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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.

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Let $r\in \delta$. Then $r<s$ (if not then $r\ge s$ and by definition of a cut, $s\in \delta$). Since $s\in \alpha$, we have $r\in \alpha$ again by definition of a cut. Thus $\delta \subset \alpha$.

Note that we proved essentially the following: let $\alpha, \beta$ be two cuts, then either one of the following holds:

$$\alpha \subset \beta, \beta \subset \alpha, \alpha = \beta.$$

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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.

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The definition of a cut is that if $q\in \delta$ then all $q' < q$ we have $q'\in \delta$. So "spreading out" $\delta$ can't work.

This property of cuts that if $w$ is in a cut then every rational $\le w$ is also in a cut makes this inevetible.

If $\delta \subsetneq \gamma$ then there is $s\in \gamma; s\not \in \delta$. Now there can not be any $r \in \delta$ where $r \ge s$ because that would mean $s \le r\in \delta$ so $s \in \delta$. And as all $r\in \delta$ have $r < s$ that means if $s \in \alpha$ we must have $r < s \in \alpha$ so $r\in \alpha$ and so $\delta \subsetneq \alpha$.

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Do you know the definition of $\alpha<\beta$ where $\alpha, \beta$ are cuts? Well, the definition is that $\alpha$ must be a proper subset of $\beta$ and you can/should confirm that this is equivalent to "there is a member in $\beta$ which is not in $\alpha$". This does not work for testing proper containment of general sets, but it does work for cuts because they are special in that they are downward closed.

Note that this fact has already been used at the beginning of the argument where you write "suppose $\delta<\gamma$. Then there is an $s$ such that $s\in\gamma$ and $s\notin\delta$". As noted earlier this is an equivalent condition and thus works in reverse and hence if $s\in\alpha$ and $s\notin\delta $ then $\delta<\alpha$.

Once you understand that order relations of cuts are nothing but proper inclusion of sets then you have no trouble understanding all of this. This simplicity is what makes the Dedekind's construction of reals the easiest to deal with (well, there is some cost in dealing with field operations like $+, -, \times, /$, but the actual goal in the construction of reals is not the field operations, but the gem of completeness).

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