A property of bounded sequences of rationals How to prove that for every bounded sequence of rational numbers there exists a subsequence $q_i$ such that for every $n$ one has 
$$ |q_i - q_j| < 1/n $$
for $i,j \ge n$.
Apparently this proves the consistency of PA, so a proof here is probably not too easy,   but I may be wrong.
 A: suppose $\{q_n\}$ is a bounded sequence of rational number so by Bolzano-Weirstrass theorem it has a limit point $l\in\mathbb{R}$ so $\forall\epsilon>0$, $\exists N\in\mathbb{N}$ such that $|q_n-l|<\epsilon \forall n\ge N$, Now you  fix $\epsilon=1$ then by definition definition you get $\exists N_1\in\mathbb{N}$ such that $|q_n-l|<1 \forall n\ge N$, chhose $q_1\in (l-1,l+1)$ so now you can choose $\epsilon=\frac{1}{2}$ and you can get $q_2\in(l-\frac{1}{2},l+\frac{1}{2})$ $\dots\dots$ for $\epsilon=\frac{1}{n}$ you get $q_n\in(l-\frac{1}{n},l+\frac{1}{n})$ so chhose $N=Max\{N_1,\dots,M_n\}$ we get $|q_i-q_j|<\frac{1}{n}$
A: By the Bolzanno-Weierstrass theorem, there is a convergent subsequence $(u_n)_{n\geq 1}$. This sequence satisfies Cauchy’s criterion, so for any $n>0$ there is an integer $\phi(n) \gt n$ such that
$$
|u_i-u_j| \leq \frac{1}{n}
$$
for any $i,j \geq \phi(n)$.
Replacing $\phi(n)$ with $\psi(n)={\sf max}_{1 \leq j \leq n}(\phi(j)+n+1-j)$, we may assume that $\phi$ is increasing.
Then, you can take the subsequence $(q_n)$ defined by $q_n=u_{\phi(n)}$. 
