Let $\Omega=${0,1,2,3,...}
Let B be the collection of subsets of $\Omega$ such that C $\in$ B if and only if either C or $C^{c}$ is a finite set.
Is B a field? Is it a $\sigma$-field?
Here is my thought process and I would appreciate if the community could offer advice.
According to pg. 19 of this book the following are 3 properties of a field:
- $\Omega \in B$
- $C \in B --> C^{c} \in B$
- $C,D \in B$ --> $C\cup D \in B$
A $\sigma$-field the same 2 properties above except the 3rd property is strengthened to only countable unions and needs to contain singletons.
Let C=$\varnothing$. Then $C^{c}=\Omega$. C is finite and $C^{c}$ is infinite. Therefore $C \in $B.
But if I let $C=\{1,3,5,...\}$ then $C^{c} = \{2,4,6,...\}$. In this case both are infinite, and therefore $C$ does not exist in $B$. So this violates property #2 above. Right? That means $B$ is not a field, nor is it a $\sigma$-field.
By definition, all finite subsets of $\Omega$ are in $B$.
Property 1 : $\Omega^C = \emptyset \in B$, so by definition $\Omega \in B$ since $\emptyset$ is finite.
Property 2 : Note that $C = C^{C^{C}} $, so if $C \in B$ is finite then by defintion $C^C \in B$. On the other hand, if $C \in B$ is infinite, then by defintion $C^C$ must be finite for otherwise $C \notin B$. So $C^C \in B$.
(A shorter proof of property 3 is provided in the EDIT at the bottom of this answer.)
Property 3 : If $C,D \in B$ then, if $C \cup D$ is finite, then $C \cup D \in B$ so by property 2 $(C \cup D)^C \in B$.
If $C \cup D$ is infinite and $(C \cup D)^C$ is finite, then by defintion $(C \cup D)^C \in B$.
This leaves the case where both $(C \cup D)$ and $(C \cup D)^C$ are infinite. Here we note De Morgan's Law $(C \cup D)^C = C^C \cap D^C$. Note that both $C^C$ and $D^C$ are in $B$ by property 2. If both $C^C$ and $D^C$ are finite, then so is their intersection and the result follows. If both $C^C$ and $D^C$ are infinite, then since $C,D \in B$ we know that both $C$ and $D$ must be finite and so must their union, and again the result follows. This just leaves the case where one of $C$, $D$ is finite and the other is infinite. Without loss of generality, let $C$ be infinite and $D$ be finite. In this case, if $C \in B$ then $C^C$ must be finite, so $C^C \cap D^C$ must also be finite and therefore in $B$.
This shows $B$ satisfies all three field properties listed. (A rather case-bound demonstration.)
To see that $B$ is a $\sigma$-field according to your definition, note that all singletons are in $B$ since, by definition, $B$ contains all finite sets. Closure under countable union follows immediately by induction.
EDIT (Sep 14): Here is a shorter and slightly more attractive proof of property 3.
If $(C \cup D) = (C \cup D)^{C^C}$ is finite, then $(C \cup D)^C \in B$ since its complement is finite.
If $(C \cup D) = (C \cup D)^{C^C}$ is infinite, then if $(C \cup D) \in B$ then it must be because its complement is finite. So $(C \cup D)^C \in B$.