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I need help with my confusion. Here is a page from Rudin's Principles of Mathematics Analysis where he will be proving the multiplication axioms:

P.19

My question is what does $\alpha > 0$* mean? Where $0$* was defined to be the set of all negative rationals.

I'm not sure how to think of real numbers at the moment: sets or just in the every day sense? See, if I were to think of them as sets $\alpha > 0$* means that $0$* is a proper subset of $\alpha$ - therefore meaning that $R^+$ contains negative rationals and $0$. But if I were to think of the inequality in the every day sense $R^+$ would only contain real numbers greater than zero which is what I'm pretty sure we want.

Any help is appreciated.

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  • $\begingroup$ Since not everyone has a copy of Rudin lying around, it would be much better if you could provide us with a complete, self-contained question - i.e. just a bit more context. $\endgroup$ – icurays1 Apr 12 '13 at 16:54
  • $\begingroup$ You might consider accepting an answer if you've found it useful, and feel it properly answers your question. $\endgroup$ – Pedro Tamaroff Apr 15 '13 at 18:45
  • $\begingroup$ @PeterTamaroff I chose your answer. Buie's comment actually answered the question I meant to ask (I voted it up), but the way the question stands at the moment I thought your answer was very enlightening. Thanks. $\endgroup$ – user70962 Apr 17 '13 at 11:35
  • $\begingroup$ I'm glad it helped. $\endgroup$ – Pedro Tamaroff Apr 17 '13 at 13:47
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If $0^*<\alpha$ then the cut $\{r<0\}$ is properly contained in $\alpha$. This means $\alpha$ contains rationals that are positive. Since we define real numbers as nonempty proper subsets of $\Bbb Q$ that are not bounded below, every real number $\alpha$ contains a "negative tail". We want to call the reals that also contain a positive "part" of $\Bbb Q$ positve, and call negative those which consist only of negative rationals (and are not $0^*$), that is those who are properly contained in $\{r<0\}$. Do you see?


What we're doing by defining $\alpha < \beta$ by saying that $\alpha\subsetneq \beta$ is producing an ordering of these cuts we have defined. If we define $\alpha\leq \beta$ by $\alpha\subseteq \beta$ (that is $\alpha <\beta$ or $\alpha=\beta$) what we obtain is a relation $ \leq $ with the following properties:

$(1)$ It is symmetric. ($\alpha\leq \alpha$ always holds.)

$(2)$ It is transitive ($\alpha\leq \beta,\beta\leq \gamma$ implies that $\alpha\leq \gamma$)

$(3)$ It is anti-symmetric. ($\alpha\leq \beta,\beta\leq \alpha\implies \alpha=\beta$)

$(4)$ It is total, in the sense that we can always show $\alpha\leq\beta$ or $\beta\leq \alpha$ hold.

(Note that $(4)$ actually steps on $(1)$ since we're saying by choosing $\alpha=\alpha$ that $\alpha\leq\alpha$ or $\alpha \leq \alpha$ hold, which means $\alpha \leq \alpha$ always holds.)

This means $\leq$ is what you're usually used to call a (total) order. $\Bbb N$, $\Bbb Z$, $\Bbb Q$ are totally ordered sets, with the usual $\leq$. What we do above is successfully extend the notion of "is less than" to the real numbers we're building up.

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  • $\begingroup$ (+1): Emphasis on the "negative tail" is a good point. $\endgroup$ – Cameron Buie Apr 12 '13 at 17:44
  • $\begingroup$ @CameronBuie & PeterTamaroff: Thanks for the two very good answers. I think though I have been slightly misunderstood because I haven't been clear in my question. I understand Dedekind cuts and what it means for a cut to be positive and both of your answers make it even easier to understand. My problem is how do I reconcile the usual way of thinking of the real numbers with Dedekind cuts. For example, in my question Rudin introduces $R^+$ to be the set of all $\alpha > 0^*$. Continued in next post.. $\endgroup$ – user70962 Apr 12 '13 at 19:03
  • $\begingroup$ To my understanding it doesn't really make sense because a cut is a set which always contains negative rationals and therefore, $R^+$ has negative rationals which isn't true - well, at least it isn't the definition of $R^+$ I'm used to. So I'm confused.. $\endgroup$ – user70962 Apr 12 '13 at 19:05
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    $\begingroup$ @Bryan $\Bbb R^+$ in this context does not contain negative rationals. The elements of $\Bbb R^+$ contain negative rationals, but $\Bbb R^+$ is a set of sets of rationals, not a set of rationals. Later, we'll identify certain elements of $\Bbb R$ (in the Dedekind cut sense) with rational numbers, using the map $\Bbb Q\to\Bbb R$ given by $q\mapsto\{r\in\Bbb Q:r<q\},$ but again, for the moment, just think of $\Bbb R$ as a set of Dedekind cuts. You'll come to see that it has all the nice properties that we expect from our "everyday" version of $\Bbb R$. $\endgroup$ – Cameron Buie Apr 12 '13 at 19:10
  • $\begingroup$ Ahh yes, you're right. I'll stick to thinking of real numbers as sets and continue reading. Thanks a lot for the help from both of you! $\endgroup$ – user70962 Apr 12 '13 at 19:49
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For the moment, think of them as sets. What Rudin is showing is that the reals (thought of this way) have all the properties that we're used to. You're absolutely right that $\alpha>0^*$ means that $0^*$ is a proper subset of $\alpha$, which (since $\alpha$ is the left side of a Dedekind cut) does mean that $\alpha$ contains $0$ (along with all the negative rationals, since it's a superset of $0^*$). In the sense of Dedekind cuts, we are saying that $\alpha$ is a positive real number, even though it is a set containing all non-positive rationals--in fact, precisely because it contains all non-positive rationals!

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  • $\begingroup$ You mean "even though it is a set containing all negative rationals, yes"? $\endgroup$ – Pedro Tamaroff Apr 12 '13 at 17:05
  • $\begingroup$ Oops! I actually meant "non-positive". Thanks, though. $\endgroup$ – Cameron Buie Apr 12 '13 at 17:24

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