# Showing a collection of sets is pairwise disjoint

$\bf Homework \; Problem.$ Suppose we have $A_1 \subseteq A_2 \subseteq A_3 \subseteq ...$. Also, let us make $A_0 = \varnothing, B_1 = A_1, B_2 = A_2 \cap A_1^c, ... B_n = A_n \cap A_{n-1}^c$. Show that $(B_n)$ is pairwise disjoint and also show $A_n = \bigcup_{k=1}^n B_k$.

Attempt:

by definition, if I can show $B_i \cap B_j = \varnothing$ for $i \neq j$, then Im done. So, we have

$$B_i \cap B_j = A_i \cap A^c_{i-1} \cap A_j \cap A^c_{j-1}$$

If $i > j$, then $A_j \subseteq A_i$ and this implies $A_j \cap A_i = A_j$. So far we have

$$B_i \cap B_j = A_j \cap A^c_{i-1} \cap A^c_{j-1}$$

Now, we know $A_{j-1} \subseteq A_j$ and $A_{i-1} \subseteq A_i$. by simple reasoning, we must have that $A_{j-1} \subseteq A_{i-1}$ and so $A_{i-1}^c \subseteq A_{j-1}^c$ which means that $A^c_{i-1} \cap A^c_{j-1} = A^c_{i-1}$. Thus, we now have that

$$B_i \cap B_j = A_j \cap A_{i-1}^c$$

Now, since $i > j$, it must be the case that $j=i-1$ in which case we obtain the desired result, otherwise $i-1 > j$ and so $A_j \subseteq A_{i-1}$ in which case it is evident that $A_j \cap A_{i-1}^c = \varnothing$ and pairwise disjointness of the sequence of sets follows. Next, we can show induction to show the equality. For $n=1$ we have $A_1 = B_1$. Assume the result is true for some $n$, then

$$\bigcup_{k=1}^{n+1} B_k = \bigcup_{k=1}^n B_k \cup B_{n+1} = A_n \cup B_{n+1}$$

And since $B_{n+1} = A_{n+1} \cap A_n^c$, then

$$\bigcup_{k=1}^{n+1} B_k =A_n \cup B_{n+1} = A_n \cup (A_{n+1} \cap A_n^c) = (A_n \cup A_{n+1} ) \cap(A_n \cup A_n^c)$$

Since $A_n \cup A_n^c$ is the whole then we only have $A_n \cup A_{n+1}$ which becomes just $A_{n+1}$ and the result follows.

My question is: Is this correct? am I overcomplicating this problem or this is correct way to do it?

$$B_i \cap B_j = A_i \cap A^c_{i-1} \cap A_j \cap A^c_{j-1}$$
The middle terms $A^c_{i-1} \cap A_j$ already give you $\emptyset$, since $j<i$ implies $A_j\subset A_{i-1}$.