Alternative subsets, whose sigma-algebra create the powerset of X So I've been stuck on this problem for a while now without any real progress and would appreciate some advice.
The problem is: Given the set X = {1,2,3,4,5,6,7} and the subset G = {{1},{2},{3},{4},{5},{6},{7}} of X, find an example of subsets A, B, C of X with $\sigma$({A,B,C})=$\sigma$(G). It is known that $\sigma$(G)=P(x).
I feel like there should be some obvious way to figure out the subsets A, B and C with the knowledge that $\sigma$({A,B,C})=$\sigma$(G)=P(X), but honestly I'm stuck to just guessing and testing different subsets of X right now.
Again, I would really appreaciate some advice/hints/techniques as to get me moving in the right direction of creating the right subsets. Surely there must be another way than just guessing? 
 A: For example,
\begin{align}
A &:= \{1, 2, 3, 5\}, \\
B &:= \{1, 4, 5, 7\}, \\
C &:= \{3, 4, 5, 6\}.
\end{align}
Indeed,
$$
\begin{align}
A \cap B & = \{1,5\} \\
\{1, 5\} \cap C &= \{5\} \\
\{1, 5\} \setminus C &= \{1\} \\
A \setminus \{1, 5\} &= \{2, 3\} \\
\{2, 3\} \cap C &= \{3\} \\
\{2, 3\} \setminus C &= \{2\} \\
B \setminus \{1, 5\} &= \{4, 7\} \\
\{4, 7\} \cap C &= \{4\} \\
\{4, 7\} \setminus C &= \{7\} \\
C \setminus \left(\{3\} \cup \{4\} \cup \{5\}\right) &= \{6\}
\end{align}
$$
The trick is to make sure that $A\cup B \cup C = \{1, 2, \dots, 7\}$, and that every pair
$$
\{i, j\} \subseteq \{1, 2, \dots, 7\},\ \#\{i,j\} = 2
$$
is separable by the triple  $\mathcal{T} = \{A, B, C\}$  in the sense that if $\{i, j\} \subseteq X$  for some  $X \in \mathcal{T}$, then there is some $Y \in \mathcal{T}$ such that $\#\left(\{i \cap j\} \cap Y\right) = 1$. There are several ways to accomplish this. I just played around a little until I found the example above.
Here's a step-by-step construction of another possible solution.


*

*Set
$$
 \begin{align}
 A &:= \{1, 2\}, \\
 B &:= \{3, 4\}, \\
 C &:= \{5, 6, 7\}.
 \end{align}
 $$
This makes sure that $A\cup B\cup C = \{1, 2, \dots, 7\}$.

*We have $\{1, 2\} \subseteq A$. So we'll add $1$ to $B$, so that $\#\left(\{1, 2\} \cap B\right) = 1$. (We could've just as well added $1$ to $C$, or added $2$ to either $B$ or $C$.)
$$
 \begin{align}
 A &:= \{1, 2\}, \\
 B &:= \{1, 3, 4\}, \\
 C &:= \{5, 6, 7\}.
 \end{align}
 $$

*We have $\{1, 3\} \subseteq B$, so we'll add $3$ to $C$, so that $\#\left(\{1, 3\} \cap C\right) = 1$. (We could've just as well added $3$ to $C$, but adding $3$ to $A$ wouldn't do. You see why?)
$$
 \begin{align}
 A &:= \{1, 2\}, \\
 B &:= \{1, 3, 4\}, \\
 C &:= \{3, 5, 6, 7\}.
 \end{align}
 $$

*We have $\{1, 4\} \subseteq B$, but $\#\left(\{1, 3\} \cap A\right) = 1$, so that's cool. We have $\{3, 4\} \subseteq B$, but $\#\left(\{3, 4\} \cap C\right) = 1$, so that's cool.

*We have $\{3, 5\} \subseteq C$, but $\#\left(\{3, 5\} \cap B\right) = 1$, so that's cool. Likewise $\{3, 6\}$ and $\{3, 7\}$.

*We have $\{5, 6\} \subseteq C$. We'll add $5$ to $A$ so that $\#\left(\{5, 6\} \cap A\right) = 1$. Now $\{5, 7\}$ is taken care of too.
$$
 \begin{align}
 A &:= \{1, 2, 5\}, \\
 B &:= \{1, 3, 4\}, \\
 C &:= \{3, 5, 6, 7\}.
 \end{align}
 $$

*We have $\{6, 7\} \subseteq C$, so we'll add $6$ to $B$ so that $\#\left(\{6, 7\} \cap B\right) = 1$.
$$
 \begin{align}
 A &:= \{1, 2, 5\}, \\
 B &:= \{1, 3, 4, 6\}, \\
 C &:= \{3, 5, 6, 7\}.
 \end{align}
 $$

*Now $\{3, 6\}$ have become non-separable, so we'll add $3$ to $A$.
$$
 \begin{align}
 A &:= \{1, 2, 3, 5\}, \\
 B &:= \{1, 3, 4, 6\}, \\
 C &:= \{3, 5, 6, 7\},
 \end{align}
 $$
and we have arrived at yet another solution (try to show that this is indeed a solution to the problem).
A: This post is a continuation of my other post. In it I will prove that the separability property I mentioned in the other post guarantees a solution.
I will first introduce some notation. Given any set $S$, any subset $T$ thereof, and any collection, $\mathcal{K}$, of subsets of $S$, denote by $\alpha(\mathcal{K}, T)$ the collection of all $E \in \mathcal{K}$ such that $T \subseteq E$, and denote by $\beta(\mathcal{K}, T)$ the collection of all $E \in \mathcal{K}$ such that $\#(E \cap T) = 1$.
Then the following holds.

