Least upper bounds and Dedekind cuts I am working an exercise in Pugh's real analysis book.

Let $b = l.u.b. S$, where $S$ is a bounded nonempty subset of $\mathbb{R}$.


(a) Given $\epsilon > 0$ show that there exists an $s \in S$ with $b - \epsilon \leq s \leq b$.


(b) Can $s \in S$ always be found so that $b - \epsilon < s < b$?


(c) If $x = A|B$ is a cut in $\mathbb{Q}$, show that $x = l.u.b. A$.

My first point of confusion is understanding the difference between (a) and (b). I can find a strict inequality immediately, which should solve both of these problems simultaneously. Here is my attempt.

Given any $\epsilon > 0$, $b - \epsilon < b$, so $b - \epsilon$ is not an upper bound of $S$. That means that there exists some element of $S$ strictly larger than it; call it $s$. Since $s \in S$ and $b$ is an upper bound of $s$, $s \leq b$. Therefore, $b - \epsilon < s \leq b$. Of course, that implies that $b - \epsilon \leq s \leq b$.

Something tells me that Pugh did not want me to solve (a) by first solving (b), but I do not know another way to solve (a) first. Contradiction, perhaps?

Suppose there dd not exist such an $s$. So for all $s \in S$. $s < b - \epsilon$ or $s > b$. The latter is impossible since $b$ is the supremum of $s$. If the former held, then $b - \epsilon$ would be an upper bound of $S$, but $b - \epsilon < b$, which contradictions the definition of least upper bound.

Is that a better proof for the purposes of this problem?
As for part (c), Pugh gives a direct proof in the textbook that I had some difficulty following, so I thought it was easier to do by contradiction. Even with that said, I am having difficult understanding exactly what a cut is or how it makes sense to talk about a cut as a supremum of a set.

Since any element $b \in B$ is an upper bound of $A$, $A$ is bounded above. By the first property of cuts, $A$ is nonempty. So $A$ has a supremum by the completeness axiom. For a contradiction, suppose that $x$ is not the least upper bound of $A$. So there exists $y = l.u.b. A$ such that $y < x$ and $y \geq a$ for all $a \in A$. By density, there exists $z$ such that $y < z < x$. Then $z > y \geq a$. Since $z > y$, $z \not \in A$, so $z \in B$. This is a contradiction.

 A: You correctly proved that there is an $s\in S$ such that $b-\epsilon<s\le b$. This does indeed imply the result in (a) and is the simplest way to prove that result, but it does not imply the result in (b): what if, for instance, $S=\{1\}$? Then $\operatorname{lub}S=1$, and no matter what $\epsilon>0$ you consider, the only possible $s\in(1-\epsilon,1]$ is $1$ itself. This sort of thing will happen whenever $S$ has an isolated maximum element. For instance, if $S=[0,1]\cup\{2\}$, then $\operatorname{lub}S=2$, and if $0<\epsilon\le 1$, the only member of $S$ in $(2-\epsilon,2]$ is $2$ itself.
I can’t judge your argument for (c) without access to Pugh’s book: it’s not clear how much of the work of constructing $\Bbb R$ you’ve already done and what properties you have available. If you’re constructing the reals from the rationals by means of Dedekind cuts, you shouldn’t really be talking about the completeness axiom: completeness of $\Bbb R$ should be a theorem. I assume that for each $q\in\Bbb Q$ you’ve identified $q$ with the cut $A_q\mid(\Bbb Q\setminus A_q)$ , where $A_q=\{r\in\Bbb Q:r<q\}$. If so, in order to show that $A\mid B=\operatorname{lub}A$ in $\Bbb R$ I would expect you to have to show that $A_q\subseteq A$ for each $q\in A$, and that if $C\mid D$ is any cut such that $A_q\subseteq C$ for each $q\in A$, then $A\subseteq C$; i.e., that $q\le x$ for each $q\in A$, and $x\le y$ for any cut $y=C\mid D$ such that $q\le y$ for all $q\in A$. (In fact you can show that $A=\bigcup_{q\in A}A_q$.)
If you’ve proved enough of the properties of $\Bbb R$ at this point, your argument can be made to work, but you have to say a bit more. Specifically, you have to use the fact that $\Bbb Q$ is dense in $\Bbb R$ (if you’ve proved it) to say that there is a rational number $z$ such that $y<z<x$; it’s not enough to know that there is a real number between $y$ and $x$. Then you know that $a\le y<z$ for all $a\in A$, so $z\notin A$ and therefore $z\in B$. But then $x<z$, which is a contradiction.
