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

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?**

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.