When a algebra is a sigma algebra

Question

I'm trying to prove:

Giving $X$ a non-empty set. Let $A \subset P(X)$, we say that $A$ it's a algebra if and only if

1. $A$ it's a ring of sets
2. $X \in A$

Prove that; if $A \subset P(X)$ it's a algebra such that for every succesion of subsets disjoint $(E_n)_{n \in N}$ with every $E_n \in A$ we have $\cup_{n=1}^{\infty} E_n \in A$. Then A it's a $\sigma$-algebra.

I already have that $X, \varnothing \in A$ and for every $E \in A, E^c \in A$. Just don't see why if $(E_n)_{n \in N} \in A$ then $\cup_{n=1}^\infty E_n \in A$

Thank you.

Answer 1

Hint: Let $E_n \in \mathcal A$ for all $n$. Let $F_1=E_1,F_n=E_n\setminus ( E_1 \cup ...\cup E_{n-1})$ for $n \geq 2$. Then $F_n \in \mathcal A$,$F_n$'s are disjoint. Now verify that $\cup_n E_n =\cup_n F_n$.