If \{A_i\}_{i \in [n]} is a finite collection or countably infinite set, then A_1 \times ... \times A_n = \prod_{i=1}^n {A_i} is a countably infinite (3 + induction)

Theorem: The union of countably infinite set \{A_n\}_{n\in \mathbb{N}} is countably infinite.

proof by double containment:

WTS |\mathbb{N}| \leq |A|. We know there exists a bijection from N to A1. Then there is a injection from N to A1. Then there is a injection from N to A.

WTS |\mathbb{N}| \geq |A|. We extract an element from A, find out its belonging set (in A_n). h: \mathbb{N}^2 \rightarrowtail A | h(n, m_ = a(n, m). Then there exists m such that y=a(n, m)=h(n, m). Since |\mathbb{N}^2| geq |A|, |A| \leq |N| TODO:???

Corollaries

If \{A_i\}_{i\in [n]} is a finite family of countably infinite sets, then the union is countably infinite

If \{A_i\}_{i\in \mathbb{N}} is a family of countable sets then the union is still countable