# Least Upper Bound Property Implies Greatest Lower Bound Property

In Rudin $1.11$ Theorem Proof he claims the following

Theorem. Suppose $S$ is an ordered set with the least upper bound 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 in $S$, and $\alpha = \inf B$.

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 about $S$ implies therefore that $L$ has a supremum in $S$ call it $\alpha$

If $\gamma < \alpha$ then $\gamma$ is not an upper bound of $L$, hence $\gamma \notin B$. It follows that $\alpha \leq x$ for every $x \in B$. Thus $\alpha \in L$

If $\alpha < \beta$ then $\beta \notin L$ since $\alpha$ is an upper bound of $L$

We have shown that $\alpha \in L$ but $\beta \notin L$ if $\beta > \alpha$. In other words, $\alpha$ is a lower bound of $B$, but $\beta$ is not if $\beta > \alpha$. This means that $\alpha = \inf B$

I am confused in the following:

I don't follow why $L \subset S$ given $S$ is an ordered set with the least upper bound property and $B \subset S$, $B$ is not empty and $B$ is bounded below.

If $L$ is not a subset of $S$, then the assumption of the proof will not follow, I think I have missed something.

Can someone help me out? This proof is in Rudin's analysis page 5

Thanks

Edit for clarification

Suppose the following let $S = (0, x]$, for which $x$ is some real positive number, we know $S$ is an ordered set with the least upper bound property, let $B = (0, y]$ for which $y < x$ and $y$ is positive real number, then $L = (-\infty, 0]$, we note that $\inf B = \sup L = 0$ however $0 \notin S$, thus we proved that an order set $S$ with the least upper bound property with $B = (0, y] \subset S \Rightarrow \inf B \notin S$

• $S$ is the "universe" here. $L$ is by definition a subset of $S$ Dec 21 '14 at 21:04
• What else do you think would be in $L$ besides members of $S$? Dec 21 '14 at 21:05
• Sorry if this question is elementary, but by the definition of ordered set, an ordered set is only a set in which order is defined, and order is defined by definition 1.5, I think you can have a set that is ordered without it being a universal set. I didn't know that all ordered set is the universal set. Dec 21 '14 at 21:07
• Had the question said $\mathbb{R}$ instead of $S$, would you argue that maybe $L$ is a subset of $\mathbb C\setminus \mathbb{R}$? Dec 21 '14 at 21:13
• If the theorem noted that $S$ is $\mathbb{R}$ or $\mathbb{C}$ I would not argue that $L$ is not in real or complex, but the theory seems to imply that $S$ is an arbitrary ordered set with least upper bound property, so I am a bit confused Dec 21 '14 at 21:16

The proof is fine. You just need to realise that everything "lives" in $S$.

So $(S,<)$ is linearly ordered and satisfies the lub property. This means that every $B \subseteq S$ that is bounded above (which means: $\exists b \in S: \forall x \in B: x \le b$) then $B$ has a least upper bound. Now he wants to prove that $S$ has the glb-property. So for every $B \subseteq S$, if $B$ has a lower bound (so $\exists b \in S: \forall x \in B: b \le x$) there exists a greatest lower bound for $B$.

So if we have such a $B$ that is non-empty and bounded below by definition of being bounded below the set $L = \{b \in S: \forall x \in B: b \le x \}$ is non-empty. This is what being bounded below means in the ordered set $S$. And as $B$ is non-empty, pick $x \in B$. Then for every $b \in L$, by definition of being in $L$: $b \le x$. So $x$ (which is in $B \subseteq S$) shows that $L$ is bounded above (in $S$), and the rest of the proof goes through.

In your example, $S = (0, 2]$ and $B = (0,1]$ (for definiteness) in their usual order, the $S$ satisfies the lub-property, but the $B$ is not bounded below in $S$ (For every $x \in S$ , with $x < 2$, $\frac{x}{2} < x$ and lies in $B$. So $x$ is not a lower bound for $B$.). So we don't have to show that $B$ has a greatest lower bound, as it has no lower bound at all. So the example is irrelevant. It's not a counterexample to $S$ also having the greatest lower bound property.

• I do not understand how \textbf{S} being universal is implied? It seems odd that this is not clarified...I spend a good hour struggling
– Zero
Apr 11 '17 at 14:01
• @Zero everything lives in $S$ because the theorem starts by suppose $S$ is a linearly ordered set etc. Apr 11 '17 at 14:44

If you look back on Definition 1.7, the lower/upper bound is defined as belonging to $S$. Thus, $L\subset S$ by definition.

Proof. Suppose that a nonempty set $$A$$ has a lower bound, call it $$ℓ$$. Define $$L$$ as the set of all lower bounds of $$A$$, then $$L$$ is nonempty ($$ℓ$$$$L$$). Observe that each member of the nonempty set A is an upper bound of $$L$$ so by the least upper bound property, $$L$$ has a least upper bound. Call this element $$α$$. First observe that $$α$$ is a lower bound for $$A$$. Otherwise, there exists an element $$b$$$$A$$ with $$b$$ < $$α$$, but each element of $$A$$ is an upper bound for $$L$$, so this element $$b$$ is an upper bound of $$L$$ which is smaller than $$α$$, the least upper bound of $$L$$. This would be a contradiction. Therefore, $$α$$$$L$$. Also, if $$ℓ$$ is any lower bound of $$A$$, then $$ℓ$$$$α$$since $$α = sup L$$. Hence α is the greatest lower bound of $$A$$.