does taking limsup preserve inclusion relation? Let $\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty}$ be sequences of sets with no further structure at that point, such that $a_n \subset b_n$ for every $n\in \mathbb{N}$, does it holds that also: $\limsup_{n\rightarrow\infty} a_n \subset \limsup_{n\rightarrow\infty} b_n $ ?  
would endowing any structure on those sets may changes this result?
 A: If $x\in\limsup_na_n$, then by definition $\{n\in\Bbb N:x\in a_n\}$ is infinite, so $\{n\in\Bbb N:x\in b_n\}$ is infinite, and $x\in\limsup_nb_n$; thus, $\limsup_na_n\subseteq\limsup_nb_n$. 
Added: This is in my opinion the most intuitive way to think about the limit superior of a sequence of sets: a point is in $\limsup_nA_n$ if and only if $x$ is frequently in $A_n$, meaning that $x$ is in infinitely many of the sets $A_n$. Similarly, $x\in\liminf_nA_n$ if and only if $x$ is eventually in $A_n$, meaning that $x$ is in all but (at most) finitely many of the sets $A_n$.
A: The limes superior for sequences of sets is defined as $$
  \limsup_{n->\infty} a_n = \bigcap_{i\in\mathbb{N}} \bigcup_{j=i}^\infty a_i
$$
If $a_i \subseteq b_i$, then $\bigcup_{j=i}^\infty a_i \subseteq \bigcup_{j=i}^\infty b_i$ and thus also $\limsup_{i\to\infty} a_i \subseteq \limsup_{i\to\infty} b_i$.
Note, however, that strict subsets can become non-strict subsets, i.e. if $a_i \subset b_i$, $a_i \neq b_i$ then it does not follow that $\limsup_{i\to\infty} a_i \neq \limsup_{i\to\infty} b_i$.
