A sigma-algebra 'generated by closure under the operation of countable union' I'm reading a paper which contains the following:

We have the following notation. If R is a partition of [Omega], and w [in] [Omega], we
  denote by R(w) the unique member of R containing w. Also, F(R) will
  denote the sigma-field generated by closing R under the operation of countable
  union. 

This is the first time I have encountered sigma-fields, but after reading up I think I'm starting to get the general concept. However, I still don't understand what is meant by "generated by closing R under the operation of countable
union". From what I understand, being closed under the operation of countable union is a requirement for being a sigma-field. 
Am I missing something here? I have tried looking for similar examples but couldn't seem to find anything.
Thanks.
 A: Could perhaps be better written. Or maybe it's clear. In any case, what the author means is this: Let $F(R)$ be the closure of $R$ under countable unions. Then $F(R)$ is the same as the sigma-field generated by $R$.
At least that's what it seems to me he's asserting. If that's what he means, what he's saying is not quite true, unless $R$ is countable. The truth is this:
Let $F(R)$ be the class of countable or co-countable unions of elements of $R$. Then $F(R)$ is the sigma-algebra generated by $R$.
To be more explicit, my definition of $F(R)$ means that $E\in F(R)$ if and only if there exists $S\subset R$ such that either $S$ or $R\setminus S$ is countable, and $$E=\bigcup_{s\in S}s.$$
It's easy to see that this $F(R)$ is indeed a sigma-field. It's closed under complements because, since $R$ is a partition, $$\left(\bigcup_{s\in S}s\right)^c=\bigcup_{s\in R\setminus S}s.$$ And it's closed under countable unions: Say $E_j\in F(R)$. If all the $S_j$ are countable then their union is countable, while if one of the $S_j$ is co-countable then the union of all the $S_j$ is co-countable.
It's also clear that $R\subset F(R)$ and that any sigma-algebra containing $R$ must contain $F(R)$. So $F(R)$ is the sigma-algebra generated by $R$.
