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At the end of chapter 1 of Principles of Mathematical Analysis, Rudin provides a proof of the construction of real numbers. The first step in the proof is to define members of $\mathbb{R}$ that are subsets of $\mathbb {Q}$ known as cuts. Rudin gives the following definition of a cut:

  1. $\alpha$ is not empty, and $\alpha \neq \mathbb{Q}$.
  2. If $p \in \alpha, q \in \mathbb {Q}$, and $q < p$, then $ q \in \alpha$
  3. If $p \in \alpha$, then $p < r$ for some $r \in \alpha$.

The letters $p, q, r, \ldots$ will always denote rational numbers, and $\alpha, \beta, \gamma, \ldots$ will denote cuts.

My question is the following: Doesn't the definition of the cut provided lead to the conclusion that each cut contains all rational numbers?

In other words, that $\alpha = \mathbb {Q}$ for all $\alpha$?

The third property of cuts provided shows that there is no maximal element in a cut. The second property implies that given an element in a cut, all rational numbers less than that element are in the cut. Doesn't this clearly imply that cuts include all the rationals?

I was hoping someone could clear this up. Thanks.

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The set of all strictly negative rationals ought to be a cut, and it doesn't equal $\mathbb Q$. How do you think it fails to satisfy the definition? –  Henning Makholm Feb 4 '13 at 21:30
    
Not every linear ordering shares the property of $\mathbb{N}$ that every nonempty subset with no greatest element is unbounded. –  Trevor Wilson Feb 4 '13 at 21:34
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@user1709828: It's clearly relevant and appropriate to tell the asker that this construction in generally credited to Dedekind (under the particular name "Dedekind cuts"). What made your initial comment sound off was that it looked like the OP was in error for not already knowing this. –  Henning Makholm Feb 4 '13 at 21:37
    
One way to read (3) is: "There is no strict maximum element of a cut." –  Thomas Andrews Feb 4 '13 at 21:40
    
I think you just have to realize that a nonempty set of numbers can be bounded above and still have no maximum. –  Michael Greinecker Feb 4 '13 at 21:48
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1 Answer

up vote 1 down vote accepted

No, of course not. For example, the set of negative rational numbers are a cut.

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But it appears to me that this is not a cut according to the three properties given to me in the question. For example, if $p = 0$ in the third property, there is no rational number $r$ that is greater than $p$ and within the set of negative rational numbers. –  Aditya Gudibanda Feb 4 '13 at 21:34
    
$p=0$ is not a negative rational number, @user60994 –  Thomas Andrews Feb 4 '13 at 21:36
    
@user60994: $0$ is not a negative number, so it is not in Chris’s cut. –  Brian M. Scott Feb 4 '13 at 21:36
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Chris, that should be "is a cut." A set is singular, it is not "are a cut." –  Thomas Andrews Feb 4 '13 at 21:37
    
Ah I see now... That should have been obvious. Thank you! –  Aditya Gudibanda Feb 4 '13 at 21:37
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