Rudin & Dedekind Cuts 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:

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.
 A: 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.
A: 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!
