I am trying to follow Rudin's proof that $\mathbb{R} = \{\text{set ot cuts}\}$ has the least upper bound property. Here is the set up:
$A$ is a nonempty subset of $\mathbb{R}$ (so it's some set of cuts) and $\beta$ is an upper bound of $A$. We define
$$\gamma := \bigcup\limits_{\alpha \in A} \alpha.$$
The claim is that $\gamma = \sup A$. I was able to easily show that $\gamma \in \mathbb{R}$ by demonstrating that it satisfies the three criteria for a cut (non-trivial, closed downward, and possesses no maximal element) and that $\gamma$ is an upper bound of $A$. I have having trouble demonstrating that $\gamma$ is the least of the upper bounds. Here is a copy of Rudin's proof.
Suppose $\delta < \gamma$. Then there is an $s \in \gamma$ and that $s \not \in \delta$. Since $s \in \gamma$, $s \in \alpha$ for some $\alpha \in A$. Hence $\delta < \alpha$, and $\delta$ is not an upper bound of $A$. This gives the desired result: $\gamma = \sup A$.
I understand that Rudin is attempting to prove the contrapositive of the statement that "if $\delta$ were an upper bound of $\gamma$, that $\delta$ is greater than (properly contains) $\gamma$" by demonstrating that if $\delta$ is less than (is properly contained in) $\gamma$, then it is not an upper bound. The first sentence about some element (rational number) $s \in \gamma$ makes sense by the definition of a proper subset. If $s \in \gamma$, where $\gamma$ is the set of $\alpha$, it also makes sense that $s \in \alpha$ for some $\alpha$. I do not understand why this gives $\delta < \alpha$. How do we know that all of $\delta$ is properly contained in some $\alpha$? Rather, what if $\delta$ is "spread out" amongst different $\alpha_i \in A$? Does this have something to do with the fact that cuts are closed downward? I am trying to prove this to myself without invoking the notion of a "maximal element" (because none exists, and I'm really trying to think of a "supremum," which I don't yet know exists.)
Any help on this would be appreciated.
Best Answer
Yes, it has to do with the fact that the cuts are downward-closed, though only a bit indirectly. You have an $\alpha\in A$ such that $s\in\alpha\setminus\delta$, so $\alpha\subsetneqq\delta$, and this is precisely the definition of $\alpha<\delta$ for cuts.