Explanation of term "closed under" in the definition of a sigma algebra? 
A sigma algebra is closed under
  countable intersections.

This is a property of a sigma algebra, but I don't understand what this means in plain English.  Can someone explain this to me in terms a n00b like me would understand?  The part I don't understand the most is what "closed" means in this quote.
Addendum:
The collection of subsets:
{0},{x},{y,z},{x,y,z}
Is a sigma algebra of:
{0},{x},{y},{z},{x,y},{x,z},{y,z},{x,y,z}
Because:


*

*The collection of subsets is not empty -> The collection contains subsets.

*Closed under complementation -> The collection of subsets contains {x} and {x,y,z}/{x} = {y,z}

*Closed under countable unions -> {x} and {y,z} are contained in our sigma algebra, so is {x} U {y,z} = {x,y,z}

*Close under countable intersections -> {x} and {y,z} are contained in our sigma algebra, so is {0}


For point 2. above do you just take the complement of the set of all elements or {x,y,z} (that's what it seems to be according to the Wikipedia page)?
 A: In general, "closed under operation * on the set $\Sigma$" means that, if you apply operation * (in this case, taking countable intersections) to elements of the set $\Sigma$ (in this case, the set is the $\sigma$-algebra of subsets of a given set $X$), then the result is still an element of $\Sigma$.
Another example which may be more familiar to you: a vector space is closed under addition (addition of vectors is still a vector)
More precisely, in this case, if $\Sigma$ is a $\sigma$-algebra of subsets of $X$, then if $\{A_i\}_{i=1}^\infty \subset \Sigma$, i.e. $A_1, A_2,\dots, A_n, \dots \in \Sigma$ then $\bigcap_{i=1}^\infty A_i = A_1 \cap A_2 \cap \dots \cap A_n \cap \dots \in \Sigma$.
As I said in a comment above, by allowing countable intersections you get finite intersections too. If $A_1,\dots,A_n \in \Sigma$, then because $X\in \Sigma$ too by definition of $\sigma$-algebra, then using the fact $\Sigma$ is closed under countable intersections, $A_1\cap \dots \cap A_n = A_1 \cap \dots \cap A_n \cap X \cap X \cap X \cap \dots \in \Sigma$.
To answer to your addendum:
There seems to be a bit of confusion. If $X=\{x,y,z\}$, then $\mathcal{P}(X)=\{\emptyset,\{x\},\{y\},\{z\},\{x,y\},\{x,z\},\{y,z\},\{x,y,z\}\}$ is the power set of $X$, and a $\sigma$-algebra is a subset of the power set with some additional properties.
So, if $\Sigma=\{\emptyset,\{x\},\{y,z\},\{x,y,z\}\} \subset \mathcal{P}(X)$, then it is a $\sigma$-algebra of subsets of $X$.
To check item 2, what "closed under complementation" means is that, whenever $A\in \Sigma$, we have $A^c=X\setminus A \in \Sigma$. So, we check that for all four elements of $\Sigma$:
$\emptyset \in \Sigma \Rightarrow \{x,y,z\} \in \Sigma$.
$\{x\} \in \Sigma \Rightarrow \{y,z\} \in \Sigma$.
$\{y,z\} \in \Sigma \Rightarrow \{x\} \in \Sigma$.
$\{x,y,z\} \in \Sigma \Rightarrow \emptyset \in \Sigma$.
All these sentences are true, so item 2 is checked.
Also, as a side note, be careful, do not confuse $\emptyset$ with $\{\emptyset\}$.
A: It means that if $A_1, A_2, \dots \in \tau$ then $A_1 \cap A_2 \cap  \dots \in \tau$ where $\tau$ is a $\sigma$-algebra on a set $X$.
A: A sigma algebra is closed under Countable Intersections means that, if suppose $\tau$ is a $\sigma$-algebra on some set $X$ and if $A_{1},A_{2},A_{3}, \cdots \in \tau$ then $\cap A_{i}$ will also be in $\tau$. Note that you can only take the countable intersections. For more information please read the wikipedia page:


*

*http://en.wikipedia.org/wiki/Sigma-algebra
