Probelm:Let $w_1\neq w_2$ in $\Omega$ and let $\mathcal{g}$ be a collection of subsets of $\Omega$ such that each $G \in \mathcal{g}$ contains either both of $w_1,w_2$ or neither of them. Let $\mathcal{F}=\sigma(g)$. Show that for every random variable $X$ defined on $(\Omega,\mathcal{F})$ we have $X(w_1)=X(w_2)$.
Solution: If we prove that every $A \in \sigma(\mathcal{g})$ contains either both of $w_1,w_2$ or neither of them, by contradiction if $X(w_1)=a$ and $X(w_2)\neq a$ then $\{w:X(w)\in \{a\}\}$ contains $w_1$ so it does contain $w_2$. But $X(w_2)\neq a$ so $w_2 \notin \{w:X(w)\in \{a\}\}$. Contradiction.
However, I couldn't prove that every $A \in \sigma(\mathcal{g})$ contains either both of $w_1,w_2$ or neither of them. Any help would be appriciated.
Hint: Let $$ \mathcal T=\{A\in \mathcal P(\Omega): \{\omega_1,\omega_2\}\subseteq A \text{ or } \{\omega_1,\omega_2\}\subseteq A^c\} $$ $\mathcal T$ consists of subsets which contain either both or neither of $\omega_1$ and $\omega_2$.
Show that $\mathcal T$ is a sigma algebra, and that $g\subseteq \mathcal T$.