I'm trying to read through and understand this proof from Rudin, but am a bit confused at just a few steps. I'm going to try to replicate the proof and pause at those particular steps.
Theorem. Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then $\alpha = \sup L$ exists in $S$, and $\alpha = \inf B$.
Proof. Let $B \subset S$, where $S$ is an ordered set and $B$ is non-empty and bounded below. Define $L$ to the set of all lower bounds of $B$, which is non-empty due to the fact that $B$ is bounded below.
(1) This is where my first question comes in: Rudin argues here that $L$ consists of all elements $y \in S$ s.t. $y \leq x$ for every element $x \in B$. I understand why this holds for all elements of $B$, as this is the definition of lower bounds. But, we aren't given the size or nature of $S$ (finite, infinite, etc.). If $S\neq \mathbb{R}$, would it noe be possible for us to have other element not in $S$ which bounds $B$ below? We also don't know whether $S$ is bounded below; if so, there exist lower bounds of $B$, and thus elements of $L$, which are in $\mathbb{R}$ but not $L$. So, might it be more accurate to say that $L = \{y \in \mathbb{R} : y \leq x, \forall x \in B\}$?
Continuing the proof, and assuming Rudin's last step as to the definition of $L$, it's clear that every $x$ is an upper bound of $L$, by definition, and thus every element of $B$ bounds $L$ above.
(2) Here, Rudin argues that the properties we assumed about $S$ tell us that $L$ has a supremum, $\alpha$, in $S$. Here, I am not quite following. Clearly, $S$ is ordered and has a least-upper bound. We don't know whether $B$ contains that least upper bound, as we don't know the nature of the subset. Does this step require — or perhaps embed, without proof — a lemma that a subset of a ordered set with a least upper bound also has a least upper bound, which is less than or equal to the least upper bound of the one contained in its superset? I'm also struggling a bit to see why this element must exist in $S$, but this may be that not only is every element of $B$ an upper bound of $L$, but that $B$ contains all such upper bounds of $L$. Is this correct? Would this not presuppose that $B$, and thus $L$ (as the largest element of $L$ and the smallest element of $B$ are the same), are composed of real numbers, and that this list is "complete"? Otherwise, we can find a real that is in neither set but bounds $B$ below.
So, again assuming this last step, we define the supremum of $L$ as some element $\alpha \in S$. We then consider some arbitrary element $\gamma$, which has the property that $\gamma < \alpha$. (Rudin does not state that this element is in $S$, but I'm not sure that it quite matters here.) Since $\gamma$ is the least-upper-bound of $L$, $\alpha$ cannot be an upper bound of $L$. But, all elements of $B$ are upper bounds of $L$, so $\gamma \not \in B$. Hence, $B$ cannot contain any element less than $\alpha$, and we have $\alpha \leq x, \forall x \in B$.
(3) I understand everything from the above except, perhaps, the final line that I have for the moment omitted. Rudin concludes from this that $\alpha \in L$. Is the argument here, again, that because $L$ contains all elements $y \leq x$ for $x \in B$, that $\alpha$ is simultaneously in line and $B$ due to the fact that we could have $y = x$? (In other words, assuming some notion of 'completeness,' the largest element of $L$ is simultaneously the smallest element of $B$? I hope I'm not assuming the conclusion of the proof here.)
So, again assuming the above that $\alpha \in L$, we consider some element $\beta$. (Rudin again makes no assumptions on whether $\beta$ lives in $S$, so I assume we don't need to make any assumptions, and can just take $\beta$ to be a real number.) Then, we let $\beta > \alpha$.
(4) From here, I am having difficulty seeing how Rudin concludes that $\beta \not \in L$. It seems that the argument is that, since $\alpha$ is the least upper bound of $S$, and every element of $L$ (per Rudin, though I questioned why this is the case above) is also in $S$, that clearly $\alpha$ must also bound this subset of $S$ above. So, clearly, a greater element, $\beta$, can't be in this set that's bounded above by $\alpha$. Is this correct? If so, this suggests that the argument that $L$ contains elements only in $S$ is even more crucial, though I still am struggling to see why this is the case.
From here, the remainder of the proof is easy: $\alpha \in L$, as we showed above, but any greater element than $\alpha$, such as $\beta$ from above, cannot be in $L$, so $\alpha$ is not only an upper bound of $L$, but the greatest such upper bound. Since $L$ contains all of the lower bounds of $B$, this implies that $\alpha$ is the greatest lower bound, so $\alpha = \inf B$.
I'd appreciate any help and insights on the questions I raised above. Hopefully I have understood and correctly framed at least parts of this proof.
Thanks in advance.
Best Answer
In your question, you mention the possibility that some element doesn't belong to $S$. That's not possible. In this context, $S$ is the whole universe. There is nothing outside of it.