This is a duplicate of questions already asked (which shows others are struggling with this concept as well) – but I don't find the answers rigorous.
When constructing the reals using Dedekind cuts, the least upper bound is proven to be the union of a (possibly infinite) set of cuts. The proof that it is, indeed, the least upper bound is straightforward.
However, I don't see how, given these proofs, that the least upper bound isn't in the union itself.
Say $S=\{(A^\prime, B^\prime)\}$ is a set of cuts, and $t =(A, B) = \bigcup S$. Clearly, we have $s \leq t, \forall s \in S$. However, $t \notin S \implies (A^\prime, B^\prime) \in S$, then $A^\prime \subsetneq A$. But that implies there are elements in $A$ not in any $A^\prime$, contradicting the union.
I know things get dicey with infinite sets of infinite sets – but can someone help clarify?
Best Answer
The least upper bound might be one of the sets we're taking the union of, or it might not. If a set has a maximum, that maximum is the least upper bound, after all.
It might help to consider an example. Consider the set $\{0,\frac12,\frac23,\dots,1-\frac1{n},\dots\}$, represented by Dedekind cuts $S_n=\{x: -\infty < x < 1-\frac1{n},x\in\mathbb{Q}\}$. The least upper bound $\bigcup_n S_n = \{x: -\infty < x < 1,x\in\mathbb{Q}\}$, of course, is the Dedekind cut representing $1$.
Any particular rational $x$ in $[0,1)$ will be in some but not all of these $S_n$ - and, conversely, for any given $S_n$, there are elements such as $1-\frac1{2n}$ not in that particular $S_n$. But it's an infinite union; all we need is some $S_n$ that contains the point, and we can get that by going farther out. None of the individual $S_n$ are equal to all of $S$, and that's fine.