= Solution
To prove the <Archimedean property>, suppose the positive integers were bounded above in $\mathbb R$. By the <least upper bound axiom> they would have a finite <supremum> $s$. Since $s-1$ is not an upper bound, some positive integer $k$ satisfies $k>s-1$. Then the positive integer $k+1>s$, contradicting the definition of $s$. Thus the positive integers are unbounded. Equivalently, for any positive real $a,b$, some positive integer $n$ satisfies $na>b$, by applying unboundedness to $b/a$.
For fixed $m$, put $r_m=\cos^2(m!\pi x)$. The standard properties needed are $0\le\cos^2 t\le1$, and $\cos^2 t=1$ exactly when $t\in\pi\mathbb Z$. If $0\le r<1$, then $r^n\to0$; for $0<r<1$ this also follows from the <Bernoulli inequality>, since $r^{-n}\ge1+n(r^{-1}-1)\to\infty$. Therefore the inner <limit> is
$$
\lim_{n\to\infty}\cos^{2n}(m!\pi x)=\begin{cases}1,&m!x\in\mathbb Z,\\0,&m!x\notin\mathbb Z.\end{cases}
$$
If $x=p/q$ is a <rational number>, with $q\ge1$, then $q$ divides the <factorial> $m!$ whenever $m\ge q$. Hence $m!x$ is an integer for all sufficiently large $m$, and the outer limit is one. This includes the endpoints $x=0,1$. If $x$ is an <irrational number>, $m!x$ cannot be an integer for any $m$, since division by $m!$ would make $x$ rational. Then every inner limit is zero. The <factorial cosine iterated limit detects rationality> result is consequently
$$
\boxed{\lim_{m\to\infty}\left[\lim_{n\to\infty}\cos^{2n}(m!\pi x)\right]=\begin{cases}1,&x\in\mathbb Q,\\0,&x\notin\mathbb Q.\end{cases}}
$$
Each inner limit was taken with $m$ fixed; the argument does not interchange the two limits or assert a joint limit as both indices grow.
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