Let $A = \cup_{i \ge 1} A_i$, $i = 1, 2, \cdots$. This is union of countably infinite sets.
Also, $A_i \subsetneq B$ for all $i$, i.e. for every $A_i$, there exists at least one element in $B$ that is not in $A_i$.
It is also the case that $A_{i-1} \subsetneq A_{i}$ for all $i > 1$.
Then is it true that $A \subsetneq B$?
Intuitively, it seems true because for every $A_i$, I can point to an element that is in $B$ but not in $A_i$. However, it is not clear to me what happens after you take union of countably infinite sets.
Best Answer
No.
Take $A_i = \{ 1,2,...,i \} \subset B = \mathbb{N}$.
However, $A = \cup_i A_i = B$.