Let $\mathcal{B}$ be a Boolean algebra, $X$ the set of ultrafilters of $\mathcal{B}$ and $\sigma:\mathcal{B}\longrightarrow\mathcal{P}(X)$ the map sending $b\in\mathcal{B}$ to the set of ultrafilters $\{\mathcal{U}:b\in\mathcal{U}\}$. I want to show that $\sigma$ is injective.
I have already proved that if $x\nleq y$ then there exists $\mathcal{U}$ such that $x\in\mathcal{U},y\notin\mathcal{U}$. Hence
$$(1) \qquad x\nleq y\Rightarrow \sigma(x)\nsubseteq\sigma(y)$$
Do (1) imply: $x\neq y\Rightarrow\sigma(x)\neq\sigma(y)$ ?
Yes, it does. If $x\ne y$, then at least one of $x\not\le y$ and $y\not\le x$ must hold, so $(1)$ implies that $\sigma(x)\nsubseteq\sigma(y)$ or $\sigma(y)\nsubseteq\sigma(x)$, and in either case $\sigma(x)\ne\sigma(y)$.