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

Theorem 1.11. Suppose π is an ordered set with the least-upper-bound property, π΅βπ, π is not empty, and π΅ is bounded below. Let πΏ be the set of all lower bounds of π΅. Then πΌ=supπΏ exists in π, and πΌ=infπ΅.

In particular, infπ΅ exists in π.

Proof. Since π΅ is bounded below, πΏ is not empty. Since πΏ consists of exactly those π¦βπ which satisfy the inequality π¦β€π₯ for every π₯βπ΅, we see that every π₯βπ΅ is an upper bound of πΏ. Thus πΏ is bounded above. Our hypothesis about π implies therefore that πΏ has a supremum in π; call it πΌ.

If πΎ<πΌ then (see Definition 1.8) πΎ is not an upper bound of πΏ, hence πΎβπ΅. It follows that πΌβ€π₯ for every π₯βπ΅. Thus, πΌβπΏ.

If πΌ<π½ then π½βπΏ, since πΌ is an upper bound of πΏ.

We have shown that πΌβπΏ but π½βπΏ if π½>πΌ. In other words, πΌ is a lower bound of π΅, but π½ is not if π½>πΌ. This means that πΌ=infπ΅.

Here is my question about the theorem.

When I followed the proof I clearly understood the proof except one thing. That is "If πΎ<πΌ then (see Definition 1.8) πΎ is not an upper bound of πΏ, hence πΎβπ΅. It follows that πΌβ€π₯ for every π₯βπ΅. Thus, πΌβπΏ.". I can't understand how is it claimed $$\alpha\in\text{L}$$. Can anyone explain it?

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 $$L$$.