For each $x \in X$, consider the lower closure $x^\preceq$: :$x^\preceq = \set {y \in X: y \preceq x}$ Let $\mathbb S \subseteq \powerset X$ be the image of $\cdot^\preceq$ considered as a mapping from $X$ to $\powerset X$, where $\powerset X$ is the power set of $X$. From Ordering is Equivalent to Subset Relation: :$\forall x, y \in X: x^\preceq \subseteq y^\preceq \iff x \preceq y$ Thus the task of finding a maximal element of $X$ is equivalent to finding a maximal set in $\mathbb S$. Thus the statement of the result is equivalent to a statement about chains in $\mathbb S$: :Let $\mathbb S$ be a non-empty subset of $\powerset X, X \ne \O$ such that every non-empty chain in $\mathbb S$, ordered by $\subseteq$, has an upper bound in $\mathbb S$. :Then $\mathbb S$ has at least one maximal set. Let $\mathbb X$ be the set of all chains in $\struct {X, \preceq}$. Every element of $X$ is included in $\map {s^\preceq} x$ for some $x \in X$. $\mathbb X$ is a non-empty set of sets which are ordered (perhaps partially) by subset relation. If $\mathcal C$ is a chain in $\mathbb X$, then: :$\displaystyle \bigcup_{A \mathop \in \mathcal C} A \in \mathbb X$ Since each set in $\mathbb X$ is dominated by some set in $\mathbb S$, going from $\mathbb S$ to $\mathbb X$ can not introduce any new maximal elements. The main advantage of using $\mathbb X$ is that the chain hypothesis is in a slightly more specific form. Instead of saying that each chain in $\mathcal C$ has an upper bound in $\mathbb S$, we can explicitly state that the union of the sets of $\mathcal C$ is an element of $\mathbb X$. This union of the sets of $\mathcal C$ is clearly an upper bound of $\mathcal C$. Another advantage of $\mathbb X$ is that, from Subset of Toset is Toset, it contains all the subsets of each of its sets. Thus we can enlarge non-maximal sets in $\mathbb X$ one element at a time. So, from now on, we need consider only this non-empty collection $\mathbb X$ of subsets of a non-empty set $X$. $\mathbb X$ is subject to two conditions: :$(1): \quad$ Every subset of each set in $\mathbb X$ is in $\mathbb X$. :$(2): \quad$ The union of each chain of sets in $\mathbb X$ is in $\mathbb X$. It follows from $(1)$ that $\O \in \mathbb X$. We need to show that there exists a maximal set in $\mathbb X$. Let $f$ be a choice function for $X$: :$\forall A \in \powerset X \setminus \O: \map f A \in A$ For each $A \in \mathbb X$, let $\hat A$ be defined as: :$\hat A := \set {x \in X: A \cup \set x \in \mathbb X}$ That is, $\hat A$ consists of all the elements of $X$ which, when added to $A$, make a set which is also in $\mathbb X$. From its definition: :$\displaystyle \hat A = \bigcup_{x \mathop \in \hat A} \paren {A \cup \set x}$ where each of $A \cup \set x$ are chains in $X$ and so elements of $\mathbb X$. Suppose there exists a maximal set $M$ in $\mathbb X$. Then, by definition of maximal, there are no elements in $x \in X \setminus M$ which can be added to $M$ to make $M \cup \set x$ another element of $\mathbb X$. Thus it follows that $\hat M = M$. From Set Difference with Self is Empty Set it then follows that if $A$ is maximal, then $\hat A \setminus A = \O$. The mapping $g: \mathbb X \to \mathbb X$ can now be defined as: :$\forall A \in \mathbb X: \map g A = \begin{cases} A \cup \set {\map f {\hat A \setminus A} } & : \hat A \setminus A \ne \O \\ A & : \text{otherwise} \end{cases}$ Thus what we now have to prove is that: :$\exists A \in \mathbb X: \map g A = A$ Note that from the definition of $g$: :$\forall A \in \mathbb X: A \subseteq \map g A$ Also note that $\map f {\hat A \setminus A}$ is a single element of $\hat A \setminus A$. Thus we obtain the ''crucial fact'' that $\map g A$ contains ''at most'' one more element than $A$. We (temporarily) define a '''tower''' as being a subset $\mathcal T$ of $\mathbb X$ such that: :$(1): \quad \O \in \mathcal T$ :$(2): \quad A \in \mathcal T \implies \map g A \in \mathcal T$ :$(3): \quad $ If $\mathcal C$ is a chain in $\mathcal T$, then $\displaystyle \bigcup_{A \mathop \in \mathcal C} A \in \mathcal T$ There is of course at least one tower in $\mathbb X$, as $\mathbb X$ itself is one. It follows from its definition that the intersection of a collection of towers is itself a tower. It follows in particular that if $\mathcal T_0$ is the intersection of ''all'' towers in $\mathbb X$, then $\mathcal T_0$ is the ''smallest'' tower in $\mathbb X$. Next we demonstrate that $\mathcal T_0$ is a chain. We (temporarily) define a set $C \in \mathcal T_0$ as '''comparable''' if it is comparable with every element of $\mathcal T_0$. That is, if $A \in \mathcal T_0$ then $C \subseteq A$ or $A \subseteq C$. To say that $\mathcal T_0$ is a chain means that '''all''' sets of $\mathcal T_0$ are '''comparable'''. There is at least one '''comparable''' set in $\mathcal T_0$, as $\O$ is one of them. So, suppose $C \in \mathcal T_0$ is '''comparable'''. Let $A \in \mathcal T_0$ such that $A \subsetneq C$. Consider $\map g A$. Because $C$ is '''comparable''', either $C \subsetneq \map g A$ or $\map g A \subseteq C$. In the former case $A$ is a proper subset of a proper subset of $\map g A$. This contradicts the fact that $\map g A \setminus A$ can be no more than a singleton. Thus if such an $A$ exists, we have that: :$(A): \quad \map g A \subseteq C$ Now let $\mathcal U$ be the set defined as: :$\mathcal U := \set {A \in \mathcal T_0: A \subseteq C \lor \map g C \subseteq A}$ Let $\mathcal U'$ be the set defined as: :$\mathcal U' := \set {A \in \mathcal T_0: A \subseteq \map g C \lor \map g C \subseteq A}$ That is, $\mathcal U'$ is the set of all sets in $\mathcal T_0$ which are comparable with $\map g C$. If $A \in \mathcal U$, then as $C \subseteq \map g C$, either $A \subseteq \map g C \lor \map g C \subseteq A$ So $\mathcal U \subseteq \mathcal U'$. The aim now is to demonstrate that $\mathcal U$ is a '''tower'''. From Empty Set is Subset of All Sets, $\O \subseteq C$. Hence condition $(1)$ is satisfied. Now let $A \in \mathcal U$. As $C$ is '''comparable''', there are three possibilities: :$(1'): \quad A \subsetneq C$ Then from $(A)$ above, $\map g A \subseteq C$. Therefore $\map g A \in \mathcal U$. :$(2'): \quad A = C$ Then $\map g A = \map g C$ and so $\map g C \subseteq \map g A$. Therefore $\map g A \in \mathcal U$. :$(3'): \quad \map g C \subseteq A$ Then $\map g C \subseteq \map g A$ Therefore $\map g A \in \mathcal U$. Hence condition $(2)$ is satisfied. From the definition of $\mathcal U$, it follows immediately that the union of a chain in $\mathcal U$ is also in $\mathcal U$. Hence condition $(3)$ is satisfied. The conclusion is that $\mathcal U$ is a '''tower''' such that $\mathcal U \subseteq \mathcal T_0$. But as $\mathcal T_0$ is the smallest '''tower''', $\mathcal T_0 \subseteq \mathcal U$. It follows that $\mathcal U = \mathcal T_0$. Consider some '''comparable''' set $C$, then. From that $C$ we can form $\mathcal U$, as above. But as $\mathcal U = \mathcal T_0$: :$A \in \mathcal T_0 \implies \paren {A \subseteq C \implies A \subseteq \map g C} \lor \map g C \subseteq A$ and so $g \left({C}\right)$ is also '''comparable'''. We now know that: :$\O$ is '''comparable''' :the mapping $g$ maps '''comparable''' sets to '''comparable''' sets. Since the union of a chain of '''comparable''' sets is itself '''comparable''', it follows that the '''comparable''' sets all form a '''tower''' $\mathcal T_C$. But by the nature of $\mathcal T_0$ it follows that $\mathcal T_0 \subseteq \mathcal T_C$. So the elements of $\mathcal T_0$ must all be '''comparable'''. Since $\mathcal T_0$ is a chain, the union $M$ of all the sets in $\mathcal T_0$ is itself a set in $\mathcal T_0$. Since the union includes all the sets of $\mathcal T_0$, it follows that $\map g M \subseteq M$. Since it is always the case that $M \subseteq \map g M$, it follows that $M = \map g M$. The result follows. {{qed}}