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Define a real number to be a subset α ⊆ Q satisfying the following four properties:

  1. ∀x ∈ α, if y ∈ Q and y < x, then y ∈ α;
  2. α ≠ ∅
  3. α ≠ Q;
  4. there is no greatest element in α: ∀x ∈ α, ∃y ∈ α so that y > x.

Given a real number α, define its negative to be the set −α = {x ∈ Q : ∃a ∈ Q \ α such that x < −a}

Show that −α is a real number (i.e., that it satisfies the four properties above)

Specifically, I'm confused about how −α was defined. Is the original 'a' supposed to be postive? Why is 'a' contained in Q \ α?

Examples of how a real number satisfies the four properties would also be very appreciated! For example, in this case is it sufficient to say α ≠ ∅ because we know it contains elements less than -a?

hf55
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1 Answers1

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  1. Assume $x\in -a$ and $y\in Q$ such that $y < x$. By definition of $-a$ there is $z\in Q\setminus a$ such that $x < -z$. But then $y<x<-z$. Therefore $y\in -a$ too, by definition of $-a$.

  2. By property 3 and 1 of $a$ there is $y\in Q\setminus a$ such that for all $x\in a$ we have that $x < y$. In fact, if there is no such $y$, then $a=Q$. Take now $z < -y$. Then $z\in a$ by definition of $-a$. Therefore $-a\neq\emptyset$.

  3. By property 2 of $a$ there is $y\in a$. If $-y\in -a$ then there is $z\in Q\setminus a$ such that $-y<-z$. But then $z<y$, which would imply that $z\in a$. This is a contradiction. Therefore $-y\notin -a$. Therefore $-a\neq Q$.

To answer your other questions:

No, the exercise has nothing to to with being positive or negative, only about defining the opposite of a real number and verifying that the definition makes sense, resulting in a real number too.

The definition of $-a$ doesn't have $a\in Q\setminus a$. The two a's have different fonts.

Example of a real number defined as a Dedekind cut:

$a=\{x\in Q:\ x^2 <2\text{ or } x < 0\}$

This is a typical simple non-trivial example. This is the Dedekind cut-way of representing the number $\sqrt{2}$ (the positive square root of $2$). You can see that $0\in a$. Therefore $a\neq \emptyset$.

Also $3\notin a$, since $3^2>2$ and $3>0$. Therefore $a\neq Q$.

Finally, if $x\in a$ and $y<x$, then either $y<0$ and $y$ is in $a$, or $y>0$ but then $y^2<x^2<2$ and therefore $y\in a$. This proves property 1.

Therefore $a$ is a real number, a Dedekind cut.

You can check that $-a=\{x\in Q:\ x^2<2\text{ or } x>0\}$

  • Btw you forgot to show why the example is simple yet non-trivial. The non-trivial part is proving the fourth property of the Dedekind cut and that's the key here. – Paramanand Singh Dec 20 '17 at 22:57
  • @ParamanandSingh Property 4 is not key to the definition of Dedekind cuts. Actually, one may define them just fine by requiring that the complement doesn't have least element. The important part is the cut part (property 1), the not $-\infty$ (property 2), and not $+\infty$ (property 3). Property 4 is just cosmetic, equivalent to making a choice in decimal notation between using $1.000...$ or $0.999...$ –  Dec 21 '17 at 18:37
  • The fact that your example is non-trivial depends on this property. And anyway its part of definition. You will also note that establishing this property is not a trivial task. – Paramanand Singh Dec 21 '17 at 19:26
  • Also in your example proving that the complement has no least member is difficult. It is not cosmetic as you might be thinking. – Paramanand Singh Dec 21 '17 at 19:27
  • @ParamanandSingh Property 4 is cosmetic. You can remove completely from the definition and you still can define the real numbers. The only thing needed is to identify cuts that have maximum with the cut in which the maximum is moved to the complement. Everybody knows that you are talking about proving that square root of 2 is not rational. Doh! This, however, is irrelevant to the question posted, and it is only an aside and the reason I gave the example above, but not for me to prove. You can try it if you think it is hard. Now, you are the one that needs to learn why property 4 isn't key. –  Dec 21 '17 at 19:47
  • Let me clarify the that is not cosmetic. You must choose either of the definitions. Moreover what needs to be proved is significantly different and difficult from proving that square root of $2$ is irrational. I only wanted to highlight that you should have given a trivial example like $x$ being set of rationals less than $1$ and easily demonstrated what is asked in current question. – Paramanand Singh Dec 22 '17 at 01:35
  • @ParamanandSingh Well, you simply don't understand well the definition of Dedekind cuts. Even more, it is not either definition. There are three posibilities: One can define the cut to contain the sup, or the complement to contain the sup or neither but then identify the two cuts. That is why property 4 is quite irrelevant. I shouldn't have given any other example, besides the one I wanted. You can write your own answer, if you are capable to do so, and rant all you want. The square root of $2$ is the perfect example of a cut. –  Dec 22 '17 at 15:16
  • This is not a rant rather a mathematical discussion on a definition given in the question itself. Instead of proving property 4 for your example you are just trying to say that it is not necessary. If it were not necessary no one would provide it. And if it were trivial to do so you would have proved it by this time. BTW I agree with all your points except that this property is unnecessary or irrelevant. At least I never seen the definition of a Dedekind cut without this property. – Paramanand Singh Dec 22 '17 at 16:19
  • @ParamanandSingh Proving that $-a$ satisfies property 4 if $a$ does is beyond trivial, and definitely not key. The fact that you rant so much about it is a reflection that you don't understand it well. Try it. –  Dec 22 '17 at 16:24
  • It seems that you are not understanding what I am trying to say. And it is also bit sad to know that you have taken my words in not so positive manner whereas you could have easily taken it in proper spirit and justified that your example indeed meets the properties of a cut outlined in the question. Anyway in order to see that this is not trivial you may have a look at https://math.stackexchange.com/q/2077096/72031 – Paramanand Singh Dec 23 '17 at 04:23
  • Also apart from the point I have mentioned your answer is good and I had indicated this by an upvote right before I started discussing this stuff. No one is coming here to rant and irritate users, but the prime motive is to learn and let learn. – Paramanand Singh Dec 23 '17 at 04:27
  • @ParamanandSingh Thank you for the link. It shows how much you don't understand what you are talking about (said explicitly in the link where they call you by name). The proof is absolutely trivial and doesn't require anything done in those links. For your information, and since I see you plastering links to that answer of yours all around, the way to prove it is just straightforwardly imposing the conditions one wants and that's all. One of the many easy ways –  Dec 25 '17 at 22:31