Limits of the sum of two sets, and the sum of two limits of sets. If I have two sets $A,B$ such that $A=\bigcap_{n\geqslant1}A_n$ and $B=\bigcap_{n\geqslant1}B_n$ where $A_n\supset A_{n+1}$ and $B_n\supset B_{n+1}$ for all $n$, and if $A_n+B_n=\{a+b:a\in A_n,b\in B_n\}=C$ for all $n\geqslant1$ and some constant bounded set $C$: can I say that $A+B=\bigcap_{n\geqslant1}(A_n+B_n)=C$?  Clearly $A=\lim_{n\to\infty}A_n$ and $B=\lim_{n\to\infty}B_n$, but can I bring this limit outside?
 A: Sadly, not always. Let me elaborate.
Choose $A_n = (0,1)$ and then $A_n$ decrease to $(0,1)$, and also choose $B_n = \left\{q_i, n\leq i \right\}$ where $q:\mathbb{N}\to \mathbb{Q}\cap(0,1)$ is a bijective enumeration of the rational numbers in the interval (0,1). Note that $B_n$ decrease to the empty set.
Now, let's observe that $(0,1)+\left\{q_i, n\leq i \right\}=(0,2)$ for all $n\in \mathbb{N}$, where the sum of the sets is immediately included in $(0,2)$ and for any real $x\in (0,2)$ we will be able to write it as $x=z+q$ in infinite ways, where $q\in \mathbb{Q}\cap(0,1)$ and $z\in (0,1)$, so there must exist $i>n$ such that $x=z + q_i$ with $z\in (0,1)$ by cardinality.
However, $(0,1)+\emptyset = \emptyset$ and both expressions do not coincide.
Remark: one can also chose $B_n' = B_n \cup \{0\}$ and in that case $B_n'$ decrease to $\{0\}$ and $A_n+B_n'=(0,2)$ but $A+B=(0,1)$ which is a less pathological example; all of this can be proven similarly to what was proved above.
That being said, there is an inclusion which does hold.
Since $A\subseteq A_n$ and $B\subseteq B_n$ for every $n\in \mathbb{N}$, $A+B\subseteq A_n+B_n$ for every $n\in \mathbb{N}$ and then $A+B\subseteq C$.
We can then conclude that $A+B \subseteq C$ but the inclusion may be strict.
