$\sigma$-ring of subsets of $X$ I am reading Kenneth Hoffman's book on Banach spaces of analytics functions. In the first chapter Measure and integration the very first paragraph starts with this:
If $X$ is a set, the collection of all subsets of $X$ forms a ring under the following operation:
$$A+B=(A\cup B)-(A\cap B),$$
$$AB=A\cap B.$$
A $\sigma$-ring of subsets of $X$ is a subring of the ring of all subset of $X$ which is closed under formation of countable unions (and, a fortiori, closed under countable intersection).
I am finding it difficult to understand. Why the "closed under countable intersection" is written in bracket. Does it mean this already follows from first two conditions and closed under formation of countable unions? If not, then why it is written in bracket? If yes, how to do it?
Thanks in advance.
Edit: 
 A: From $A+B=(A\setminus B)\cup (B\setminus A)$, you get $A + (A\cap B)=A\setminus B$. So a ring is closed under difference of sets. Also $(A+B)+(A\cap B)=A\cup B$, so a ring is closed under finite unions. Conversely, a collection of sets that is closed under  differences and finite unions is a ring: for example $A\setminus(A\setminus B)=A\cap B$.

*

*A $\sigma$-ring is a ring that is closed under countable unions.


*A $\delta$-ring is a ring that is closed under countable intersections.
If $\mathcal{R}$ is a $\sigma$-ring and $(A_n:n\in\mathbb{N})\subset\mathcal{R}$, then
$$A_1\setminus\bigcup_n(A_1\setminus A_n)\in\mathcal{R}$$
Notice that
$$A_1\setminus\bigcup_n(A_1\setminus A_n)=A_1\cap\bigcap_n(A^c_1\cup A_n)=\bigcap_n(A_1\cap(A^c_1\cup A_n)=\bigcap_nA_1\cap A_n=\bigcap_nA_n$$
That is, a $\sigma$-ring is also a $\delta$-ring. The converse is not necessarily true. For example, the collection of all Borel sets in the real line that have finite Lebesgue measure is a $\delta$-ring, but not a $\sigma$-ring.
