Way to explicitly give a sigma algebra Let $\Omega =\{ w_1,w_2,w_3,w_4,w_5,w_6,w_7\}$.
Is there a smart way to explicitly state the $\sigma$-algebra $\Sigma\subset \mathcal{P}(\Omega)$ containing $\{\{ w_1,w_2\},\{w_3\},\{w_4,w_5\},\{w_6,w_7\}\}$?
I've tried doing the combinations in my head and writing it out: $\Sigma = \{\{ w_1,w_2\},\{w_3\},\{w_4,w_5\},\{w_6,w_7\}, \{ w_1,w_2,w_3,w_4,w_5\},\{w_1,w_2,w_3\},\{w_1,w_2,w_4,w_5\},\{w_3,w_4,w_5\},\{w_3,w_6,w_7\},... \}$,
 but considering I have to consider all unions and complements it doesn't seem to be the ideal way to do this. Is there some sort of program that can do it for me?
Or maybe since I can't split $\{w_1,w_2\}$ into $\{w_1\},\{w_2\}$, there is a way to write $\Sigma = \mathcal{P}(\Omega)\setminus \{...\} ?$
 A: In general, if $\mathcal{C}$ is a finite or countable partition of $\Omega$, then $\Sigma = \sigma(C)$ consists of all unions of sets from $\mathcal{C}$:
$$\sigma(\mathcal{C}) = \left\{\bigcup \mathcal{C}' : \mathcal{C}' \subseteq \mathcal{C}\right\}$$ This is a nice exercise to prove.  Moreover, since the sets from $\mathcal{C}$ were pairwise disjoint, we have $\bigcup \mathcal{C}' = \bigcup \mathcal{C}''$ iff $\mathcal{C}' = \mathcal{C}''$.  So you can simply enumerate all the subsets of $\mathcal{C}$ and take the union of each one.
Here, $\mathcal{C} = \{\{ w_1,w_2\},\{w_3\},\{w_4,w_5\},\{w_6,w_7\}\}$, so $|\mathcal{C}| = 4$.  It has $2^4 = 16$ subsets and so $\Sigma$ will contain 16 distinct sets.
A simple way to enumerate all the subsets of a set is to count in binary.  Here I associate a binary number between 0000 and 1111 with a subset of $\mathcal{C}$, where each bit tells you whether or not to include the corresponding set.  I will think of the most-significant bit as corresponding to the set $\{\omega_1, \omega_2\}$, the second-most significant bit as corresponding to $\{\omega_3\}$, and so on.


*

*0000: $\emptyset$

*0001: $\{\omega_6, \omega_7\}$

*0010: $\{\omega_4, \omega_5\}$

*0011: $\{\omega_4, \omega_5, \omega_6, \omega_7\}$

*...

