Conditions: $A,B \subset \mathbb Q$ such that $A,B \neq \emptyset$ moreover $A\cup B = \mathbb Q$ with the condition $\forall a \in A$ and $ \forall b \in B$ we have $a\leq b$.

I reckon I can describe the situation more plainly as; show that there is a unique real number between any two rational numbers.

My work: I have been bouncing the idea of using the fact that there exists a supremum for $A$ which lies in $B$, although that's rather obvious and I've found no prevail from there. From my experience a problem like this that involves proving uniqueness can be usually solved easily using a proof by contradiction. But again... I'm not progressing. Any help/pointers will be much appreciated, many thanks.

  • $\begingroup$ A unique real between any two different rational numbers? Seems unlikely $\endgroup$ – Laurent Duval Aug 2 '18 at 9:11
  • $\begingroup$ "there is a unique real number between any two rational numbers" is not true of course. $\endgroup$ – Mark Aug 2 '18 at 9:13
  • $\begingroup$ Sorry, but why not? Could you counter it. (unless you're provoking my use of in-between and $\leq$ or $\geq$ $\endgroup$ – Florian Suess Aug 2 '18 at 9:16
  • 1
    $\begingroup$ Between any two distinct rationals there are infinitely many real numbers. $\endgroup$ – Kavi Rama Murthy Aug 2 '18 at 9:26
  • 1
    $\begingroup$ See en.wikipedia.org/wiki/Dedekind_cut $\endgroup$ – lhf Aug 2 '18 at 11:27

Let $x=\sup A$. Since $a \leq b$ for all $a \in A$, for all $b \in B$ we get $a \leq b$ for all $b \in B$. Of course $a \leq x$ for all $a \in A$ by definition of supremum. If $y$ is another real number with the same properties then $ a\leq y$ for all $a \in A$ so $x \leq y$. If possible let $x<y$ . Let $r$ be a rational number in $(x,y)$ Then $r \in A$ or $r \in B$. In the first case $x<r \in A$ contradicting the definition of $x$. In the second case there is a member of $b$ (namely $r$) less than $t=y$ which is again a contradiction.

  • $\begingroup$ But how is the supA $\leq \forall b \in \mathbb B$ $\endgroup$ – Florian Suess Aug 2 '18 at 9:22
  • 2
    $\begingroup$ By definition of supremum. Since each member of $A$ is less than or equal to $b$ ( which means $b$ is an upper bound for $A$ we see that $x \leq b$ because $x$ is the least upper bound. $\endgroup$ – Kavi Rama Murthy Aug 2 '18 at 9:24
  • $\begingroup$ Oh yes, by definition. Okay, I'm still ravelling through this. $\endgroup$ – Florian Suess Aug 2 '18 at 9:25
  • $\begingroup$ This is awesome, you've helped me twice today. I really appreciate it. Question though, given an open set of distinct x and y, how can you be sure there exists a rational in between? Where is this reasoning coming from? $\endgroup$ – Florian Suess Aug 2 '18 at 9:29
  • $\begingroup$ I've gone to follow this math.stackexchange.com/questions/444681/… $\endgroup$ – Florian Suess Aug 2 '18 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.