Let $X$ be the set of Cauchy sequences of rational numbers.

  • Then $(q_n)\sim (r_n)$ means that for every $0<\epsilon \in \mathbb{Q}$ there exists $N\in \mathbb{N}$ such that $|q_n-r_n|<\epsilon$ whenever $n\ge N$.

  • $(q_n)\le (r_n)$ means that for every $0<\epsilon \in \mathbb{Q}$ there exists $N\in\mathbb{N}$ such that $q_n < r_n +\epsilon $ whenever $n\ge N$.

  • $(q_n) <(r_n)$ means that $(q_n)\le (r_n)$ but $(q_n)\not\sim (r_n)$.

Suppose that $(q_n)<(r_n)$ in $X$. Show that there exists $q\in\mathbb{Q}$ such that $(q)=(q,q,\dots)$ satisfies $(q_n)<(q)<(r_n)$.

My approach:

$(q_n)<(r_n)$ implies the following:

  • $\forall\epsilon>0,\exists N\in\mathbb{N}$ such that $n\ge N\implies q_n\le r_n+\epsilon$;

  • $\exists \epsilon_0>0$ such that $\forall N\in\mathbb{N}, n\ge N \implies |q_n-r_n|\ge \epsilon_0$.

Let $\epsilon = \epsilon_0$, let $\epsilon_0 := q$. Then we have

  • $q_n \le r_n + q$ (*)

  • $-q\le q_n-r_n \le q$ (**)

But I'm stuck here and have no idea how to proceed. I've tried to manipulate the inequalities, but didn't achieve much success.

I'd appreciate if you could help me with this. Or is my approach not so good?

  • $\begingroup$ Can you use $\mathbb{R}$ ? If yes, then $q_n\to L_1\in \mathbb{R}$, $r_n\to L_2$, it is easy to show from your hypothesis that $L_1<L_2$, and hence any rational $q$ such that $L_1<q<L_2$ do the job. $\endgroup$ – Kelenner Sep 17 '17 at 5:36
  • $\begingroup$ The definition we use for $\mathbb{R}$ is that it is the set of all equivalence classes of $X$. But I'm not sure if this means that it is allowed to use $\mathbb{R}$ in this proof. $\endgroup$ – sequence Sep 17 '17 at 5:45

Since $(q_n)\nsim (r_n)$, there exists $\epsilon_0>0$ such that for any $N$, there will always be some $n\geq N$ such that $|q_n - r_n| \geq \epsilon_0.$

Next, since $(q_n)$ and $(r_n)$ are Cauchy, there exist $N_q$ and $N_r$ such that $|q_n - q_{N_q}| <\epsilon_0/6$ for $n \geq N_q$ and $|r_n - r_{N_r}| <\epsilon_0/6$ for $n\geq N_r$.

Next since $(q_n) \leq (r_n)$, there exists $N^*\geq \max\{N_q,N_r\}$ such that $q_n \leq r_n + \epsilon_0/6$ for all $n\geq N^*$.

Since $(q_n) \nsim (r_n)$ there is still some $m>N^*$ such that $|r_m - q_m| \geq \epsilon_0.$

Since $|r_m - q_m| \geq \epsilon_0$ and $q_m \leq r_m + \epsilon_0/6$, it follows that $$r_m - q_m > \epsilon_0.$$

Further since $m>N^*$, $|q_n - q_m| \leq |q_n - q_{N_q}| + |q_{N_q} - q_m| \leq \epsilon_0/3 $, and $|r_n - r_m| \leq |r_n - r_{N_r}| + |r_{N_r} - q_n| \leq \epsilon_0/3 $ for all $n>m$.

Let $q = (q_{m} + r_m)/2$. Then $q\in \mathbb{Q}$ as the rationals are closed under addition and multiplication,and $q_n < q_m +\epsilon_0/3 < q < r_m - \epsilon_0/3 < r_n$ for all $n>m$.

  • $\begingroup$ Can you please clarify: when you say "Combining this with the previous inequality", which inequality are you referring to? Also, can you please clarify how you deduced that $q_m+\epsilon_0/3<q<r_m-\epsilon_0/3$? $\endgroup$ – sequence Sep 17 '17 at 15:11
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    $\begingroup$ I hope that your first question was answered with my edit. For the second question, we deduced that $q_m$ is strictly less than & at least $\epsilon_0$ away from $r_m$. Also, by choice of $N^*$, all the $q_n$ are within $\epsilon_0/3$ of $q_m$, as are the $r_n$ w.r.t. $r_m$ ("Further since..."). So there's an interval between $(q_n)$ and $(r_n)$ of width at least $\epsilon_0/3$ where there are no $q_n$ or $r_n$. We picked the average of $q_m$ and $r_n$ for $q$ so that $r_m - q_m > \epsilon_0$, $q_m + \epsilon_0/2 < q < r_m - \epsilon_0/2$. $\endgroup$ – user474330 Sep 17 '17 at 18:48
  • $\begingroup$ It's still not clear how $r_m-q_m>\epsilon_0$ implies that $q_m+\epsilon_0/2<\frac{q_m+r_m}{2}<r_m-\epsilon_0/2$. Would you please clarify? $\endgroup$ – sequence Sep 17 '17 at 19:12
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    $\begingroup$ It might help to draw the points on a number line. In general, if any two points are more than $\epsilon$ apart, then their midpoint must be at least $\epsilon/2$ away from each of them. You can also see this algebraically (but the physical understanding is especially useful). Here I'll just use $r$, $q$, and $\epsilon.$ The first equation says $q + \epsilon < r$. Now add $q$ to both sides and divide by 2. Then $q+ \frac{\epsilon}{2} < \frac{q+r}{2}$. Then if you start again with $q + \epsilon < r$, add $r$ to both sides, and divide by 2, you'll get the second half of the inequality. $\endgroup$ – user474330 Sep 18 '17 at 12:44
  • $\begingroup$ Maybe this is a nicer explanation: If any two points $x$ & $y$ are more than $d$ apart, then their midpoint must be more than $d/2$ away from each of them. Algebraically, if $x+d< y$, then $y = x +d + s$ for some $s>0$. So the midpoint of $x$ & $y$ is $q = \frac{x+y}{2} = x + \frac{d+s}{2} = y - \frac{d+s}{2}.$ So $x + \frac{d}{2}< x + \frac{d+s}{2} = q = y - \frac{d+s}{2} < y - \frac{d}{2}$. $\endgroup$ – user474330 Sep 19 '17 at 14:52

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