I have asked this question before but I am rephrasing it here ( and deleting the previous post) as I slightly more nuanced take on the question. In Rudin Principles of Mathematical Analysis we have:
Theorem 1.11. 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 $S$.
Proof. Since $B$ is bounded below, $L$ is not empty. Since $L$ consists of exactly those $y \in S$ which satisfy the inequality $y \leq x$ for every $x \in B$, we see that every $x \in B$ is an upper bound of $L$. Thus $L$ is bounded above. Our hypothesis about $S$ implies therefore that $L$ has a supremum in $S$; call it $\alpha$.
If $\gamma < \alpha$ then (see Definition 1.8) $\gamma$ is not an upper bound of $L$, hence $\gamma \not\in B$. It follows that $\alpha \leq x$ for every $x \in B$. Thus, $\alpha \in L$.
If $\alpha < \beta$ then $\beta \not\in L$, since $\alpha$ is an upper bound of $L$.
We have shown that $\alpha \in L$ but $\beta \not\in L$ if $\beta > \alpha$. In other words, $\alpha$ is a lower bound of $B$, but $\beta$ is not if $\beta > \alpha$. This means that $\alpha = \inf B$.
I understand the following portion of the proof as follows below:
"If $\gamma \lt \alpha$ then $\gamma$ is not an upper bound of L, hence $\gamma \notin B$. It follows that $\alpha \le x$ for every $x \in B$. Thus $\alpha\in L.$"
We have proven the there is a supremum of $L$ somewhere in $S$ but for all we know this supremum, which we call $\alpha$, could be in $B$. So to prove this point Rudin uses the definition of the least upper bound to remind us that $\gamma$ cannot be in $B$ because $\gamma$ must always be less than the supremum $\alpha$. But note also that since $\alpha$ must be a lower bound of $B$ so it follows from this that $\alpha \le x$ for every $x \in B$.
In other words Rudin uses the line above to cement the place of $\alpha $ as being between $L$ and $B$.
I have two questions dear reader:
After staring at this question as an undergrad novice for 12 hours is my explanation above some what close to correct?
Rudin ends this paragraph obliquely :"Thus $\alpha\in L.$"
How is this the case? Could it not be that $\alpha$ be in $B$ if in the previous line $\alpha = x$ for some $x \in B$. I am asking how does this follow from line reasoning previous in the paragraph? I am really passionate about understanding this proof so any effort you put forth will be greatly appreciated. Thanks in advance!