How to show that C is a Boolean algebra but not a sigma-algebra if I have already demonstrated some additional properties?

Question

The exercise asks:

Let be $Ω = \mathcal{N}$ and $\mathscr{C} = \{A , A ⊆ \mathcal N : A\text{ is finite or }A^c\text{ is finite}\}$.

Prove that $\mathscr{C}$ is an $Algebra$ $Of$ $Boole$, but not an $σ-Algebra$.

My proof is:

Properties of $\mathscr{C}$, a $Boolean$ algebra:

1. $Ω∈\mathscr{C}$
2. If $A∈\mathscr{C}⇒A^c∈\mathscr{C}$
3. If $A,B∈\mathscr{C}⇒A∪B∈\mathscr{C}$

Demostrated properties of $\mathscr{C}$:

4. $∅∈\mathscr{C}$
5. If $A,B∈\mathscr{C}⇒A\cap B∈\mathscr{C}$
6. If $A_{1},...,A_{n}∈\mathscr{C}⇒\left(\bigcup_{i=1}^{i=n}A_{i}\right)∈\mathscr{C}$ and $\left(\bigcap_{i=1}^{i=n}A_{i}\right)∈\mathscr{C}$ with $n∈N^*$

Every $σ-Algebra$ has $(1)$, $(2)$, $(3)$ and the following:

7. If $A_{1},...,A_{n},...∈\mathscr{C}⇒\left(\bigcup_{i=1}^{i=\infty}A_{i}\right)∈\mathscr{C}$

1. $Ω=\mathcal{N}⇒\mathcal{N}⊆Ω⊆\mathcal{N}⇒^{(1)}Ω∈\mathscr{C}$
2. $A⊆\mathcal{N}\Leftrightarrow A^c$ is finite $⇒^{(2)}A^c∈\mathscr{C}$
3. $A⊆\mathcal{N}\Leftrightarrow A$ is finite $⇒A_{1},...,A_{n}∈\mathscr{C}⇒^{(6)}\left(\bigcup_{i=1}^{i=n}A_{i}\right)∈\mathscr{C}$ with $n∈N^*$
4. $A⊆\mathcal{N}\Leftrightarrow A$ is finite $⇒A_{1},...,A_{n},...\notin\mathscr{C}⇒^{(7)}\left(\bigcup_{i=1}^{i=\infty}A_{i}\right)\notin\mathscr{C}$

By $(1,2,3)$, $\mathscr{C}$ is an $Algebra$ $Of$ $Bool$. But by $(4)$, is not an $σ-Algebra$.

Is correct?

Answer 1

For each even number $n \in \mathcal{N}$, $\{ n\} \in \mathscr{C}$. But the countable union of all such $\{n\}$ is not in $\mathcal{N}$ because this union is not finite, nor is its complement. So $\mathscr{C}$ is not closed under countable unions and is therefore not a $\sigma$-algebra.

On the other hand, $\mathscr{C}$ is a Boolean algebra. Clearly, $\mathscr{C}$ is close under complementation. Suppose $A$ is the finite union of sets $\{A_i \}_{i=1}^n$ in $\mathscr{C}$. If each $A_i$ is finite, then so is $A$, and so $A \in \mathscr{C}$. If some $A_i$ is not finite, then $A_i^c$ is finite and $$(\cup_{i=1}^n A_i)^c = \cap_{i=1} ^n A_i^c$$ is finite, so $A \in \mathscr{C}$. This shows that $\mathscr{C}$ is closed under finite unions, and we are done.