Measure space, definition of $C_\sigma$ I have read about a definition these days but found it's hard for me to understand. The definition is like this:
Let X be a set. For any set C of subsets of X, we write $C_\sigma =\{\bigcup_{k=1}^{\infty}A_k |$ each $A_k\in C\}$ and $C_\delta =\{\bigcap_{k=1}^{\infty}A_k |$ each $A_k\in C\}$. We have $C_{\sigma\sigma} =C_\sigma $ and $C_{\delta\delta}=C_\delta$
Suppose X is {1, 2,3}, and C is $\{ \emptyset, \{1\}, \{1,2\}\}$, can I know what is $C_\sigma$ and $C_\delta$ under this circumstance, and why $C_{\sigma\sigma} =C_\sigma $ and $C_{\delta\delta}=C_\delta$
?
Thanks.
 A: Given the $X$ and $C$, the set $C_{\sigma}$, in words, consists of all possible countable unions of elements of $C$ (which are sets).
That is, we take countably many elements of $C$ (which includes taking finitely many elements), then we take their union. A new set will come out of this. We lump all that together, and create $C_{\sigma}$.
For the sake of illustration I will take a different $C$.  You may compute $C_{\sigma}$ in your case.
Let us take $C = \{\{1\},\{2,3\},\{2\},\{1,3\}\}$.
Take the elements $\{1\} \in C$ and $\{1,3\} \in C$. Their union is $\{1,3\}$.
Take the elements $\{1\},\{2,3\},\{2\} \in C$. Their union is $\{1,2,3\}$.
Take just the element $\{2,3\} \in C$. Its union (the union involving just one set) is just $\{2,3\}$.
Take absolutely no element of $C$. This forms the empty union, which is equal to the empty set.
Take elements of $C$ with repetition. For example, take $\{1\},\{2,3\},\{1\},\{2\},\{2,3\},\{1,3\}\in C$ , their union is $\{1,2,3\}$.
The set of all possible unions that we have obtained on the right, are the elements of $C_{\sigma}$. So, for example, in our case, $C_{\sigma}$ contains the set $\{1,3\}$ as an element. It contains $\{1,2,3\}$  as an element. It contains $\{2,3\}$ as an element. It contains the empty set as an element.
Can you show that $\{1\} \in C_{\sigma}$? How about $\{1,2\}$? To show that something is an element of $C_{\sigma}$, you have to exhibit it as a union of sets which are in $C$.
Slightly more involved, but try to show that $\{3\} \notin C_{\sigma}$.

Now, $\delta$ is the same thing , but with intersection. We need to take a better example to understand this.
Take $C=\{\{1,2,3\},\{1,2\},\{3\},\{2,3\}\}$.
Now, take the elements $\{1,2\}, \{1,2,3\}\in C$. Their intersection is $\{1,2\}$.
Take the elements $\{1,2,3\},\{3\},\{2,3\} \in C$. Their intersection is $\{3\}$.
Take just the element $\{2,3\}\in C$. Its intersection (the intersection involving just one set) is $\{2,3\}$.
Take absolutely no element of $C$. This is the empty intersection, which is equal to the universal set , in this case $ X= \{1,2,3\}$.
Take elements of $C$ with repetition : for example, $\{1,2\},\{3\},\{1,2\},\{1,2,3\},\{1,2\},\{3\} \in C$, their intersection is the empty set.
The set of all possible intersections that we have obtained on the right, are the elements of $C_{\delta}$. For example, in our case $C_{\delta}$ contains the set $\{1,2\}$ as an element. It contains $\{3\}$ as an element. It contains $\{2,3\}$ as an element. It contains $\{1,2,3\}$ as an element. It contains the empty set as an element.
Can you show that $\{2\}\in C_{\delta}$? 
Try to show that $\{1\} \notin C_{\delta}$.

Now, there is the most important thing that we must mention at this point : infinite unions and intersections are permitted. We took only like $2,3,$ maybe at most $6$ elements of $C$ above and took their union/intersection to get an element of $C_{\sigma}$. In general, however, we may take infinite intersections as well. However, for this , $C$ will at least have to be infinite to bring some more clarity.

For example, consider the set $X = [0,1]$, and consider the collection of subsets $C = \{[0,\frac 1n)\}$. Then,the set $\{0\}$ is an infinite intersection $\cap_{k=0}^{\infty} [0,\frac 1k)$ of subsets of $C$, but not a finite intersection. That is, $\{0\} \in C_{\delta}$ as an infinite intersection, not a finite one.

The fact that $C_{\sigma\sigma} = C_{\sigma}$ and $C_{\delta\delta} =C_{\delta}$ follows like this : $C_{\sigma\sigma}$ consists of the countable union of elements of $C_{\sigma}$ which consists of the countable union of elements of $C$. Combining thee unions, we retain countability from a well known fact that a countable union of countable unions remains countable, and therefore $C_{\sigma \sigma}$ is a subset of $C_{\sigma}$. The reverse inclusion is near obvious. Similarly for $\delta$.

If you have NOT understood the infinite case yet, then do not worry : just remember what finite unions/intersections look like, and this will give you a good idea of what $C_{\sigma / \delta}$ looks like.
