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


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

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


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 $m,n ∈\Bbb R$


$M = \{q_1 ∈\Bbb Q : q_1 < m\}$

$N = \{q_2 ∈\Bbb Q : \text{for some rational }r ∈\Bbb Q\setminus M, q_2 < -r\}$

$M + N = \{q_1 + q_2 : q_1 ∈ M\text{ and }q_2 ∈ N\}$

$Z = \{q_3 ∈\Bbb Q : q_3 < 0\}$

$\Bbb Q\setminus M,$ 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.

  • $\begingroup$ 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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