Proof of rational density using Dedekind cuts

1. Aug 7, 2014

loops496

The Problem

Let $x$ and $y$ be real numbers such that $y<x$, using the Dedekind cut construction of reals prove that there is always a rational $q$ such that $y<q<x$

What I've done

Since I can associate a cut to every real number, let $x^∗$ be the cut associated to $x$ and $y^∗$ the one associated with $y$.
Since $y<x \implies y^* \subsetneq x^*$ then $\exists q \in \Bbb Q$ such that $q \in x^*$ and $q\not\in y^*$. Next I associate a cut $q^∗$ to $q$. Now how can I deduce from there that $q^* \subsetneq x^*$ and $y^* \subsetneq q^*$, thus proving $y<q<x$

Any help will be appreciated,
M.

Last edited: Aug 7, 2014
2. Aug 7, 2014

HallsofIvy

Staff Emeritus
If you are simply asking if that is a valid proof, yes, it looks good to me.

3. Aug 7, 2014

loops496

The thing is that my professor told me that its not immediate from what I previously wrote that $q^* \subsetneq x^*$ and $y^* \subsetneq q^*$ so I can't conclude solely on my argument $y<q<x$. But I can't see what I'm missing, I was hoping you guys would guide me towards finding that subtlety to complete the proof.

4. Aug 7, 2014

jbunniii

I'm assuming that in your definition of Dedekind cut, if $q$ is rational then $q$ is NOT an element of $q^*$ but rather is the smallest element of $\mathbb{Q}\setminus q^*$. If you use the opposite convention, then adjust my argument below appropriately.

If $x$ and $y$ are irrational, then your proof is fine.

If $x$ and $y$ are rational, then you can simply choose $q = (x+y)/2$.

If $y$ is irrational and $x$ is rational, then your proof is fine: $q$ cannot be $x$ because you chose $q \in x^*$ whereas $x \not\in x^*$.

Finally, if $y$ is rational and $x$ is irrational, then your proof does not exclude the possibility that $q=y$. In other words, $q$ might be the smallest element of $x^* \setminus y^*$. But if this is the case, note that $x^*$ does not have a largest element, so you can simply replace $q$ with a larger rational in $x^*$.

5. Aug 8, 2014

disregardthat

Let the real numbers x and y be represented by cuts $(X_1,X_2)$ and $(Y_1,Y_2)$, where a cut $(A,B)$ is a partition of $\mathbb{Q}$ such that any element of A is less than any element of B, and A has no greatest element. To say that $y < x$, is to say that $Y_1 \subset X_1$ by the ordering of cuts. Since this is a proper subset, we may find a rational $p \in X_1 - Y_1$. But since $X_1$ has no greatest element, we may find another rational $q \in X_1-Y_1$ such that $q>p$. It is easily seen that its cut $(Q_1,Q_2)$ is such that $Y_1 \subset Q_1 \subset X_1$, because if we let the cut $(P_1,P_2)$ represent p, we have the following: $Y_1 \subseteq P_1 \subset Q_1 \subset X_1$, where the last inclusion is true because q cannot be the greatest element of $X_1$.

6. Aug 9, 2014

loops496

Oh, now I get it. Thanks for the replies and the help guys, it was very helpful.