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Least Upper Bound Property and Order Relations

July 9, 2026

I was reading the beginning section of Munkres' Topology on Set Theory and Logic since the early chapters (and some of the subsequent ones) of Probability and Measure by Billingsley uses a decent amount of set theory. Here I prove the following theorem (which was two exercises in Munkres' book):

An ordered set $A$ has the least upper bound property if and only if $A$ has the greatest lower bound property.

Before proceeding, we will need the following lemma:

Let $C$ be a relation on $A$. Let $D$ be a relation on $A$ that is defined so that $(b, a) \in D$ only if $(a, b) \in C$. Then:

$C$ is symmetric if and only if $C = D$.
If $C$ is an order relation, $D$ is an order relation.

Suppose $C$ is symmetric. Let $(a, b) \in C$. We show $(a, b) \in D$. The symmetry of $C$ implies that $(b, a) \in C$. By hypothesis, this means that $(a, b) \in D$. Hence, $C \subset D$. Next, let $(a, b) \in D$. We show that $(a, b) \in C$. By the definition of $D$, $(b, a) \in C$. By symmetry, $(a, b) \in C$. Therefore, $D \subset C$. Since $C \subset D$ and $D \subset C$, $C = D$. Now, suppose that $C = D$. We show $C$ is symmetric. If $(a, b) \in C$, then $(a, b) \in D$ since $C = D$. However, $(a, b) \in C$ implies that $(b, a) \in D$. Since $C = D$, $(b, a) \in C$. So, $C$ is symmetric.
First, we show comparability. Let $x \ne y$. Then, $(x, y) \in C$ or $(y, x) \in C$. If $(x, y) \in C$ and $(y, x) \n C$, then (1) implies that $C = D$, which would mean $D$ is an order relation. However, if this is not the case, then $(x, y) \in D$ or $(y, x) \in D$. Now we show nonreflexivity. For contradiction, suppose that $(x, x) \in D$. Then this would imply $(x, x) \in C$. But $C$ is an order relation, so $(x, x) \not\in C$. This is a contradiction, so $(x, x) \not\in D$. Suppose $(x, y) \in D$ and $(y, z) \in D$. Then, by the definition of $D$, $(y, x) \in C$ and $(z, y) \in C$. Because $C$ is an order relation, $C$ is transitive, so $(z, x) \in C$. But this is true if and only if $(x, z) \in D$, meaning $D$ is transitive. $D$ is an order relation.

Now, we can proceed with the proof of Theorem 1

First, we show that $A$ having the least upper bound property implies that it has the greatest lower bound property. Let $A_0 \subset A$, $A_0 \ne \emptyset$, and $A$ is bounded below. Let $L = \{a ∈ A: \text{$a$ is a lower bound for $A_0$}\}$. Since $A_0 \ne \emptyset$, there exists $a_0 \in A_0$ such that $l \le a_0$ for all $l \in L$. Then, $L \subset A$, $L \ne \emptyset$, and $L$ is bounded above. By the least upper bound property of $A$, $L$ has a least upper bound: there exists $l^* \in A$ so that $l \le l^*$ for all $l \in L$. Every $a_0 \in A_0$ satisfies $l^* \le a_0$ since $a_0$ are all upper bounds for $L$. This means that $l^* \in L$. Hence, $l^*$ is the greatest element of all the lower bounds for $A_0$, implying that it is the greatest lower bound of $A_0$. Next, we show that $A$ having the greatest lower bound property implies that it has the least upper bound property. Suppose $A$ is an ordered set having the greatest lower bound property. We show that it has the least upper bound property. Suppose the order relation on $A$ is given by $C$. We can define a relation $D$ so that $(x, y) \in D$ if and only if $(x, y) \in C$. By Lemma 2, $D$ is an order relation. This leads to the following series of implications:

\begin{align*} &\text{$A$ has the greatest lower bound property w.r.t $C$} \\ &\implies \text{$A$ has the least upper bound property w.r.t $D$} \\ &\implies \text{$A$ has the greatest lower bound property w.r.t $D$} & \text{ the first direction of Theorem 1 } \\ &\implies \text{$A$ has the least upper bound property w.r.t $C$} \end{align*}