Prove $\mathfrak{B}$ is a $\sigma$-field on $\mathfrak{R}$?

Question

Let $\Omega = \mathfrak{R}$. Define a collection of subsets of $\mathfrak{R}$ as follows: $$ \mathfrak{B} = \{A \subseteq \mathfrak{R} \,\,|\,\, A \text{ or } A' \text{ is countable}\}$$

Prove $\mathfrak{B}$ is a $\sigma$-field on $\mathfrak{R}.$

I know you need to follow the definition of a $\sigma$-field to prove this.

Definition: $\sigma$-field - A collection $\mathscr{F}$ of subsets of $\Omega$ is called a $\sigma$-field if it satisfies the following conditions:

(a) $\phi \in \mathscr{F}$

(b) If $A_1, A_2, \cdots \in \mathscr{F}$ then $\cup_{i=1}^{\infty}A_i \in \mathscr{F}$

(c) If $A \in \mathscr{F}$ then $A^c \in \mathscr{F}$

but how does the statement given $A$ or $A'$ being countable affect the proof?

Thanks in advance!

Answer 1

a) and c) are obvious. For b) let $A_1, A_2, \cdots \in \mathscr{F}$. Then either $A_n $ is countable for all n or no. If yes, the countable union of countable sets is countable. If no, then $A_n$ is uncountable for some n, then $A'_n$ is countable. Since the complement of a union is the intersection of the complement, the complementar of the union is contained in $A'_n$, so it's countable, so the initial union belongs to the collection.

Answer 2

a. and c. are clear from the definition. b. follows from two simple facts: a countable union of countable sets is countable and $(\sum_{i=1}^\infty A_i)' \subseteq A'_j$ for any $j \in \Bbb N$.