Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 8E (Numbers and Sets)

A set is countable set if it is finite or admits an enumeration by natural numbers. If $A_n$ are countable, choose enumerations $a_{n,m}$. Since ${\mathbb{N}}^2$ is countable by diagonal enumeration, the array $(n,m)\mapsto a_{n,m}$ enumerates ${\bigcup }_n{A}_n$ after repetitions are removed. This proves that a countable union of countable sets is countable.

The integers are countable, so $\mathbb{Z}\times {\mathbb{N}}_{>0}$ is countable. Mapping $(p,q)$ to $p/q$ surjects onto $\mathbb{Q}$, hence $\mathbb{Q}$ is countable. The same diagonal argument shows that $A\times B$ is countable whenever $A,B$ are.

Finally, if the reals in $(0,1)$ had decimal expansions $x_1,x_2,\dots$, choose a decimal whose $n$th digit differs from the $n$th digit of $x_n$, avoiding digits $0$ and $9$ to remove expansion ambiguity. This number differs from every listed number. The Cantor diagonal argument proves that $\mathbb{R}$ is uncountable.

No. For every $m\in \mathbb{R}$, the line $y=mx$ contains the origin. These lines are distinct, so the uncountable familyhas nonempty common intersection.

Yes. The set ${\mathbb{Q}}^2$ is countable, hence the set of unordered pairs of distinct rational points is countable. Two distinct points determine at most one line. Mapping each such pair to its line therefore has countable image, and every line in the stated collection lies in that image. Thus the collection is countable.