# 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.

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## 2 Answers

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 $q_n=u_{\phi(n)}$.

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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 $|q_i-q_j|<\frac{1}{n}$

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