Partition of space I have a problem with the following exercise:
Let $\left(A_{k}\right)_{k=1...n}$ be a sequence of subsets of space $\Omega$. Introduce the notation $A^{0} = \Omega \setminus A$ and $A^{1}=A$.
For $\epsilon \in \{0,1\}^{n}$, we put $$A_{\epsilon} = \bigcap^{n}_{k=1} A_{k}^{\epsilon_{k}}.$$
1. Show that if $\epsilon$, $\eta\in \{0,1\}^{n}$ and $\epsilon \neq \eta$ then $A^{\epsilon}\cap A^{\eta} = \emptyset$.
2. Show that $\bigcup \{A_{\epsilon}:\epsilon \in \{0,1\}^{n}\} = \Omega$
3. Conclude that $\{A_{\epsilon}:\epsilon \in \{0,1\}^{n}\}$ is a partition of the space $\Omega$.  
I have succesfully proven part 1 of the exercise by considering $\epsilon$ and $\eta$ such that $\left(\exists i\in \{1,2,...,n\}\right) \left(\eta_{i} \neq \epsilon_{i}\right)$ 
. The second part seems intuitive but hard for me to write formally. I have tried a direct computation but it failed. I would be grateful for any hint that will point me in the right direction.
 A: To prove 2: Let $x \in \Omega$. For any $k$ in $\{1,\ldots,n\}$ define $\delta_k = 1$ iff $x \in A_k$, otherwise (when $x \notin A_k$), we define $\delta_k =0$.
Then $\delta=(\delta_0,\ldots, \delta_n) \in \{0,1\}^n$ and by construction $x \in A_\delta$ as defined.
So $x \in \bigcup \{A_\varepsilon: \varepsilon \in \{0,1\}^n \}$ as witnessed by taking $\varepsilon=\delta$. 
A: An attempt with a direct computation was a step in the right direction.  I suspect, however, you may have tried to do the computation for the general case.  Generally, what is recommended is first to try the computation on elementary special cases.
So, first, let 
$$n = 2, \quad \Omega = \{ \omega_1, \omega_2, \omega_3 \}, \quad
A_1 = \{\omega_1, \omega_2\}, \quad A_2 = \{\omega_2, \omega_3\},
$$
and try proving the required statements for this case by direct computation.
Once you are done, if the result is not yet instructive enough to enable you to "see" how to do the general case, try a slightly bigger special case; e.g., $n = 3$.
And so on.
