4

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 $\exists q$ such that $q\in x^*$ and $q\not\in y^*$. Next I associate A cut $q^*$ to $q$ but I can't seem to get a valid argument for $q^* \subsetneq x^*$ and $y^* \subsetneq q^*$ thus proving $y<q<x$.

How can I procced from there to get the desired result?

Thanks in advance,
M.

Matt Rest
  • 41

1 Answers1

2

I will proceed with the case where $x$ and $y$ are irrational. let $x^*$ and $y^*$ the associated cut to $x$ and $y$ respectively such that $x < y$, this is equivalent to $x^* \subset y^*$. Hence there exist some rational number $q$ such that $q \in y^*$ and $q \notin x^*$. Let $q^* = \left \{ p \in \mathbb{Q} | p < q\right \}$ be the associated cut to $q$.

  1. First notice that $q \notin x^* \Leftrightarrow \forall p \in x^*/ p < q$ $(*)$
    We will prove it by contradiction, Suppose that the LHS is true and the RHS is false, then $\exists p \in x^* /p \ge q$ then by definition of Dedekind cut this implies that $q \in x^*$ contradicting the LHS.
    Now suppose that the RHS hold and the LHS is false. Hence $q \in x^*$ and again by definition of Dedekind cut $\exists r \in x^* / q < r$, contradicting the RHS, hence the result.
  2. Let p some rational number such that $p \in x^*$, since $q \notin x^*$ then by $(*)$ we have $p < q$, thus $p \in q^*$, and since $p$ was arbitrarily chosen then $\forall p \in x^* \rightarrow p \in q^*$, hence $x^* \subset q^*$, note that $q^*$ and $x^*$ are note equal since $x$ is irrational.
  3. Let $r \in q^*$, since $q \in y^*$ then $r \in y^*$, and because $r$ was chosen arbitrarily then $\forall r \in q^* \rightarrow r \in y^*$, thus $q^* \subset y^*$, similarly $q^*$ and $y^*$ are not equal since $y$ is irrational.
    We conclude that if $x^* \subset y^*$ then there exist an rational $q$ such that $x^* \subset q^* \subset y^*$ which is equivalent to if $x < y$ then there exist some rational $q$ such that $x < q < y$
Predator Monarch
  • 108