# Regarding proving a $\sigma$ algebra is equal smallest $\sigma$ algebra containing a algebra.

My instructor on measure theory solved this problem in class but I am having serious problem in understanding it's proof although I am clear with definitions.

Problem is - Let $$P$$ be a set and any set $$A$$ belongs to $$P$$ iff either cardinality of $$A$$ is finite or cardinality of Compliment of A is finite. ( it is easy to see $$A$$ is an algebra) . Define $$Q$$ to be a set such that any set $$B$$ belongs to $$Q$$ iff either Cardinality of $$B$$ is countable or cardinality of complement of $$B$$ is countable . ( I proved it to be $$\sigma$$ algebra. ) Define $$\sigma ( P)$$ = smallest $$\sigma$$ algebra containing P.

I have to prove that $$\sigma (P) = Q$$.

Can somebody give explanation on how to prove it ?

Prove that $$\mathcal{P} \subset \mathcal{Q}$$ and since $$\mathcal{Q}$$ is a sigma algebra $$\sigma({\mathcal{P}})\subset \mathcal{Q}$$
Also if $$B \in \mathcal{Q}$$ and $$B$$ countable the $$B$$ is a countable union of finite sets so $$B \in \sigma({\mathcal{P}})$$
If $$B \in \mathcal{Q}$$ and $$B^c$$ is countable then as before $$B^c \in \sigma({\mathcal{P}})$$ and since $$\sigma({\mathcal{P}})$$ is a sigma algebra then $$B=(B^c)^c \in \sigma({\mathcal{P}})$$