This is a question on what is written here.
If $A_i$ is a sequence of sets, define$$\liminf_i A_i = \bigcup_{j = 1}^\infty \bigcap_{i = j}^\infty A_i, \quad \limsup_i A_i = \bigcap_{j = 1}^\infty \bigcup_{i = j}^\infty A_i.$$What is an example where $\liminf_i A_i \neq \limsup_i A_i$?
Take $A_{2i} = \emptyset$ and $A_{2i+1} = \{1\}$. Why is it that the $\liminf$ is $\emptyset$ while the $\limsup$ is $\{1\}$? How do I see these two things?
For any $n$, $A_{2n} \cap A_{2n+1} = \emptyset$ and $A_{2n}\cup A_{2n+1}=\{1\}$.
So $\cap_{i\geq j}A_i\subset A_{i}\cap A_{i+1}=\emptyset$ hence $\cap_{i\geq j}A_i=\emptyset$ so $\cup_{j\geq1} \cap_{i\geq j}A_i=\cup_{j\geq1}\emptyset= \emptyset$.
Again, $\cup_{i\geq j}A_i\supset A_i\cup A_{i+1}=\{1\}$ and since each $A_i$ is either $\emptyset$ or $\{1\}$ we have that $\cup_{i\geq j}A_i=\{1\}$ Thus $\cap_{j\geq1}\cup_{i\geq j}A_i=\cap_{j\geq1}\{1\}=\{1\}$.