Rudin Theorem 1:11: understanding why $L \subset S$

Question

Rudin's theorem 1.11 states:

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 $\alpha = \inf B$. In particular, $\inf B$ exists in $S$.

I am having trouble understanding why $L \subset S$. This has to be the case so that we can invoke the LUB property to conclude that $\sup L$ exists in $S$. After a lot of searching, the answer seems to be that "$S$ is the universe; nothing exists outside of $S$." I'm struggling to understand why that follows from the problem. What if, for example, $S$ is the set $\mathbb{Z}$, $B$ is the positive integers, and $L$ is the set of all lower bounds in $\mathbb{R}$? That is, $L = (-\infty, 0)$. Surely, $L \not \in S$.

If we went a step further and defined $L$ as the set of all lower bounds of $B$ in $S$, the proof would make much more sense to me. Is that what Rudin, implicitly, means?

Answer 1

Simply the sentence 'Let $L$ be the set of all lower bounds of $B$.' should be implicitly understood as

Let $L$ be the set of all lower bounds $s\in S$ of $B$.

Answer 2

I struggled with this myself, and I was looking for a bit more rigorous explanation. Here's what I thought.

To answer this question more explicitly without relying on any "intuition" or contextual meanings, observe the following.

Recall 1.7 Definition in the textbook (changed to bounded below, as suggested in the textbook): Suppose S is an ordered set, and $E\subset S$. If there exists a $\beta \in S$ such that $x \ge \beta$ for every $x \in E$, we say that $E$ is bounded below, and call $\beta$ a lower bound of $E$.

(I am not quoting above as I have changed the original text of the textbook to suit the definition we are needing)

We are going to transform this into two logical pieces.

Let $p = $ "There exists a $\beta \in S$ such that for every $x \in E$, $x \ge \beta$"

Let $q = $ "$\beta$ is a lower bound of $E$"

Therefore $p \implies q$. But $q \implies p$, therefore $p \Leftrightarrow q$

In other words, If $\beta$ is a lower bound of $E$, then $\beta \in S$ such that for every $x \in E$, $x \ge \beta$.

Meaning that it is in the definition of a "lower bound" to exist in the $E$'s superset, which is $S$ in this case. And same can be said about the Theorem. By definition, the lower bounds of $B$ exist in it's superset which is $S$.