# Baby Rudin Def. 1.1o and Theorem 1.11, confused

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.

• For starters, $E$ and $L$ are not exactly the same thing. The set $E$ in 1.10 represents any number of sets, not just one specific set. $E$ represents all sets that satisfy the following conditions: $E \subseteq S$, $E$ is nonempty, and $E$ is bounded above. In 1.11, the set $L$ is a very specific set: it's the set of all lower bounds of $B$. – user307169 Sep 12 '17 at 2:28
• So $L$ is an instance of $E$? And if it is so, was it implied from the beginning? – CrypticParadigm Sep 12 '17 at 2:37
• Rudin states, "we shall now show that there is a close relation between greatest lower bounds and least upper bounds, and that every ordered set with the least-upper-bound property also has the greatest-lower-bound property." Therefore the significance of assuming S has that property is made explicit, right before the theorem. It's bottom of page 4 third ed. – smokeypeat Sep 12 '17 at 11:16
• $L$ in 1.11 is indeed an instance of what $E$ represents in 1.10. I'm not sure what you mean by implied from the beginning, but a big part of Baby Rudin, which many hate and many others love, is filling in the gaps on your own. I think in this case it's expected that the reader will verify $L$ in 1.11 has all those properties that $E$ has in 1.10. In other words, the reader must verify three things: $L \subseteq S$, $L$ is nonempty, and $L$ is bounded above. – user307169 Sep 12 '17 at 12:41

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?

• I corrected a silly misprint... – Angina Seng Sep 12 '17 at 2:33
• @LordSharktheUnknown Thanks! – TomGrubb Sep 12 '17 at 2:35
• So the conclusion that can be drawn is that the premise that $S$ satisfies the least-upper-bound property is proven true? – CrypticParadigm Sep 12 '17 at 2:45
• I have a homework problem where I have to prove the following theorem: 'Suppose that $S$ is an ordered set with the least-upper-bound property, then $S$ has the greatest-lower-bound property.' What is the difference between this theorem and theorem 1.11? – CrypticParadigm Sep 12 '17 at 2:45
• @physicsmajor I don't think there is a difference – TomGrubb Sep 12 '17 at 2:52

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.

Key question

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



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.