Understanding the steps in Pugh's proof about the Least Upper Bound Property

Question

2 Theorem

The set $\Bbb R$, constructed by means of Dedekind cuts, is complete in the sense that it satisfies the

Least Upper Bound Property:If $S$ is a nonempty subset of $\Bbb R$ and is bounded above then in $\Bbb R$ there exists a least upper bound for $S$.

Proof: Easy! Let $\mathcal C\subset\Bbb R$ be any nonempty collection of cuts which is bounded above, say by the cut $X | Y$. Define $C:=\{a\in\Bbb Q:\text{ for some cut } A | B\in\mathcal C\text{ we have } a\in A\}$ and $D=$ the rest of $\Bbb Q$. source: Real Mathematical Analysis by Pugh.

I'm trying to make sense of the significance of the sentence beginning with "By the assumption that ...". How does Pugh go from that sentence to making the conclusion that $C$ is a subset of $C'$? How does this show that $C$ is a subset of $C'$? Doesn't the sentence still hold if we replaced every instance of $C'$ with $C$? That is, "By the assumption that $A|B\leq C|D$ for all $A|B\in ℭ$, we see that the $A$ for every member of $ℭ$ is contained in $C$? Maybe im misunderstanding the sentence's purpose but I'm not seeing its significance nor how it ties into the his conclusion that $C$ is a subset of $C'$.

Answer 1

The proof has two steps:

1. $\text{Show that } z \text{ is an upper bound}.$

2. $\text{Show that } z \text{ is less than any other upper bound.}$

It's useful to note that by the construction of $z = C|D$ we have $C = \bigcup_{A|B\in \cal{C}}A$. In words: $C$ is the union of all the $A$'s from $\cal{C}$. Recall that $A|B \leq A^{\prime}|B^{\prime}$ if and only if $A \subseteq A^{\prime}$. It follows that $A|B \leq z$ for all $A|B \in \cal{C}$, which is exactly what it means for $z$ to be an upper bound. So this takes care of step 1.

For step 2 we have to show that $z$ in addition to being an upper bound in fact is a least upper bound. We do this by considering any other upper bound $z^{\prime}$ and showing that $z \leq z^{\prime}$. We write $z^{\prime} = C^{\prime}| D^{\prime}$ and what we need to show is that $C \subseteq C^{\prime}$. We have assumed that $z^{\prime}$ is an upper bound for $\cal{C}$ hence $A|B \leq z^{\prime}$ for all $A|B \in \cal{C}$, which means that $A \subseteq C^{\prime}$ for all $A|B \in \cal{C}$. Since $C^{\prime}$ contains all the $A$s, $C^{\prime}$ also contains their union: $\bigcup_{A|B\in \cal{C}}A \subseteq C^{\prime}$. But as we noted above this union is equal to $C$. So in fact we have shown $C \subseteq C^{\prime}$, that is : $z\leq z^{\prime}$. This takes care of step 2.