Dedekind Cut additive inverse Let $\alpha$ be Dedekind cut and define $\alpha^* :=\{x\in\mathbb{Q}|\exists r>0\space \text{such that} -x-r\notin\alpha\}$. I need to show that $\alpha^*$ is a Dedkind Cut and the additive inverse of $\alpha$. Note the the additive identity is defined as $0^*:=\{x\in\mathbb{Q}|x<0\}$.  Any help is greatly apprecyed and needed. Thanks in advance  
For the 1st part I can show that $\alpha^*$ is downwardly closed as if $y\in\mathbb{Q}$ and $x\in\alpha^*$ with $y<x$ then for some $r>0$, $-x-r>a$,  $\forall a\in \alpha$, but as $-y>-x$, we have $-y-r>-x-r>a$ $\forall a \in\alpha $ i.e. $-y-r\notin \alpha$ so $y\in\alpha^*$.  But I am stuck showing that $\alpha^*$ has no top element and that $\alpha^*\neq\emptyset$ and $\alpha^*\neq\mathbb{Q}$, so any help please.
For the second part I can show that $\alpha+\alpha^*\subset 0^*$ as if $w\in\alpha+\alpha^*$ then $w=u+v$ where $u\in\alpha$  &  $v\in\alpha^*$. Therefore for some (rational) $s>0$ 
$-v-s>u$ that is $u+v<-s<0$.  But I am stuck with the reverse inclusion.  So please any help will be greatly appreciated and please let me know if what I've done is correct and on the right track.  Thanks
 A: Well, for the first, suppose that $x_0\in\alpha^*.$ Then by definition, there is some rational $r>0$ such that $-x_0-r\notin\alpha.$ But then $\frac r2$ is also a positive rational, and $x_1=x_0+\frac r2$ is rational and greater than $x_0,$ and $-x_1-\frac r2=-x_0-r\notin\alpha.$ Thus, for any $x\in\alpha^*,$ we can find a $y\in\alpha^*$ with $x<y$.
For the second, we will need the following result:

Lemma:  Given a Dedekind cut $\alpha$ and any $r\in\Bbb Q$ with $r>0,$ there exist $a,b\in\Bbb Q$ with $a\in\alpha$ and $b$ a non-least element of $\Bbb Q\setminus\alpha$ such that $0<b-a<r$.

To prove this we let $z_0$ be the least element of $\Bbb Q\setminus\alpha$--if there is such an element. We define a function $m:\Bbb Q\times\Bbb Q\to\Bbb Q$ by $$m(x,y)=\frac{x+y}2,$$ we fix any $a_0\in\alpha,$ and any non-least $b_0\in\Bbb Q\setminus\alpha.$ Then we define sequences $\langle a_n\rangle_{n=0}^\infty$ and $\langle b_n\rangle_{n=0}^\infty$ recursively as follows:
(1) If $m(a_n,b_n)=z_0$ for some $n,$ then for all integers $k\ge n$ we let $a_{k+1}=m(a_k,z_0)$ and $b_{k+1}=m(z_0,b_k).$
(2) If $m(a_n,b_n)\in\alpha,$ let $a_{n+1}=m(a_n,b_n)$ and let $b_{n+1}=b_n.$
(3) If $m(a_n,b_n)$ is a non-least element of $\Bbb Q\setminus\alpha,$ then let $a_{n+1}=a_n$ and $b_{n+1}=m(a_n,b_n).$
It can be shown that $m$ is well-defined, that $\langle a_n\rangle_{n=0}^\infty$ is a well-defined sequence of elements of $\alpha,$ that $\langle b_n\rangle_{n=0}^\infty$ is a well-defined sequence of non-least elements of $\Bbb Q\setminus\alpha,$ and that for all integers $n\ge 0$ we have $$0<b_0-a_0=2^n\cdot(b_n-a_n).$$ By the Archimedean Property of the rationals, there exists some positive integer $n$ such that $n\cdot r>b_0-a_0,$ so since $2^n>n>0,$ we have $$2^n\cdot r>n\cdot r>b_0-a_0=2^n\cdot(b_n-a_n)>0,$$ whence $$0<b_n-a_n<r,$$ as desired.
I leave the details to you (unless, of course, you already have that result). Now, to show that $0^*\subseteq\alpha+\alpha^*,$ take any $q\in0^*,$ and put $r=-q,$ so $r\in\Bbb Q$ and $r>0.$ By the Lemma, there exist some $a\in\alpha$ and some non-least $b\in\Bbb Q\setminus\alpha$ such that $0<b-a<r.$ Then $q=-r<a-b=a+-b.$ It remains only to show that $-b\in\alpha^*,$ which again I leave to you.
A: We first show that $\alpha^*$ has no maximal element: Take $x\in\alpha^*$. 
Then there exists $r\in\Bbb{Q}_{>0}$ such that $-x-r\notin\alpha$. But $\frac r2>0$ and $y=x+\frac r2>x$ 
 and $-y-\frac r2=-x-r\notin\alpha$, hence $y\in\alpha^*$ and $x$ is not a maximal element.
$\alpha^*$ is non empty: Take any element $z\in \Bbb{Q}\setminus \alpha$ and $r>0$, then $x=-z-r$ satisfies $-x-r=z\notin\alpha$ and so $x\in\alpha^*$.
