Show $\exists x \in \Bbb R$ that is unique, such that $ \forall a \in A$ and $ \forall b \in B$ $a\leq x\leq b$.

Question

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.

A unique real between any two different rational numbers? Seems unlikely — Laurent Duval, Aug 02 '18 at 09:11
"there is a unique real number between any two rational numbers" is not true of course. — Mark, Aug 02 '18 at 09:13
Sorry, but why not? Could you counter it. (unless you're provoking my use of in-between and $\leq$ or $\geq$ — Florian Suess, Aug 02 '18 at 09:16
Between any two distinct rationals there are infinitely many real numbers. — Kavi Rama Murthy, Aug 02 '18 at 09:26

score 1 · Accepted Answer · edited Aug 10 '18 at 08:24

1

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.

edited Aug 10 '18 at 08:24

Taroccoesbrocco

16,964

answered Aug 02 '18 at 09:12

Kavi Rama Murthy

311,013

But how is the supA $\leq \forall b \in \mathbb B$ – Florian Suess Aug 02 '18 at 09:22
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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. – Kavi Rama Murthy Aug 02 '18 at 09:24
Oh yes, by definition. Okay, I'm still ravelling through this. – Florian Suess Aug 02 '18 at 09:25
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? – Florian Suess Aug 02 '18 at 09:29
I've gone to follow this https://math.stackexchange.com/questions/444681/proof-that-for-any-interval-a-b-with-ab-in-the-real-numbers-contains-both-rat – Florian Suess Aug 02 '18 at 09:32
Fine. I hope my answer is convincing enough now. – Kavi Rama Murthy Aug 02 '18 at 09:35

Show $\exists x \in \Bbb R$ that is **unique**, such that $ \forall a \in A$ and $ \forall b \in B$ $a\leq x\leq b$.

1 Answers1

Show $\exists x \in \Bbb R$ that is unique, such that $ \forall a \in A$ and $ \forall b \in B$ $a\leq x\leq b$.