= Solution
For the <cardinality of integer-periodic function spaces>, fix a positive period $n$ first. A <periodic function> $\mathbb Z\to\mathbb Z$ with that period is completely and uniquely determined by its $n$ values at $0,\ldots,n-1$, so this class is in <bijection> with $\mathbb Z^n$. The integers are a <countable set>, and repeated pairing shows that a finite <Cartesian product> of countable sets is countable. Taking the union over positive $n$ and using the <countable union of countable sets> proves that \b[the integer-domain periodic <functions> form a <countable set>].
For the rational domain, even period one leaves infinitely many independent choices. Let $q_k=1/(k+2)$ for $k\in\mathbb N_0$; these are distinct members of $\mathbb Q\cap[0,1)$. For every subset $A\subseteq\mathbb N_0$, define
$$
f_A(q)=
\begin{cases}
1,&q-\lfloor q\rfloor=q_k\text{ for some }k\in A,\\
0,&\text{otherwise}.
\end{cases}
$$
The <fractional part> is unchanged by adding one, so $f_A$ is an integer-valued <periodic function> on $\mathbb Q$. Distinct subsets give distinct <functions> because $f_A(q_k)$ records whether $k\in A$. Thus the uncountable <power set> of $\mathbb N_0$ injects into these <functions>. \b[The rational-domain periodic <functions> form an <uncountable set>.] The difference is that $\mathbb Z$ has only finitely many residue classes modulo a fixed integer period, whereas $\mathbb Q$ has infinitely many.
Back to article page
