For every $x\in\Bbb R$ and $n\in\Bbb N\setminus \{0\}$, there exists $r,s\in\Bbb Q$ such that $r< x<s$ and $s-r< 1/n$.
This is a lemma in Chapter 10 from my textbook Introduction to Set Theory by Hrbacek and Jech. Although the authors have provided a proof, I have figured another one. Please help me verify my proof! Thank you so much!
My attempt:
By recursion theorem, we define a strictly increasing sequence $(l_i \mid i\in\Bbb N)$ of rationals converging to $x$. Similarly, we define a strictly decreasing sequence $(u_i \mid i\in\Bbb N)$ of rationals converging to $x$.
We define a sequence $(s_i \mid i\in\Bbb N)$ by $s_i=u_i-l_i$ for all $i\in\Bbb N$. It is easy to verify that $(s_i \mid i\in\Bbb N)$ is a strictly decreasing sequence converging to $0$.
Let $i_0=\min \{i\in\Bbb N \mid s_i<1/n\}$. Since $(s_i \mid i\in\Bbb N)$ is a strictly decreasing sequence converging to $0$, there exists $s_i$ such that $0<s_i<1/n$ for all $n\in\Bbb N$ and thus such $i_0$ exists.
It is clear that $l_{i_0}<x<u_{i_0}$ and $s_{i_0}=u_{i_0}-l_{i_0}<1/n$. Let $r=l_{i_0}$ and $s=u_{i_0}$.
With a similar reasoning, I have proved the below theorem:
For every $x,k\in\Bbb R$ such that $0<x,1<k$. There exists $r,s\in\Bbb Q$ such that $0<r<x<s$ and $s/r<k$.
My attempt:
By recursion theorem, we define a strictly increasing sequence $(l_i \mid i\in\Bbb N)$ of positive rationals converging to $x$. Similarly, we define a strictly decreasing sequence $(u_i \mid i\in\Bbb N)$ of rationals converging to $x$.
We define a sequence $(k_i \mid i\in\Bbb N)$ by $k_i=u_i/l_i$ for all $i\in\Bbb N$. It is easy to verify that $(k_i \mid i\in\Bbb N)$ is a strictly decreasing sequence converging to $1$.
Let $i_0=\min \{i\in\Bbb N \mid k_i<k\}$. It is clear that $l_{i_0}<x<u_{i_0}$ and $k_{i_0}=u_{i_0}/l_{i_0}<k$. Let $r=l_{i_0}$ and $s=u_{i_0}$.
