Dedekind Cut additive inverse

Question

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

Answer 1

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.

Answer 2

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 $0^*\subset \alpha+\alpha^*$.

Answer 3

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

$$v=nw+p\in \alpha + \alpha ^*.$$.

Answer 4

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.