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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

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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.

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    $\begingroup$ If we had $q-y\in\alpha,$ then there would be $x\in\alpha$ such that $q-y=x,$ right? But then what could we say about $q$? $\endgroup$ – Cameron Buie Mar 5 '14 at 20:48
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    $\begingroup$ As for why $\alpha^*$ cannot be empty or all of $\Bbb Q,$ you can show that among downward-closed subsets $A\subseteq\Bbb Q,$ we have $A^*=\emptyset$ (respectively, $A^*=\Bbb Q$) if and only if $A=\Bbb Q$ (respectively, $A=\emptyset$). As hint for one direction of one of these results: if $A^*=\emptyset,$ then for all $x,r\in\Bbb Q$ with $r>0,$ we have $-x-r\in A.$ From this, it follows that $A$ is unbounded above (why?), and so.... $\endgroup$ – Cameron Buie Mar 5 '14 at 20:59
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    $\begingroup$ Your reasoning for $\alpha\subseteq 0^*$ isn't quite right. Rather, for all $y\in\alpha,$ we have $q-y\notin\alpha,$ so since $q<0$ then for all $y\in\alpha$ we have $q-y<-y,$ and since $\alpha$ is downward-closed, it follows that $-y\notin\alpha$ whenever $y\in\alpha.$ Can you see from there how to conclude that $\alpha\subseteq0^*$? $\endgroup$ – Cameron Buie Mar 5 '14 at 21:11
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    $\begingroup$ I have updated my answer with an alternative approach. The result $0^*\subseteq\alpha+\alpha^*$ is non-trivial. You'll run into similar issues when dealing with multiplicative inverses, and again the Lemma above will be useful. $\endgroup$ – Cameron Buie Mar 5 '14 at 22:04
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    $\begingroup$ Since $q<0,$ then $q-y<0-y=-y.$ That reasoning (for why $0>y$ for all $y\in\alpha$) is just fine. The Archimedean property of reals does indeed follow from the completeness property, but completeness isn't actually necessary. Your proof is just fine, and shows that for all positive $p,q\in\Bbb Q$ there exists some $n\in\Bbb N$ such that $np>q$. $\endgroup$ – Cameron Buie Mar 6 '14 at 13:02
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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 $0^*\subset \alpha+\alpha^*$.

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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.

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  • $\begingroup$ Really? Why is this being downvoted? $\endgroup$ – David Jun 28 '14 at 5:07
  • $\begingroup$ I have taken the liberty of improving the formatting of your post, though I tried to avoid changing the text or thrust of your approach at all. Please look it over to make sure I've been faithful to your intent. $\endgroup$ – Cameron Buie Aug 14 '15 at 3:47
  • $\begingroup$ As for why your post was downvoted, I can't say for sure. I only just discovered it, myself, but I see several issues with your proof. For example, what is $n$? It doesn't seem to have anything to do with anything, except that you refer to it in your reference to the "simple lemma," so I must suspect that it is supposed to be connected somehow. For another example, you seem to be attempting to define $M,N,$ and $Z$ twice--the first time, $M$ and $Z$ seem to be arbitrary Dedekind cuts, and $N$ defined in terms of $M$. (cont'd) $\endgroup$ – Cameron Buie Aug 14 '15 at 3:53
  • $\begingroup$ The second time, $M$ is still arbitrary, but $Z$ is very specific, and $N$ is still defined in terms of $M,$ but in a non-equivalent way! Another issue: this result is attempting to prove that the real numbers exist (in the set-theoretic sense), so what is $m$? Another issue is in your statement that $\Bbb Q\setminus M>M.$ What does that even mean? The obvious interpretation is that every element of $\Bbb Q\setminus M$ is greater than every element of $M,$ but then you go on to say that this is implied by the fact that $\Bbb Q\setminus M>M,$ so it's rather confusing. (cont'd) $\endgroup$ – Cameron Buie Aug 14 '15 at 3:58
  • $\begingroup$ Another issue is that $q_2$ shows up rather randomly in your arguments. One suspects that it's probably intended to be an arbitrary element of $N,$ but it's good practice to make this explicit. Your biggest issue is when you conclude that since $q_1+q_2<0$ for all $q_1\in M$ and all $q_2\in N,$ then "we can redefine $M+N$ as $M + N = \{q_1 + q_2 ∈\Bbb Q : q_1 + q_2 < 0\}.$" This is faulty reasoning (though the conclusion is true). For example, if we take $A=\{q\in\Bbb Q:q<-1\},$ then $q+r<0$ for all $q\in A$ and all $r\in Z.$ (cont'd) $\endgroup$ – Cameron Buie Aug 14 '15 at 4:04

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