Let $\Omega$ be a countable set, and let $\mathcal{C}$ be a collection of finite subsets of $\Omega$ that covers $\Omega$. Then $\sigma(\mathcal{C}) = \mathcal{P}\Omega$ iff every $\{a, b\} \subseteq \Omega$ with $a\neq b$ is $\mathcal{C}$-separable in the sense that $\alpha(\mathcal{C}, \{a,b\}) \neq \emptyset$ implies $\beta(\mathcal{C}, \{a,b\}) \neq \emptyset$.

Proof


*

*Firstly suppose that every $\{a, b\} \subseteq \Omega$ with $a\neq b$ is $\mathcal{C}$-separable. We will show that $\sigma(\mathcal{C}) = \mathcal{P}\Omega$.
By $\Omega$'s countability, it suffices to show that, for every $\omega \in \Omega$,
$$
 \{\omega\} \in \sigma(\mathcal{C}). \tag{1}\label{Apple}
 $$
Let $\omega \in \Omega$, and define
$$
 A := \bigcap \alpha\left(\sigma(\mathcal{C}), \{\omega\}\right). \tag{2}\label{Banana}
 $$
Since $\{\omega\}\subseteq A$, showing \eqref{Apple} is tantamount to showing that
$$
 A \subseteq \{\omega\} \tag{3}\label{Cucumber}
 $$
and that
$$
 A \in \sigma(\mathcal{C}). \tag{4}\label{Dill}
 $$
To show \eqref{Dill}, choose any $B \in \alpha\left(\mathcal{C},\{\omega\}\right)$. Then
$$
 A = \bigcap_{D \in \alpha\left(\sigma(\mathcal{C}),\{\omega\}\right)} (B \cap D).
 $$
Since $B$ is finite, this is a finite intersection. Additionally, every set in this intersection belongs to $\sigma(\mathcal{C})$.
Now, suppose \eqref{Cucumber} is false. We will obtain a contradiction. Choose any $\omega' \in A\setminus\{\omega\}$, and any $D \in \alpha\left(\mathcal{C},\{\omega\}\right)$. Then $\{\omega, \omega'\} \subseteq A \subseteq D$, so that $\alpha\left(\mathcal{C},\{\omega,\omega'\}\right) \neq \emptyset$. Hence, by assumption, $\beta\left(\mathcal{C}, \{\omega,\omega'\}\right) \neq \emptyset$. Choose some $B \in \beta\left(\mathcal{C}, \{\omega, \omega'\}\right)$. If $\omega \in B$, then $\omega' \notin B$, and we have $A\cap B \in \alpha\left(\sigma(\mathcal{C}),\{\omega\}\right)$ and $A\cap B \subsetneq A$, in contradiction to \eqref{Banana}. So we must have $\omega \notin B$ and $\omega' \in B$, but then $A\setminus B \in \alpha\left(\sigma(\mathcal{C}),\{\omega\}\right)$ and $A\setminus B \subsetneq A$, in contradiction to \eqref{Banana}.

*Suppose that it is not the case that every $\{a, b\} \subseteq \Omega$ with $a\neq b$ is $\mathcal{C}$-separable. We will show that $\sigma(\mathcal{C}) \neq \mathcal{P}\Omega$.
Let $\{a,b\} \subseteq \Omega$ with $a\neq b$ be such that $\alpha(\mathcal{C}, \{a,b\}) \neq \emptyset$ and such that $\beta(\mathcal{C}, \{a,b\}) = \emptyset$. In other words, for every $A \in \mathcal{C}$, either $\{a,b\} \in A$, or $\{a,b\}\cap A = \emptyset$. We will see that $\{a\} \notin \sigma(\mathcal{C})$.
It suffices to show that, for every $A \in \sigma(\mathcal{C})$, either $\{a,b\} \subseteq A$ or $\{a,b\}\cap A = \emptyset$. Define
$$
 \mathcal{F} := \left\{A \in \sigma(\mathcal{C})\ |\!:\ \{a,b\} \subseteq A\text{ or }\{a,b\}\cap A = \emptyset\right\}.
 $$
Then $\mathcal{C} \subseteq \mathcal{F}$, and therefore it suffices to show that $\mathcal{F}$ is a $\sigma$-algebra.


*

*$\Omega \in \mathcal{F}$ since $\{a,b\} \subseteq \Omega$.

*Let $A \in \mathcal{F}$. If $\{a, b\} \subseteq A$, then $\{a,b\}\cap A^c = \emptyset$, which implies that $A^c \in \mathcal{F}$. Otherwise, $\{a, b\} \cap A = \emptyset$, and therefore $\{a, b\} \cap A^c = \emptyset$, which again implies that $A^c \in \mathcal{F}$.

*Let $A_1, A_2, \dots \in \mathcal{F}$. If, for some $n \in \{1, 2, \dots\}$, $\{a, b\} \subseteq A_n$, then $\{a, b\} \subseteq \cup_{k=1}^\infty A_k$, whence $\cup_{k=1}^\infty A_k$. Otherwise, for every $n \in \{1, 2, \dots\}$, $\{a,b\} \cap A_n = \emptyset$, whence
$$
 \{a, b\} \cap \cup_{n=1}^\infty A_n = \cup_{n=1}^\infty \left(\{a,b\} \cap A_n\right) = \emptyset,
 $$
so that, again, $\cup_{k=1}^\infty A_k$.
Q.E.D.
