I am struggling to prove the following:
If $(x_n)$ is a Cauchy sequence in $\mathbb Q$ with $x_n\nrightarrow0$, then the sequence $(1/x_n)$ is also Cauchy.
My attempt at proof:
Let $\epsilon>0$. Since $x_n\nrightarrow0$, it grants us some $\epsilon'>0$ so that for all $N\in\mathbb N$, $|x_n|\ge\epsilon'$ for some $n>N$. [……] The sequence $(x_n)$ is bounded below by some $M>0$. Since $(x_n)$ is Cauchy, there is $N'\in\mathbb N$ s.t. $|x_m-x_n|<M^2\epsilon$ for all $m,n>N'$. Choose $N=N'$. Then for $m,n>N$,
$$\displaystyle\left|\frac{1}{x_m}-\frac{1}{x_n}\right|=\left|\frac{x_n-x_m}{x_mx_n}\right|=\frac{|x_n-x_m|}{|x_m||x_n|}<\frac{M^2 \epsilon}{M^2}=\epsilon.$$
I am struggling to fill in the blank [……] part. I want to show that $(x_n)$ is bounded below by some positive real number $M>0$, i.e. $x_n\ge M$ for all $n$. I am thinking about constructing a positive subsequence of $(x_n)$ but I am not sure where to start. If $(x_n)$ is a Cauchy sequence in $\mathbb R$, then I can easily use the completeness of $\mathbb R$ to show that $(x_n)$ is bounded below.
Best Answer
You can't claim that on the sequence $x_n$ itself, but you can prove that for the sequence of absolute values $|x_n|$. (which is what you actually need) Also, that is not necessary true for all $n$, but only when $n$ is sufficiently large.
Hint: As you pointed out, there is some $\epsilon'>0$ such that $|x_n|\geq\epsilon'$ for infinitely many values of $n$. But note that $|x_n|$ is a Cauchy sequence (why?), and so if $m$ and $n$ are sufficiently large then $|x_m|$ and $|x_n|$ are very close to each other. Try to conclude that if $n$ is large enough, we will always have $|x_n|>\frac{\epsilon'}{2}$.