Question from a theorem in Baby Rudin (Re: Least-Upper-Bound Property)

[PieceOfPi]

Hi,

I am reading Baby Rudin on my own to get prepared for the analysis class that I will be taking in the fall. I got up to this theorem, and I was wondering if someone can clarify me the proof of this (i.e. Thm 1.11, the first theorem).

1.11 Theorem: Suppose S is an ordered set with the lub 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, and $$\alpha$$ = inf B. In particular, inf B exists in S.

(The Part of) 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 is abound S implies therefore that L has a supremum in S; call it $$\alpha.$$

The part I underlined is where I want to be clarified. I understand that S has an lub property, and that implies that if E $$\subset$$ S, E is not empty, and E is bounded above, then sup E exists in S (EDIT: Originally it said "... then sup E exists in E, until rasmhop pointed out the misprint.). While I understand why L is bounded above, I am not clear on how this proves the existence of sup L. My guess is that we're concerned about the set K $$\subset$$ S $$\cap$$ L, and by definition sup K exists, and thus sup L exists, but my guess could be wrong.

Please let me know if you can help me out. Notice that I only put the part of the proof here since I believe the rest is irrelevant, but please let me know if you want to read the whole proof--I can always edit it later.

Thanks.

 

[rasmhop]

I understand that S has an lub property, and that implies that if E $$\subset$$ S, E is not empty, and E is bounded above, then sup E exists in E.

This is not the lub property. We can only guaranteee that supE exists in S. Consider for instance [0;1) which is bounded above by 1 and $\sup[0;1) = 1$, but $1 \notin [0;1)$.

I am not clear on how this proves the existence of sup L. My guess is that we're concerned about the set K $$\subset$$ S $$\cap$$ L, and by definition sup K exists, and thus sup L exists, but my guess could be wrong.

We have $L \subset S$, $L \not= \emptyset$ and L has an upper bound in S. These are precisely the necessary conditions in the least upper bound property and therefore by the lub property there exists an element $x \in S$ that is a least upper bound for L which proves the existence of $\sup L$ in S.

I don't really know what this set K you talk about is (an arbitrary subset of $S \cap L$ perhaps?) Anyway you don't need to introduce additional sets.

 

[PieceOfPi]

Thanks. I am assuming that when it says "L consists of exactly those y $$\in$$ S . . . " it already implies L $$\subset$$ S. I just wasn't clear that L is a subset of S.

And thanks of clarifying lub property--I think that was my typo, but nevertheless that changed the meaning of the statement a lot.

 

[rasmhop]

Thanks. I am assuming that when it says "L consists of exactly those y $$\in$$ S . . . " it already implies L $$\subset$$ S. I just wasn't clear that L is a subset of S.

And thanks of clarifying lub property--I think that was my typo, but nevertheless that changed the meaning of the statement a lot.

Yes it's implied that we only care about the lower bounds in S. Perhaps he should have been a bit more clear and stated explicitly that "Let L be the set of all lower bounds of B in S" in the statement of the theorem; especially as this chapter is kind of an introduction to rigorous mathematics for many readers. However he doesn't really need to because it's already implied that when we speak of lower bounds we speak of lower bounds relative to the order imposed on the ordered set S, but for an element not in S there is no way to talk about order. Also in the proof he remarks "L consists of exactly those $y \in S$ such .." so here he actually states that L is a subset of S (well actually he states that all elements of L are in S, but that's the same).