Union of infinite sets 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.
 A: No.
Take $A_i = \{ 1,2,...,i \} \subset B = \mathbb{N}$.
However, $A = \cup_i A_i = B$.
A: " i.e. there exists at least one element in B that is not in Ai for every Ai."
Not quite.  The element of $B$ that is not in $A_i$ could be an entirely different b that is not in $A_j$.  In fact you don't even need infinite sets.
Let $A_1 = \{1,3,5\}; A_2 = \{2, 4, 6\}; B = \{1,2,3,4,5,6\}$. $A = A_1 \cup A_2$, $A_i \subsetneq B$ but $A = B$
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With the extra condition $A_i \subset A_{i-1}$ the answer is not so trivial but again the elusive element that isn't in $A_i$ can keep slipping away.
$A_i = (0, i) \subsetneq (0, \infty)$ so $A = \cup_{i} A_i = (0, \infty)$.
Let $b_i \notin A_i$ but $b \in (0, \infty)$ there is some $A_j; j > i$ that $b \in A_j$.  And then the $b_2$ that isn't in $A_j$ is in some further $A_k$ and so on.  
There is no one element that is not in all.  Just for every $A_i$ there is some element not in it and for $A_j$ there is another.
A: Unless I am extremely confused, the assumption $A_i \subset A_{i-1}$ implies that, for example, $A_4$ is a subset of $A_3$. Then clearly, the principle of induction can be used to show that $\forall i, A_i \subset A_1,$ then $A=\cup_i A_i = A_1.$ Since by assumption $A_1 \subsetneq B$, we have $A\subsetneq B$. 
If the assumption was actually that each subsequent set in the sequence contains the previous, then the other answers show that in fact it is possible to have $A=B$. 
