# σ -algebra by choosing sets, where either the set or its complement is countable: is the complement countable?

I am reading Schilling's “Measures, integrals and martingales”, where on page 15 he constructs a $σ$-algebra, according to: $$\mathcal{A} = \lbrace A \subset X: \# A \leq \# \mathbb{N} \quad \text{or} \quad \# A^c \leq \# \mathbb{N} \rbrace,$$ i.e. those sets which are countable/have cardinality smaller than or equal to the natural numbers.

He then goes on to show that this set fulfills the properties of $σ$-algebras; the one I am struggling with is:

• $A \in \mathcal{A} \implies A^c \in \mathcal{A}$.

The author's argument is:

• If $A\in\mathcal{A}$, either $A$ or $A^c$ is by definition countable, so $A^c \in \mathcal{A}$.

I am not sure I can agree with this being so simple: if $A$ is indeed countable (so we choose the $\#A \leq \#\mathbb{N}$ condition), then it is already part of the set; its complement $A^c$ need not be countable as per the definition.

I am not sure I understand how the author relates $A$ being countable to its complement being countable? In fact, we don't know anything about its complement, other than it being the complement with respect to $X$ (which we also don't know if it is countable or not).

• How do we know that all countable unions are in the σ-algebra? – Jos van Nieuwman Jun 1 '19 at 17:15

$(A^c)^c = A$. If $A$ is countable then $(A^c)^c$ is countable, and if $A^c$ is countable then $A^c$ is countable. Either way, $A^c$ or its complement is countable.