A field generated by a countably infinite class. Let $\mathbb{A}$ be a countably class. Then the smallest field that contains $\mathbb{A}$ is countably.
How to prove it?
 A: We can give an explicit characterization of the field generated by a class $\mathcal C=\{A_i,i\in I$. Denote $\mathcal A$ the field (algebra) generated by $\mathcal C$. We have 
$$\mathcal A(\mathcal C)=\{\bigcup_{i\in F}\bigcap_{j\in J_i}B_j, F\mbox{ finite }, F\subset \mathbb N, J_i\subset I, J_i\subset I, B_j=A_k \mbox{ or }A_k^c\mbox{ for some }k\},$$
so in fact the generated algebra consists of finite unions of finite intersection of elements of $\mathcal C$ with their complement. The RHS, which will be call $\mathcal R$, contains $\mathcal C$, so we have to show that it's indeed an algebra. $\emptyset\in\mathcal R$, and we show that $\mathcal R$ is stable for finite intersections. Take $A_1:=\bigcup_{i\in F_1}\bigcap_{j\in J_i^{(1)}}B_j$ and $A_2:=\bigcup_{i\in F_2}\bigcap_{j\in J_i^{(2)}}B_j$; then 
$$A_1\cap A_2=\bigcup_{(i_1,i_2)\in F_1\times F_2}\bigcap_{j_1\in J_{i_1}^{(1)}}
\bigcap_{j_2\in J_{i_2}^{(2)}}B_{j_1}\cap B_{j_2}.$$
Since $F_1\times F_2$ is finite, and so are $J_{i_1}^{(1)},J_{i_2}^{(2)}$, $A_1\cap A_2$ can be rewritten as an element of $\mathcal R$. Now we have to show that $\mathcal R$ is stable by complementation. Let $A:=\bigcup_{i\in F}\bigcap_{j\in J_i}B_j\in\mathcal R$, so $A^c=\bigcap_{i\in F}\bigcup_{j\in J_i}B_j^c$ and for each $i$, $\bigcup_{j\in J_i}B_j^c\in\mathcal R$ so $A^c\in\mathcal R$.
Now, if $\mathcal C$ is infinite, of cardinal $|\mathcal C|$, since $|\mathcal C|=|\mathcal C\times\mathcal C|$, the set $\{A, A\in\mathcal C\}\cup \{A^c,A\in\mathcal C\}$ has the same cardinality as $\mathcal C$, and the collections of the finite intersections of such sets has also the same cardinality as $\mathcal C$. So the generated algebra has the same cardinality. 
Note that it's not true for $\sigma$-algebras. Indeed, take $\mathcal N$ the set of natural numbers, and consider $\mathcal A:=\{\{x\},x\in\mathbb N\}$. Then this class is countable, but the $\sigma$-algebra generated by $\mathcal A$ is the collection of all subsets of $\mathcal A$, which is not countable. 
