Countable union of equivalence classes

Question

Let $(X, \mathbb A)$ be a measurable space. For any $x \in X$, we define $x* = \cap A$.

In addition, we have a binary relation ~ on X defined as: x ~ y <=> $\{ x \in A\} = \{y \in A\}$.

If $\mathbb A$ is countable:

(a) why is $x* \in \mathbb A$ for every $x \in X$?

(b) and why can $A \in \mathbb A$ be written as a countable union of equivalence classes of ~ ?

Answer 1

you may need to clarify that the intersection defining $x^*$ is taken over all $A\in \mathbb A$ that contain $x$.

That done, if $\mathbb A$ is countable, the intersection is a countable intersection of element of $\mathbb A$ thus, by definition of $\sigma$-algebra, is an element of $\mathbb A$.

You can check that $$ \forall A \in \mathbb A, \quad A = \bigcup_{x\in A} x^*$$ then you could check that $x^*$ is the equivalence classe of $x$ for $\sim$.

Putting these two things together, you can conclude.