$\alpha^*\ne \Bbb{Q}$: For any $y\in \alpha$, the element $x:=-y\notin \alpha^*$, since $-x-r<y\in \alpha$ implies $-x-r\in\alpha$, for all $r>0$.
Finally we prove that $0^*\subset \alpha+\alpha^*$. Assume by contradiction that 
there is $r_0<0$ such that $r_0\notin \alpha+\alpha^*$. Then for all $a\in\alpha$, $x_0\in \Bbb{Q}\setminus\alpha$ we have $r_0-a+x_0\ge 0$. In fact, if $r_0-a+x_0< 0$
then $r_0=a-x_0-r$ with $a\in\alpha$, $x_0\in\Bbb{Q}\setminus\alpha$ and 
$r=-r_0+a-x_0>0$, hence $-x_0-r\in\alpha^*$ and $r_0\in \alpha+\alpha^*$.
From $r_0-a+x_0\ge 0$ it follows that for all $x_0\in\Bbb{Q}\setminus \alpha$, we have $x_0+r_0\ge a$ for all $a\in \alpha$, hence $x_0+r_0\notin\alpha$ (otherwise it would be a maximal element). 
So we have that for all $x_0\in\Bbb{Q}\setminus \alpha$, $x_0+r_0\in\Bbb{Q}\setminus \alpha$. But then, by induction, we have that for all $x_0\in\Bbb{Q}\setminus \alpha$ and $n\in\Bbb{N}$, $x_0+nr_0\in\Bbb{Q}\setminus \alpha$.
This is a contradiction, since for any fixed $x\in \alpha$, $x_0\in\Bbb{Q}\setminus \alpha$, there exists an $n\in\Bbb{N}$ such that $x_0+nr_0<x$ which implies $x_0+nr_0\in \alpha$. This contradiction shows that $r_0\in \alpha+\alpha^*$ and so $0^*\subset \alpha+\alpha^*$.
A: Here is a short proof of the inclusion $0^*\subseteq \alpha + \alpha$ from Rudin's Principles of Mathematical Analysis which uses the Archimedean Property on the rationals:
Let $v\in 0^*$, so that $v<0$. Let $w=-v/2>0$. There exists an integer $n$ such that $nw\in \alpha$, but $(n+1)w\notin \alpha$ (Archimedean property). Let $p=-(n+2)w$, so that there exists $r>0$, namely $r=w$, such that $-p-r\notin \alpha$. Thus $p\in \alpha ^*$, and
$$v=nw+p\in \alpha + \alpha ^*.$$.
A: I actually found an easier proof of this without requiring the above lemma.  So, let $M,$ $N,$ and $Z$ be Dedekind cuts.  Let $N$ be defined as containing the elements of $\Bbb Q$ such that the elements are negative and not in $M.$  It is easy to show the $M + N$ consists of rationals that are negative and less than zero.  
$\Bbb Q$ and $\Bbb R$ are the respective number systems
Let $m,n ∈\Bbb R$
Define:
$M = \{q_1 ∈\Bbb Q : q_1 < m\}$
$N = \{q_2 ∈\Bbb Q : \text{for some rational }r ∈\Bbb Q\setminus M, q_2 < -r\}$
$M + N = \{q_1 + q_2 : q_1 ∈ M\text{ and }q_2 ∈ N\}$
$Z = \{q_3 ∈\Bbb Q : q_3 < 0\}$
$\Bbb Q\setminus M,$ by the way, is another way of saying $\Bbb Q-M,$ or subtract set $M$ from $\Bbb Q$ 
It can be shown that $\Bbb Q\setminus M$ is $> M$ (just do a simple proof by contradiction).  So, this implies that for any $r ∈\Bbb Q\setminus M,$ $r > m > q_1,$ for any $q_1$ in $M.$  
Thus, $q_2 < -r \implies -q_2 > r > m > q_1.$  So, $-q_2 > q_1$ and $0 > q_2 + q_1.$
So, for any $q_1 + q_2 ∈ M + N,$ $q_1 + q_2 < 0.$  That is, we can redefine $M + N$ as 
$M + N = \{q_1 + q_2 ∈\Bbb Q : q_1 + q_2 < 0\}$
Now, the strategy is this, to show that any element in $Z$ (which contains negative rationals less than zero) must also be in $M + N.$  This is now pretty easy.  So, let 
$z ∈ Z.$  So, $z < 0.$  If $z \leq q_1 + q_2,$ then we know $Z$ is a proper subset of $M + N$ (this comes from either the definition of Dedekind cuts or a simple lemma to show that $M$ is a subset of $N$ if $m < n,$ and vice versa).  So, for this case, we are done.  Suppose, on the contrary, that $q_1 + q_2 < z.$  Then, since $M + N$ is also a Dedekind cut, there must be a larger element $p = q_3 + q_4$ in $M + N$ such that $z < p.$  $\frac{q_1 + q_2}{2}$ is one such example.  It is less negative, and therefore larger, than $q_1 + q_2.$ It is also another quotient and belongs in $M + N.$  Continuing this train of thought, $\frac{q_1 + q_2}{2^k}$ must be larger than the $\frac{q_1 + q_2}{2^{k-1}}.$  So, there must be a $k$ such that $z \leq \frac{q_1 + q_2}{2^k}.$  So, for any $z,$ $z \leq p,$ for some $p$ in $M + N.$  Thus, $Z$ must be a proper subset of $M + N,$ and we are done.
I personally found this proof to be easier to follow that Cameron's proof.     
