Hi everybody I'd like to know if the next argument is sound.

Definitions: A real number x is said to be positive if can be written as a formal limit of a Cauchy sequence of rational numbers $(x_n)$ which is positively bounded away from zero, i.e., $\,x_n\ge c$ for each n, where $c$ is a positive rational number.

Proposition: Given any two real numbers $x<y$, we can find a rational number $q$ such that $x<q<y$.


It will suffice to find two rational numbers such that $x\le q_1<q_2 \le y$, because we can define $q= \frac{q_1+q_2}{2}$ and we're done.

Let $(a_n)_{n=1}^{\infty}$, $(b_n)_{n=1}^{\infty}$ be Cauchy sequences such that their formal limits are the real number $y$ and $x$ respectively.

Since $y-x$ is a positive real number there exists a positive rational number $c$ such that $y-x>c$. Let $\,\epsilon= c/6$. Then there exists sufficient large $N_{\epsilon}\in \mathbb{N}$ such that: $|a_n-a_m|\le \epsilon$, $|b_n-b_m|\le \epsilon$ occurs simultaneously for all $n,m\ge N_{\epsilon}$. Moreover the sequence $(a_n-b_n)$ is eventually greater than $c$, let $M$ be the natural number such that $ a_n-b_n \ge c$ for all $n\ge M$. We set $N =max(N_{\epsilon}, M)$. Now let us fix some $n_0$ such that $n_0\ge N$. Then clearly $|y - a_{n_0}|\le c/6$ and $|x-b_{n_0}|\le c/6$ at the same time.

We claim that $a_{n_0}-\frac{c}{6}$ and $b_{n_0}+\frac{c}{6}$ have the desired property. It is not difficult to see that both are rational numbers, follows immediately by construction so, we will show that $a_{n_0}-\frac{c}{6}\le y$, $x\le b_{n_0}+\frac{c}{6}$ and $a_{n_0}-\frac{c}{6}>b_{n_0}+\frac{c}{6}$.

We already know that $|y - a_{n_0}|\le c/6$, so $a_{n_0}-c/6\le y \le a_{n_0}+c/6$ as desired. A similar argument shows that $x\le b_{n_0}+c/6$. To conclude the proof our task is to show $b_{n_0}+c/6<a_{n_0}-c/6$. We argue by contradiction, suppose $b_{n_0}+c/6\ge a_{n_0}-c/6$ then $c/3\ge a_{n_0}-b_{n_0}$. But since $n_0\ge N$, it follows that $a_{n_0}-b_{n_0}\ge c$ and then $c/3\ge c$ a contradiction.

We define $q$ as the average of $a_{n_0}-\frac{c}{6}$ and $b_{n_0}+\frac{c}{6}$ and we're done.

I would appreciate for any suggestion. Thanks in advance. :)

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    $\begingroup$ Seems fine, but you are assuming that $\mathbb{Q}$ is (topologically) dense in $\mathbb{R}$ to prove this. Usually, one just uses the Archimedean property, and then uses this to prove that $\overline{\mathbb{Q}} = \mathbb{R}$. $\endgroup$ – Prahlad Vaidyanathan Oct 13 '13 at 3:45
  • $\begingroup$ Well in the book we construct the real numbers by Cauchy sequences in some way that the author called metric completion. So this is the only thing which I've already known. In the next exercise I need to prove the Archimedean property. $\endgroup$ – Jose Antonio Oct 13 '13 at 3:53

This should be a comment, but I lack the power (insert Scotty reference).

You have stated that $y - x$ is positive, but no where did you assert $y > x$. Of course, you can do that, without loss of generality, but, if you want to precise, you should. Otherwise, looks fine.

Since I'm writing an answer anyway: If you wanted to simplify the final argument and avoid the proof by contradiction, it isn't hard. You've chosen $n_0$ so that $a_{n_0} - b_{n_0} \geq c$. This means that

$$ b_{n_0} + c \leq a_{n_0}.$$

Subtracting $\frac{c}{2}$ from both sides gives the obvious inequalities:

$$ b_{n_0} + \frac{c}{6} < b_{n_0} + \frac{c}{2} \leq a_{n_0} - \frac{c}{2} < a_{n_0} - \frac{c}{6}. $$

Indeed, this shows that either of $b_{n_0} + c/2$ or $a_{n_0} - c/2$ could be the quantity $q$ you sought in the first place. Anyway, your proof is correct.

  • $\begingroup$ Thanks. I like it your suggestion to avoid the proof by contradiction. But if I'm honest with you I really love to argue by contradiction, I don't know why, but I find it so beautiful and in almost any situation my first thought is: maybe I should be use contradiction :P $\endgroup$ – Jose Antonio Oct 13 '13 at 5:32
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    $\begingroup$ I have no issues with proof by contradiction. But I have, on occasion, argued by contradiction only to find that I actually proved the statement in order to obtain the contradiction. It's a good way to start reasoning, but if I can remove it in the end, I usually prefer that. $\endgroup$ – Jeremy West Oct 13 '13 at 5:36
  • $\begingroup$ Thanks for you suggestion :) $\endgroup$ – Jose Antonio Oct 13 '13 at 5:38

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