I'm currently reading through Pugh's Analysis, and just finished reading the proof of the Least Upper Bound property of the reals. The proof is pretty straight forward, there's just a little subtlety that I think I'm missing. Here's the proof.
Let $\mathcal{C}\subset \mathbb{R}$ with an upper bound, let's say $X|Y$. We define $C$ to be the set $\{a\in\mathbb{Q}: \text{for } A|B\in\mathcal{C}\text{, } a\in A\}$ and the set $D$ to be the remainder of $\mathbb{Q}$. Clearly $C|D$ is a cut, and since for all $A|B\in \mathcal{C}$ $A\subset C$, $C|D$ is an upper bound of $\mathcal{C}$. Let $C'|D'$ be any cut that bounds $\mathcal{C}$ above. Clearly, $C\subset C'$ and $C|D$ is the least upper bound.
I mostly understand the above proof. Here's what's confusing me. Doesn't $C$ being the union of all of the left halves of the cuts in $\mathcal{C}$ imply that the cut $C|D$ is in $\mathcal{C}$? Clearly this can't be true, because then the set of all numbers less than three would contain three for example. This is where I'm confused. Can someone explain like I'm six?
"Doesn't $C$ being the union of all of the left halves of the cuts in $\mathcal{C}$ imply that the cut $C|D$ is in $\mathcal{C}$?"
No, it doesn't. In general when you have a (countably) infinite amount of sets $H_0 \subseteq H_1 \subseteq ...$, then $\bigcup_{n \in \mathbb{N}} H_n=H$ doesn't have to be equal to $H_n$ for any $n$. (for example $H_n := \{0,1,2,...,n\};$ then $H = \mathbb{N}$)
To take your example, the set of all numbers less than three doesn't contain the cut $C|D$ where $C=\{a \in \mathbb{Q}| a<3\}$ and $D = \mathbb{Q}\setminus C$ (since that corresponds to $3$); but $C$ is exactly what you get by taking the union of all the left halves of the cuts in this set.
Hope this answers anything. :)