Baby Rudin Def. 1.1o and Theorem 1.11, confused

Question

In baby Rudin we have the definition and theorems:

Definition 1.10 An ordered set $S$ is said to have the $\it{least}$-$\it{upper}$-$\it{bound \hspace{1mm}property}$ if the following is true: If $E\subset S$, $E$ is not empty, and $E$ is bounded above, then $sup(E)$ exists in $S$.

Theorem 1.11 Suppose $S$ is an ordered set with the $\it{least}$-$\it{upper}$-$\it{bound \hspace{1mm}property}$, $B \subset S$, $B$ is not empty, and $B$ 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 $S$.

My question is what is the significance of assuming $S$ has the LUB property? Particularly, since the theorem assumes that $S$ has the LUB property does that imply that there exists another set $E$ (the same one from the definition) in addition to the sets $B$ and $L$-- is the set $L$ the same as the set $E$, or is the set $B$ the same as the set $E$? What confuses me is that the theorem references the set $S$ having the LUB property-- the property being that the set $E$ is bounded $\it{above}$ while in the theorem makes a reference to the set $B$ with almost identical properties to the set $E$ like except the last one, that is $B$ is bounded $\it{below}$. My best guess is that sets $E$ from the definition is the set $L$ in the theorem, can someone confirm or clarify this? Thanks in advance.

Answer 1

The point of the Theorem is to show that a set has the least upper bound property if and only if it has the greatest lower bound property (technically it only shows one direction; the reverse direction is almost identical).

The goal is, given an arbitrary set $B\subset S$ that has any lower bound, to show that $B$ has a greatest lower bound. To do this, consider the set $$ E=\{s\in S: s\text{ is a lower bound for }B\}. $$ What can we say about $E$? We know it is not empty, since by assumption $B$ was bounded below. Further, it is bounded above by any point of $B$. Thus we can apply the least upper bound property to the set $E$. What conclusion can we draw from there?

Answer 2

You are confusing yourself by considering E from the definition and how it relates in the theorem.

In the definition, E can be any subset of an ordered set which we call S. I want you to forget about E, and think, any arbitrary subset of S, but also those that meet the criteria laid down in the definition.

Now, In the theorem, B is a subset of S, and S is an ordered set. Claim: L consists of points in S also, L is a subset of S which is not empty. Explanation: B is not empty, by assumption, by ordering there is a smallest element, therefore B is bounded below. Now we see L is not empty. I think you can take it from here.

Answer 3

Key question

My question is what is the significance of assuming S has the LUB property?

$$ $$

The answer

What we are doing in this theorem is this. We have two sets L and B, with two points $$\sup{L} \quad \text{and} \quad \inf{B} \tag{A}$$

Least upper bound property of set S means that these two points exists in S as elements. In other words, imagine set S as a box with pencils. You pick up one pencil with your left hand out of this box and call it $\sup{L}$, then you pick another pencil with your right hand and call this pencil $\inf{B}$.

As box S is an ordered set you may now compare pencils, say, by one of their properties. Are they both have equal sharpness, or one pencil is definitely sharper than the other.

The whole point of this theorem is to show these two elements you pick have equal properties, given particular assumptions.

The rigid proof from my notes.

Sketch of the proof.

$L$ is non-empty. $L$ is bounded above.
$\alpha = \sup{L}$ is a lower bound of $B$
$\alpha \in L$,

so $\alpha = \inf{B}$.

Important note: $\inf{B} = \alpha = \sup{L}$, so both $B$ and $L$ may intersect in one point $\alpha$.

Proof

We construct $$ L = \{y \in S \mid \forall x \in B : y \leq x\} $$ Since $B$ is bounded $L$ is not empty. We see that every $x \in B$ is an upper bound of $L$, so $L$ is bounded above. Point $\alpha$ is a lower bound of B. We prove it by contradiction. Suppose $\alpha$ is not a lower bound of B, then there exists $$\overline{x} \in B \quad| \quad \overline{x} < \alpha,$$

but then $$ L = \{y \in S \quad | \quad \forall x \in B : y \leq \overline{x} < \alpha \} ,$$
so $\alpha$ is not $\sup{L}$, a contradiction. As $\alpha$ is a lower bound of $B$, $\alpha \in L$. As $\alpha$ is $\sup{L}$ then for all $y \in L$ $y \leq \alpha$, $\alpha$ is $\inf{B}$.

Again, as with pencils, while reading this stuff you have to keep in mind that these two points are "in the box", really exist in your set.