Is it true that if $$A \subseteq \bigcup_{n=1}^{\infty}A_n$$ then $$ \sup A - \inf A \le \sum_{n=1}^{\infty} \sup A_n - \sum_{n=1}^{\infty} \inf A_n \quad (\star)$$
I suppose that it is true (I need it to other proof). I suppose that we have $$\sup A \le \sup \left( \bigcup_{n=1}^{\infty}A_n \right) \le \sum_{n=1}^{\infty} \sup A_n$$ What can I say about infimum? We have $$\inf A \ge \inf \left( \bigcup_{n=1}^{\infty}A_n \right)$$ but of course $$\inf \left( \bigcup_{n=1}^{\infty}A_n \right) \le \sum_{n=1}^{\infty} \sup A_n $$
Could you give me some hints how can I get $(\star)$?
Take $A = \{{0,1,10,11\}}$, $A_1 = \{{0,1\}}$, $A_2 =\{{10,11\}}$, $A_3 = A_4 = \dots = \{{0\}}$. Then the LHS of $(\star)$ is $11$ and the RHS is $2$, so $(\star)$ is not satisfied.