A question about the number of possible $\sigma$-algebras

Question

Let $F$ be an event space. Show that the total number of events in $F$ cannot be exactly six. What integers can be the number of events in a finite event space?

MY ATTEMPT

Let us suppose there are exactly six events in $F$. Precisely speaking, if $\Omega$ denotes the sample space, we should have \begin{align*} F = \{\emptyset,\Omega,A,B,C,D\} \end{align*}

Since it is a $\sigma$-algebra, if $A\in F$, then $A^{c}$ must belong to $F$ as well. That means $C = A^{c}$, for instance. Similarly, if $B\in F$, then $B^{c}$ must belong to $F$ as well, which means that $D = B^{c}$. Finally, $A\cup B$ must belong to $F$, but it doesn't happen, because $F = \{\emptyset,\Omega,A,A^{c},B,B^{c}\}$.

However I am unable to answer the second question. Could someone help me? Any help is appreciated. Thanks in advance.

Answer 1

Since it is a $\sigma$-algebra, if $A\in F$, then $A^{c}$ must belong to $F$ as well. That means $C = A^{c}$, for instance. Similarly, if $B\in F$, then $B^{c}$ must belong to $F$ as well, which means that $D = B^{c}$. Finally, $A\cup B$ must belong to $F$, but it doesn't happen, because $F = \{\emptyset,\Omega,A,A^{c},B,B^{c}\}$.

Your solution starts out well but has a flaw at the end -- why do you claim $A \cup B$ does not belong to $F$? It is totally possible, for example, that $A \subseteq B$, which means that $A \cup B = B$, which does belong to $F$.

You should be more careful about reasoning that the different sets are actually distinct. Why is $A$ distinct from $A^c$? Why are $A$ and $A^c$ distinct from $B$ and $B^c$?

Then, to continue: so far we know that $F$ is as follows: $$F = \{\varnothing, \Omega, A, A^c, B, B^c\}.$$ Now note that $A \cup B$ is a set containing $A$ and containing $B$. So it can't equal $\varnothing,$ it can't equal $A^c$, and it can't equal $B^c$. The remaining possibilities are

1. $A = A \cup B$
2. $B = A \cup B$,
3. $\Omega = A \cup B$.

In the first case, that means that $B \subseteq A$. Now consider $A^c \cup B$; maybe you can show that it is not equal to any of the sets in $F$, for a contradiction.

In the 3rd case ($\Omega = A \cup B$), I would consider $A^c \cup B^c$. Try to show that it is not equal to any of the sets in $F$.

However I am unable to answer the second question.

Try some small integers $1, 2, 3, 4, 5$. Out of those, which are possible? Is there any pattern?

Answer 2

Say $A$ is a finite $\sigma$-algebra on $\Omega$. For $x\in\Omega$ let $$C(x)=\bigcap\{E\in A:x\in E\}.$$Show that for every $x$ and $y$ either $C(x)=C(y)$ or $C(x)\cap C(y)=\emptyset$, so the $C(x)$ form a partition of $\Omega$.

Show that if $E\in A$ then $$E=\bigcup_{x\in E}C(x).$$So the elements of $A$ are precisely the unions of subsets of the collection of all the $C(x)$. So if there are $n$ distinct $C(x)$ then $A$ has $2^n$ elements.