Proof of rational density using Dedekind cuts

[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.

 

[HallsofIvy]

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

 

[loops496]

Thank HallsofIvy for replying.
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.

 

[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 $x^*$.

 

[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.

 

[loops496]

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