I have to basically prove the following, but I am having a difficult time (i'll elaborate more in the problem). The problem starts below:
For any family $\{A_i\}_{i \in I}$ of subsets of a set $X$ and any subset $J \subseteq I$ of the index set, let
\begin{equation} A_J:={(x \in X \ | \ x \in A_i \quad \mathrm{whenever} \ i \in J \quad \mathrm{and} \ x \notin A_i \quad \mathrm{whenever} \ i \notin J) } \end{equation}
Question: Show that the family of sets $\displaystyle\{A_J\}_{J \in \mathcal{P}(I)}$ consists of pairwise disjoints subsets of $X$: \begin{align*} A_J \cap A_J = \emptyset \quad \mathrm{if} \quad J \neq K \end{align*} and \begin{align*} X = \displaystyle\bigcup_{J \in \mathcal{P}(I)} A_J \end{align*}
Attempt: To better understand this problem, I assigned certain elements in $X$, $I$, and so on. Here they are:
Let $X = \{a,b,c,d,e\}$, $I = \{1,2\}$, $\mathcal{P}(I) = \{ \emptyset, {1}, {2}, {(1,2)} \}$, also, let $J = \{1\}$ and $K = \{2\}$. We assign elements to the following sets: $A_1 = {a,b}$, $A_2 = {c,d}$.
Here's what I got for $A_{\{1\}} = {a,b}$. $A_{\{2\}} = {c,d}$. $A_{\{1,2\}} = \{a,b,c,d\}$ (not sure if this set is right).
Overall, my problem is in trying to show that $X = \displaystyle\bigcup_{J \in \mathcal{P}(I)}A_J$, I run into a problem with that specific example in showing that all $A_J$ is disjoint and is the union which makes up $X$. Maybe there is a typo or something off in the assumption. I would appreciate it if someone points me in the right direction.