \le a_k\} \subseteq \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} \{x \le a_k\}$ share something in common.