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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 $\mathscr{C}$:

  1. $∅∈\mathscr{C}$
  2. If $A,B∈\mathscr{C}⇒A\cap B∈\mathscr{C}$
  3. 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:

  1. 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?

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    So, what are the properties of an Algebra of Boole, and a sigma-algebra, and more importantly, what properties are not in common? – 2017-01-24
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    In my edition I wrote the basics and the ones I already showed. Then try to demonstrate what the exercise called for. Is it well proven? Thanks. – 2017-01-24

1 Answers 1

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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.

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    Why "_For each even number_" only? – 2017-01-24
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    Not only. I give just one counterexample. Odd numbers $n$ are also such that $\{ n\} \in \mathscr{C}$ and the same argument shows that $\mathscr{C}$ is not a $\sigma$-algebra. – 2017-01-24