A sigma field question This is real basic, but a question that is annoying me:
$\Omega = \left \{ 1,2,3 \right \}$
$T = \left \{  \emptyset  , \Omega , \left \{ 1,2 \right \}, \left \{ 2,3 \right \}, \left \{ 1,3 \right \}\right \}$
Is $T$ a sigma-field?
Apparently the answer is no because $\left \{ 2,3 \right \}^{c} = \{1\}$ and $\{1\}\notin T$.
My question is: isn't $1 \in \Omega$ and therefore $1 \in T$ and $T$ is a sigma-field?
Does each specific set need its own specific complement?
 A: Why is the complement of $\left\{1,2\right\}$ with respect to $\left\{1,2,3\right\}$ equal to $\left\{1\right\}$? It should be equal to $\left\{3\right\}$, which is not in $T$ either. Hence $T$ cannot be a sigma field.
Yes, the complement of each set in $T$ needs to be in $T$ for $T$ to be considered a sigma field. Of course, this condition alone is not sufficient; you need the countable unions too. 
A: A $\sigma$-algebra that consists of a finite number of events (subsets of $\Omega$) must have exactly $2^m$ events for some $m \geq 1$. If 
$|\Omega| = n$, then we also have that $m \leq n$. So, since your purported
$\sigma$-algebra has $5$ events in it, it cannot be a $\sigma$-algebra.
Note that this test does not work the other way: the collection $ \left \{  \emptyset,  \Omega, \left \{ 1,2 \right \}, \left \{ 2,3 \right \} \right \}$ of subsets of $\{1,2,3\}$ has $4 = 2^2$ members but it is not a $\sigma$-algebra. On the other hand, 
$ \left \{ \emptyset, \Omega, \left \{ 1,2 \right \}, \left \{ 3 \right \} \right \}$ also consists of $4$ subsets but is a $\sigma$-algebra. That is, having $2^m$
members is a necessary condition but not a sufficient condition
for a collection of subsets of $\Omega$ to constitute a $\sigma$-algebra.
