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 $\boldsymbol r$:
- $\boldsymbol r$ is a proper, non-empty subset of $\mathbb{Q}$
- for every $p,q\in\mathbb{Q}$, if $q\in \boldsymbol r$ and $p<q$, then $p\in\boldsymbol r$
- 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.
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, $0$.