About the proof of Rudin's Principles of Mathematical Analysis Theorem 1.11

Question

Theorem 1.11. Suppose 𝑆 is an ordered set with the least-upper-bound property, 𝐵⊂𝑆, 𝑆 is not empty, and 𝐵 is bounded below. Let 𝐿 be the set of all lower bounds of 𝐵. Then 𝛼=sup𝐿 exists in 𝑆, and 𝛼=inf𝐵.

In particular, inf𝐵 exists in 𝑆.

Proof. Since 𝐵 is bounded below, 𝐿 is not empty. Since 𝐿 consists of exactly those 𝑦∈𝑆 which satisfy the inequality 𝑦≤𝑥 for every 𝑥∈𝐵, we see that every 𝑥∈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 𝛼.

If 𝛾<𝛼 then (see Definition 1.8) 𝛾 is not an upper bound of 𝐿, hence 𝛾∉𝐵. It follows that 𝛼≤𝑥 for every 𝑥∈𝐵. Thus, 𝛼∈𝐿.

If 𝛼<𝛽 then 𝛽∉𝐿, since 𝛼 is an upper bound of 𝐿.

We have shown that 𝛼∈𝐿 but 𝛽∉𝐿 if 𝛽>𝛼. In other words, 𝛼 is a lower bound of 𝐵, but 𝛽 is not if 𝛽>𝛼. This means that 𝛼=inf𝐵.

Here is my question about the theorem.

When I followed the proof I clearly understood the proof except one thing. That is "If 𝛾<𝛼 then (see Definition 1.8) 𝛾 is not an upper bound of 𝐿, hence 𝛾∉𝐵. It follows that 𝛼≤𝑥 for every 𝑥∈𝐵. Thus, 𝛼∈𝐿.". I can't understand how is it claimed $\alpha\in\text{L}$. Can anyone explain it?

Answer 1

If $\alpha \leq x$ for every $x\in B$, then by definition of the term "lower bound", $\alpha$ is a lower bound of $B$, hence belongs to the set of lower bounds of $B$, which is $L$.