2
$\begingroup$

Let $S\subseteq \mathbb{R}$, and $\{x_n\}$ and $\{y_n\}$ be two Cauchy sequences in $S$, with $x_n,y_n\geq 10$. Prove that $z_n:=\left\{\frac{2x_n}{y_n}\right\}$ is also Cauchy in $S$. [It is assumed that $z_n\in S$.

My proof:

Since $x_n, y_n$ are Cauchy in $S$, they are convergent in $S$ and thus bounded in $S$. Let $x_n$ converge to real number $L$ and $y_n$ converge to real number $M$. Since $x_n, y_n$ are Cauchy, $\left|x_n-x_m\right|<\varepsilon$ for $N_1(\varepsilon)<n,m$ and $\left|y_n-_m\right|<\varepsilon$ for $N_2(\varepsilon)<n,m$. WLOG, suppose that $y_n > y_m$. Let $N:=\max\{N_1, N_2\}$, then $$\begin{align}\left| \frac{2x_n}{y_n}-\frac{2x_m}{y_m}\right| & =\left|\frac{2x_ny_m-2x_my_n}{y_ny_m}\right| \\&\leq \left|\frac{x_ny_m-x_my_n}{5y_m}\right|\\&< \left| x_n-x_m\frac{y_n}{y_m} \right|\\&\leq \underbrace{\left| x_n-x_m \right|}_{\text{since } \left| y_n-y_m\right| <\varepsilon\implies \frac{y_n}{y_m}\ge 1}\\ &<\varepsilon.\end{align}$$ Hence $z_n$ is Cauchy.

End of proof.

But... I'm somewhat concerned that the argument under the underbrace is too handwavy. Would appreciate some help.

$\endgroup$
11
  • $\begingroup$ is it $a_n, b_n$ or $x_n ,y_n$? $\endgroup$ Commented Sep 29, 2016 at 23:07
  • $\begingroup$ Sorry, my mistake. Fixed. $\endgroup$
    – sequence
    Commented Sep 29, 2016 at 23:11
  • $\begingroup$ False. Let $S = \{10\}.$ Define $x_n,y_n = 10$ for all $n.$ Then $2x_n /y_n = 2 \not \in S$ for all $n.$ $\endgroup$
    – zhw.
    Commented Sep 30, 2016 at 0:05
  • $\begingroup$ Well for starters your underbrace assertion is false if $y_n=1-\epsilon/2$ and $y_m=1$. $\endgroup$
    – Alex R.
    Commented Sep 30, 2016 at 0:05
  • 1
    $\begingroup$ Here's what I did now: $2\left|\frac{x_ny_m-x_my_n}{y_my_n}\right|=2\left|\frac{(x_n-x_m)(y_m+y_n)-(x_ny_n+x_my_m)}{y_my_n}\right|<2\left| x_n-x_m \right|\left| y_m+y_n \right| < 4M\left| x_n-x_m \right| < 4M\varepsilon<\varepsilon'$ (where $M:=\sup\{y_n\}$ and $\varepsilon'=\frac{\varepsilon}{4M}$). $\endgroup$
    – sequence
    Commented Sep 30, 2016 at 2:13

1 Answer 1

2
$\begingroup$

Since they are Cauchy in $S$, they are also Cauchy in $\mathbb R$. Hence they are convergent in $\mathbb R$. Since $y_n\ge10$ we have that $\lim_{n\to\infty}\limits y_n = M\ge 10$, in particular $\lim_{n\to\infty}\limits y_n \not=0$. Hence $z_n=\frac{2x_n}{y_n}$ is convergent in $\mathbb R$, hence it is Cauchy in $\mathbb R$, hence Cauchy in $S$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .