Equivalence of Dedekind cuts and Dedekind left sets

Question

I am currently working on the book "Classic Set Theory" by Goldrei. Goldrei is using Dedekind left cuts or left sets, i.e. the subset $L$ of a Dedekind cut. He gives the following definition of a left cut $\boldsymbol r$:

1. $\boldsymbol r$ is a proper, non-empty subset of $\mathbb{Q}$
2. for every $p,q\in\mathbb{Q}$, if $q\in \boldsymbol r$ and $p<q$, then $p\in\boldsymbol r$
3. for every $p\in \boldsymbol r$ there exists some $q\in \boldsymbol r$ with $p<q$

In the book there is given a excercise to show that the above defined sets $\boldsymbol r$ and $\mathbb{Q}\setminus \boldsymbol r$ suffice the definition of a Dedekind cut, i.e.

A. $\boldsymbol r$ and $\mathbb{Q}\setminus\boldsymbol r$ are non-empty

B. $\boldsymbol r \cup \mathbb{Q}\setminus\boldsymbol r =\mathbb{Q}$

C. $\boldsymbol r \cap \mathbb{Q}\setminus\boldsymbol r =\emptyset$

D. every $x\in\boldsymbol r$ is less than every $y\in\mathbb{Q}\setminus\boldsymbol r$

I think I managed the excercise, but now I'm trying to prove equivalence of both definitions, i.e. a Dedekind cut suffices the definition of a Dedekind left set.

Now, the properties 1. and 2. were fairly easy but I'am stuck on 3. My main problem I think is, that I can't assume anything on the "border" between left and right set, that is the number dividing both sets. So can anybody give my a hint on how to prove property 3. given A., B., C. and D.?

Please let me know, if more context is needed. Thanks in advance.

Answer 1

You can't prove this because the two sets of properties are not equivalent. Let $R= \Bbb Q^+, L = \Bbb Q \setminus R$. Then $L$ and $R$ satisfy properties A through D, but $L$ is not a Dedekind left cut because $L$ has a greatest element; namely, $0$.