I am reading Rudin's Principles of Mathematical Analysis and I have some questions regarding the proof of Theorem 1.11.
Theorem 1.11. Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $S$ 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 $\alpha = \inf B$.
Question 1: In the proof above, I do not understand the reasoning behind "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 $\alpha\in L.$" In other words I interpret this as saying that if there was an element less than the supremum of $L$ then it couldn't be an upper bound of $L$ nor could it be an element of $B$, which seems like an obvious thing to say why state this?
Question 2: So it goes on to say that $\alpha$ is less than or equal to every element of B but how does this show $\alpha \in L$?
Question 3: Also can someone explain how the sentence "In other words, $\alpha$ is the lower bound of $B$ but $\beta$ is not if $\beta\gt \alpha.$" implies "$\alpha=inf B$"