A property of Dynkin system Given the Dynkin system with the property that if $A_1, A_2, A_3, \dots$ is a sequence of subsets in $B$ and $A_{n}\subset A_{n+1}$ for $n ≥ 1$, then ${\bigcup _{n=1}^{\infty }A_{n}\in B}$. Prove that this is equivalent to the property that if ${A_n}$ are disjoint sets, and $A_n\in B$, $\bigcup _{n=1}^{\infty }A_{n}\in B$.
My attempt: I use induction on $n$. The base case is with $n=2$, as $n=1$ is trivially true. Now, for $n=2$, since $A_1$ and $A_2$ are disjoint subsets of $B$, $A_{1}\cup (B-A_2)\in B$ (since $A_1\subset {(B-A_2)}\subset B$). By distribution law, $A_{1}\cup (B-A_2) = B-(A_{1}\cup A_2)\in B$ (since $A_1\subset B= B$). This implies $A_{1}\cup A_2\in B$, and we are done with the base case.
The rest follows easily since for the case $n=k+1$, we just substitute $Z = \bigcup_{n=1}^{k} A_n$ and by inductive hypothesis, $(B- Z)\in B$. Now use the same trick as with the base case with $B-Z$ and $A_{n+1}$, we are done.
Question Could anyone help verify whether my proof above is correct? If not, please help point out the mistake. 
 A: Let us first assume that for every sequence $A_i \in B$ such that $A_i \subset A_{i+1}$, it follows that $\displaystyle\bigcup_{i=1}^{\infty} A_i \in B$.
Let $X_i$ be a sequence of disjoint sets. Then, define the sequence $Y_n =\displaystyle\bigcup_{i=1}^n X_i$. Note that $Y_i \subset Y_{i+1}$, and by this property, for all $n$, $\displaystyle\bigcup_{i=1}^n Y_i =\displaystyle\bigcup_{i=1}^n X_i$.  By our assumption, $\displaystyle\bigcup_{i=1}^\infty Y_i \in B$. But then, taking $n \to \infty$, $\displaystyle\bigcup_{i=1}^\infty Y_i =\displaystyle\bigcup_{i=1}^\infty X_i$, whence  $\displaystyle\bigcup_{i=1}^\infty X_i \in B$.
For the other way round, we will assume that for every disjoint countable collection $X_i\in B$, we know that $\displaystyle\bigcup_{i=1}^\infty X_i \in B$.
Let $Y_i$ be a sequence of sets such that $Y_i \subset Y_{i+1}$. Then, define the sequence $Z_n =Y_{n+1} $\ $ Y_n$. Note that $Z_i \cap Z_j = \phi$ for all $i$ and $j$ not equal (because if, WLOG, $i>j$, then $Z_j \subset Y_i$ while $Z_i \in (Y_i)^c$),and for all $n$, $\displaystyle\bigcup_{i=1}^n Y_i =\displaystyle\bigcup_{i=1}^n Z_i$.  By our assumption, $\displaystyle\bigcup_{i=1}^\infty Z_i \in B$. But then, taking $n \to \infty$, $\displaystyle\bigcup_{i=1}^\infty Y_i =\displaystyle\bigcup_{i=1}^\infty Z_i$, whence  $\displaystyle\bigcup_{i=1}^\infty Y_i \in B$.
Hence the two properties are equivalent.
