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


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 $B = (0, y] \subset S \Rightarrow \inf B \notin S$

  • $\begingroup$ $S$ is the "universe" here. $L$ is by definition a subset of $S$ $\endgroup$
    – David P
    Dec 21, 2014 at 21:04
  • $\begingroup$ What else do you think would be in $L$ besides members of $S$? $\endgroup$ Dec 21, 2014 at 21:05
  • $\begingroup$ 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. $\endgroup$
    – Kevin
    Dec 21, 2014 at 21:07
  • $\begingroup$ Had the question said $\mathbb{R}$ instead of $S$, would you argue that maybe $L$ is a subset of $\mathbb C\setminus \mathbb{R}$? $\endgroup$
    – David P
    Dec 21, 2014 at 21:13
  • $\begingroup$ 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 $\endgroup$
    – Kevin
    Dec 21, 2014 at 21:16

3 Answers 3


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.

  • $\begingroup$ 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 $\endgroup$
    – Zero
    Apr 11, 2017 at 14:01
  • 1
    $\begingroup$ @Zero everything lives in $S$ because the theorem starts by suppose $S$ is a linearly ordered set etc. $\endgroup$ Apr 11, 2017 at 14:44

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 $A$.


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

  • $\begingroup$ This is absolutely right. Notice that Definition 1.7 states that when a set E is bounded above, then the upper bound is in S. Eventually this works in the same way for lower bounds. Thus, when saying that B is bounded below and that L is the set of all lower bounds of B, it is necessarily the case that all elements of L are in S. $\endgroup$
    – Alex Ruiz
    Dec 24, 2021 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.